FS
O
PA
G
E
PR
O
10
U
N
C
O
R
R
EC
TE
D
Further
applications
of integration
10.1 Kick off with CAS
10.2 Integration by recognition
10.3 Solids of revolution
10.4 Volumes
10.5 Arc length, numerical integration and
graphs of antiderivatives
10.6 Water flow
10.7 Review
c10FurtherApplicationsOfIntegration.indd 494
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10.1 Kick off with CAS
U
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C
O
R
R
EC
TE
D
PA
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<To come>
Please refer to the Resources tab in the Prelims section of your eBookPlUs for a comprehensive
step-by-step guide on how to use your CAS technology.
c10FurtherApplicationsOfIntegration.indd 495
08/07/15 2:33 AM
10.2 Integration by recognition
Deducing an antiderivative
Because differentiation and integration are inverse processes, differentiating a
function f(x) with respect to x will result in another function g(x). It follows that
an antiderivative of the function g(x) with respect to x is the function f(x) with the
d
addition of an arbitrary constant c. In mathematical notation, if ( f(x)) = g(x), then
dx
g(x)
dx
=
f(x)
+
c.
However,
functions
may
not
be
exactly
the
derivatives of other
3
O
FS
functions, but may differ by a constant multiple.
d
Let k be a non-zero constant. If ( f(x)) = kg(x), it follows that
dx
3kg(x) dx = k3g(x) dx = f(x), since constant multiples can be taken outside
1
WritE
1 Write the equation in index form.
EC
2 Express y in terms of u and u in terms of x.
R
O
R
dx
using the chain rule.
C
dy
5 Substitute back for u and cancel factors.
6 Write the result as a derivative of
one function.
7 Use integration by recognition.
496
dx.
y=
(25x2
+ 16)
1
2
1
2
y = u where u = 25x2 + 16
1
U
N
4 Find
"25x2 + 16
dy 1 − 2
du
= u and
= 50x
du 2
dx
1
=
2"u
dy dy du
=
dx du dx
1
=
× 50x
2 !u
dy
25x
=
dx "25x2 + 16
3 Differentiate y with respect to u and u with
respect to x.
x
Let y = "25x2 + 16.
TE
D
tHinK
Differentiate "25x2 + 16 and hence find 3
PA
G
WorKeD
eXaMPLe
E
PR
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integral signs. Then, dividing both sides by the constant multiple k, we obtain
1
3g(x) dx = k f(x) + c, as we can add in the arbitrary constant at the end. This is
known as integration by recognition.
25x
d
c "25x2 + 16 d =
dx
"25x2 + 16
3
25x
"25x2
+ 16
dx = "25x2 + 16
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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253
8 Take the constant factor outside the
integral sign.
9 Divide by the constant factor, add in the
arbitrary constant +c, and state the final result.
3
x
"25x + 16
2
x
"25x2 + 16
dx = "25x2 + 16
dx =
1
"25x2 + 16 + c
25
•
WorKeD
eXaMPLe
Differentiate cos−1 a
R
tHinK
R
O
C
U
N
2 Differentiate y with respect to u and u with
respect to x.
dy
dx
using the chain rule.
4 Substitute back for u.
8
!x
1
dx.
b and hence evaluate 3
4
"16x − x2
1 Express y in terms of u and u in terms of x.
3 Find
E
a
x
d
for x ∈ R.
atan−1 a b b = 2
a
dx
a + x2
EC
2
x
d
−1
for ∣x∣ < a
acos−1 a b b =
a
2
2
dx
"a − x
TE
D
•
x
d
1
for ∣x∣ < a
asin−1 a b b =
a
2
dx
"a − x2
PA
G
•
PR
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Integrating other types of functions
For the example above, integration techniques developed in earlier topics could
have been used to find the antiderivative. However, there are functions for which an
antiderivative could not have been obtained using the rules developed so far. Also,
this method of deducing an antiderivative can be used to evaluate definite integrals or
find areas.
To understand the examples that follow, it is necessary to review the differentiation
techniques developed in earlier topics. Recall the results for the derivatives of the
inverse trigonometric functions. If a is a positive constant, then:
0
WritE
Let y = cos−1 a
=
cos−1
!x
b
4
1
u
a b where u = "x = x 2
4
1
du 1 − 2
1
=
and
= x =
dx 2
du "16 − u2
2"x
dy
dy
dx
dy
dx
dy
dx
−1
=
=
=
dy du
1
−1
=
×
du dx 2 !x "16 − u2
−1
since 0 < x < 16
2 !x!16 − x
−1
2"16x − x2
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 497
497
08/07/15 2:33 AM
!x
d
−1
c cos−1 a
bd =
dx
4
2"16x − x2
5 Write the result as a derivative of
one function.
3
6 Use integration by recognition.
−1
2"16x − x2
8
3
9 However, in this case we are required to
evaluate a definite integral.
0
"16x − x2
1
"16x − x2
a−2 cos
−1 !8
a
bb
4
dx = −2 cos−1 a
!x
b+c
4
FS
arbitrary constant +c.
1
O
3
8 Multiply by the constant factor and add in the
dx = c −2 cos−1 a
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O
integral sign.
and simplify.
!x
b
2
!x
1
1
− 3
dx = cos−1 a
b
2 "16x − x2
4
7 Take the constant factor outside the
10 Substitute in the upper and lower terminals
dx = cos−1 a
8
"x
bd
4
0
− (−2 cos−1 (0))
π
π
+2×
4
2
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= −2 ×
11 Evaluate.
E
−1
= a−2 cos−1a !2
bb − (−2 cos (0))
2
8
3
TE
D
12 State the final value of the definite integral.
0
1
"16x −
x2
dx =
π
2
Differentiate x cos(4x) and hence find 3x sin(4x) dx.
C
3
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WorKeD
eXaMPLe
O
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R
EC
using the product rule
Some problems require a combination of integration by recognition and the product
rule. This is typical when integrating products of mixed types of functions. Recall that
the product rule states that if u = u(x) and v = v(x) are two differentiable functions of
dy
dv
du
x and y = u.v, then
=u
+v .
dx
dx
dx
tHinK
1 Write the equation.
Let y = x cos(4x) = u . v.
2 State the functions u and v.
u = x and v = cos(4x)
3 Differentiate u and v with
dv
du
= 1 and
= −4 sin(4x)
dx
dx
respect to x.
498
WritE
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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4 Find
dy
dx
dy
using the product rule.
Substitute for u,
dx
dy
dv
du
, v and .
dx
dx
dx
=u
dv
du
+v
dx
dx
= −4x sin(4x) + cos(4x)
d
[x cos(4x)] = −4x sin(4x) + cos(4x)
dx
5 Write the result as a derivative of
one function.
3 (−4x sin(4x) + cos(4x)) dx = x cos(4x)
7 Separate the integral on the left
−43x sin(4x) dx + 3cos(4x) dx = x cos(4x)
FS
6 Use integration by recognition.
which can be performed on the right
hand side.
−43x sin(4x)dx = x cos(4x) − 3cos(4x) dx
E
8 Transfer one part of the integral
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into two separate integrals and
take the constant factor outside
the integral sign.
PA
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−43x sin(4x) dx = x cos(4x) − 14 sin(4x)
9 Perform the integration on the
term on the right hand side.
1
1
3x sin(4x) dx = −4 x cos(4x) + 16 sin(4x) + c
10 Divide by the constant factor, add
TE
D
in the arbitrary constant +c, and
state the final result.
R
EC
Integrating inverse trigonometric functions
The technique used in the last example can also be used to find antiderivatives of
inverse trigonometric functions.
d[
x cos−1 (2x) ] and hence find 3cos−1 (2x) dx.
dx
C
O
Find
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4
R
WorKeD
eXaMPLe
WritE
1 Write the equation.
Let y = x cos−1 (2x) = u . v.
2 State the functions u and v.
u = x and v = cos−1 (2x)
3 Differentiate u and v with
respect to x.
4 Find
dy
dx
using the product rule.
dv
du
Substitute for u, , v and .
dx
dx
dv
du
−2
= 1 and
=
dx
dx "1 − 4x2
dy
dx
dy
dx
=u
=
dv
du
+v
dx
dx
−2x
"1 − 4x2
+ cos−1 (2x)
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 499
499
08/07/15 2:33 AM
left into two separate integrals.
−3
8 Transfer one part of the integral
−2x
"1 −
4x2
2x
"1 − 4x
2
+ cos−1 (2x)r dx = x cos−1 (2x)
dx + 3cos−1 (2x) dx = x cos−1 (2x)
−1
−1
3cos (2x) dx = x cos (2x)
which can be performed on the
right-hand side.
9 Consider now just the integral
on the right.
+3
3
2x
"1 − 4x2
= 32x(1 − 4x )
11 Use a non-linear substitution.
Let t = 1 − 4x .
2
12 The integral cannot be done
R
13 Substitute for dx, noting
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R
that the terms involving x
will cancel.
C
14 Transfer the constant factors
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N
outside the front of the
integral sign.
15 The integral can now be done.
Antidifferentiate using
n+1
n dt = t
t
with n = −12,
3
n+1
so that n + 1 = 12.
positive indices.
500 1
2
dx
dx
t = 1 − 4x2
dt
= −8x
dx
dx −1
=
8x
dt
−1
dx =
dt
8x
EC
in this form, so differentiate.
Express dx in terms of dt by
inverting both sides.
= 32xt
−
1
2
dx
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power using index laws.
−
"1 − 4x2
TE
D
10 Write the integrand as a
16 Write the expression with
dx
2
2x
FS
7 Separate the integral on the
3q
O
6 Use integration by recognition.
PR
O
of one function.
−2x
d3
x cos−1 (2x) 4 =
+ cos−1 (2x)
dx
"1 − 4x2
E
5 Write the result as a derivative
= 32xt
=
−
1
2
×
−1
dt
8x
1
−
1
−43t 2 dt
1
= −12 t 2 + c
= −12 !t + c
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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3
17 Substitute back for x and
state the result for this part of
the integral.
2x
1
dx = − "1 − 4x2
2
"1 − 4x
2
1
−1
−1
2
3cos (2x) dx = x cos (2x) − 2"1 − 4x + c
18 Substitute for the integral, add
in the arbitrary constant +c,
and state the final result.
WE1
Differentiate "(9x2 + 16) 3 and hence find 3x"9x2 + 16 dx.
2 Determine
O
1
d
[cos5 (3x)] and hence find 3sin(3x)cos4 (3x) dx.
dx
PR
O
PRactise
FS
Exercise 10.2 Integration by recognition
4
Differentiate sin−1 a
4 If f(x) = arctana
5
WE3
!x
1
dx.
b and hence evaluate 3
2
"4x − x2
0
E
WE2
48
4
1
dx.
b, find f′(x) and hence evaluate 3
!x
!x(x + 16)
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3
d
Find [x sin(3x)] and hence find 3x cos(3x) dx.
dx
16
WE4
Find
d
[x sin−1 (5x)] and hence find 3sin−1 (5x) dx.
dx
1
4
EC
7
TE
D
6 Differentiate xe−2x and hence find 3xe−2x dx.
R
8 Differentiate x tan−1 (4x) and hence evaluate 3arctan(4x) dx.
9 a Differentiate x sina b and hence find 3x cosa b dx.
2
2
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x
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Consolidate
0
x
b Differentiate x cos(2x) and hence find 3x sin(2x) dx.
c Differentiate xe
10a Determine
−
x
2
and hence find 3xe
−
x
2
dx.
x
x
d
c x sin−1 a b d and hence find 3sin−1 a b dx.
dx
3
3
b Determine
d
[x cos−1 (4x)] and hence find 3cos−1 (4x) dx.
dx
c Determine
x
x
d
c x tan−1 a b d and hence find 3tan−1 a b dx.
dx
5
5
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 501
501
08/07/15 2:33 AM
π
8
11 a If f(x) = x sin(4x), find f ′(x)and hence evaluate 3x cos(4x) dx.
0
π
12
x
3
0
c If f(x) = xe , find f ′(x). Hence determine the shaded
y
8
6
4
2
x
3
area bounded by the graph of y = xe , the origin, the
x-axis and the line x = 3, as shown.
2!2
3
2
1
O
–0.50 0.5 1 1.5 2 2.5 3 x
–2
d
4
c cos−1 a 2 b d and hence evaluate
dx
x
x"x4 − 16
PR
O
12 a
Determine
FS
b If f(x) = x cos(6x), find f′(x) and hence evaluate 3 x sin(3x) cos(3x)dx.
dx.
PA
G
E
!x
d
1
b Determine
dx.
c tan−1 a
b d and hence evaluate 3
dx
3
"x(9
+
x)
9
27
TE
D
d
2
1
c Determine
dx.
c sin−1 a b d and hence evaluate 3
x
2
dx
2 x"x − 4
4
13aDifferentiate x cos−1 (2x). Hence determine the
2
R
R
EC
shaded area bounded by the graph of y =
the coordinates axes and x = 12, as shown.
b Differentiate x sin−1 (3x) and hence find the area
bounded by the curve y = sin−1 (3x), the x-axis, the
origin and the line x = 13.
–0.5
0
0.5 x
x
4
c Differentiate x tan−1 a b and hence find the area
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C
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y
cos−1 (2x),
x
bounded by the graph of y = tan−1 a b, the x-axis,
4
the origin and the line x = 4.
14 Assuming that a is a positive real constant:
a differentiate xeax and hence find 3xeax dx
b differentiate x sin(ax) and hence find 3x cos(ax) dx
c differentiate x cos(ax) and hence find 3x sin(ax) dx.
502 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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08/07/15 2:33 AM
π
8
15 a Use your results from question 14a to evaluate 3x cos2 (2x)dx.
0
π
9
b Use your results from question 14b to evaluate 3x sin2 (3x)dx.
0
2
c Determine
d
[(2x + 1) loge (2x + 1)] and hence evaluate 3 loge (2x + 1) dx.
dx
FS
0
a differentiate x sin−1 a b and hence find 3sin−1 a b dx
a
a
x
PR
O
x
O
16 Assuming that a is a positive real constant:
b differentiate x cos−1 a b and hence find 3cos−1 a b dx
a
a
x
x
c differentiate x tan−1 a b and hence find 3tan−1 a b dx.
a
a
x
PA
G
17 a Determine
E
x
d
[x loge (4x)] and hence find 3 loge (4x) dx.
dx
d 2
[x loge (4x)] and hence find 3x loge (4x) dx.
dx
c Determine
d 3
[x loge (4x)] and hence find 3x2 loge (4x) dx.
dx
EC
TE
D
b Determine
R
d Hence deduce 3xn loge (4x) dx. What happens if n = −1?
U
N
C
O
R
18a If f(x) = e2x (2 cos(3x) + 3 sin(3x)), find f ′(x) and hence find 3e2x cos(3x) dx.
b Determine
d −3x
[e (2 cos(2x) + 3 sin(2x))] and hence find 3e−3x sin(2x) dx.
dx
c Let f(x) = eax (a cos(bx) + b sin(bx)) find f ′(x) and hence find 3eax cos(bx) dx.
d Let g(x) = eax (a sin(bx) − b cos(bx)) find g′(x). Hence find the area enclosed
between the curve y = e−2x sin(3x), the x-axis, the origin and the first intercept
it makes with the positive x-axis.
19 Assuming that a is a positive real constant:
Master
x
x
a find the derivative of x2 tan−1 a b and hence find 3x tan−1 a b dx
a
a
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 503
503
08/07/15 2:33 AM
biuse the substitution x = a sin(θ)to show that
3
x2
"a2 − x2
dx =
a2
x
1
sin−1 a b − x"a2 − x2
a
2
2
x
a
iifind the derivative of x2 sin−1 a b and hence show that
2
2
x
1
−1 x dx = x − a
x
sin
a
b
a
bsin−1 a b + x"a2 − x2 + c.
3
a
a
4
2
4
d
[ loge (tan(x) + sec(x))]. Hence, find the area enclosed between the
dx
π
curve y = sec(x), the coordinate axes and the line x = .
4
d
b Given that a is a positive real constant, find [ loge (cosec(ax) + cot(ax))].
dx
Hence, find the area enclosed between the curve y = cosec(2x), the x-axis and
π
π
the lines x = and x = .
6
3
dy
sin(2x) + cos(2x)
c If y = loge
, show that
= 2 sec(4x). Hence, find
Å sin(2x) − cos(2x)
dx
the area enclosed between the curve y = sec(4x), the x-axis and the
π
π
lines x = and x = .
6
4
Solids of revolution
Rotations around the x-axis
TE
D
10.3
PA
G
E
PR
O
O
FS
20aFind
y
y = f(x)
y
O
C
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N
0
y = f(x)
P(x, y)
R
R
EC
Suppose that the curve y = f(x) is continuous on the closed
interval a ≤ x ≤ b. Consider the area bounded by the
curve, the x-axis and the ordinates x = a and x = b.
If this area is rotated 360° about the x-axis, it forms a solid
of revolution and encloses a volume V.
a
b
x
y
0
a
b
x
∆x
Consider a point on the curve with coordinates P(x, y). When rotated it forms
a circular disc with a radius of y and a cross-sectional area of A( y) = π y2. If the disc
has a width of Δx, then the volume of the disc is the cross-sectional area times the
504 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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08/07/15 2:33 AM
width, that is A( y)Δx. The total volume, V, is found by adding all such discs between
x = a and x = b and taking the limit as Δx → 0. This is given by
b
V = lim a π y2 Δx = π3y2 dx.
Δx→0
x=b
x=a
a
We can use the above results to verify the volumes of some common geometrical shapes.
Volume of a cylinder
The volume of a cylinder of height h and radius r is given by πr2h. To verify this
result, consider a cylinder lying on its side.
y
FS
y
h
x
0
h
PR
O
0
O
r
r
r
x
h
PA
G
E
To form the cylinder, rotate the line y = r by 360° about the x-axis between x = h
and the origin. Since h is a constant and r is a constant, they can be taken outside the
integral sign.
h
V = π3y2dx
0
h
TE
D
= π3r2dx
0
h
EC
= πr231 dx
0
h
R
= πr2[x]
0
R
= πr2 (h − 0)
U
N
C
O
= πr2h
Volume of a cone
The volume of a cone of height h and radius r is given by 13πr2h. To verify this result,
it is easiest to consider the cone lying on its side.
Consider a line which passes through the origin, so its equation is given by y = mx.
Now rotate the line 360° about the x-axis, between x = h and the origin, so that it
forms a cone.
y
y
r
0
h
h
x
0
r
x
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 505
505
08/07/15 2:33 AM
rx
r
The gradient of the line is m = , so the line has the equation y = . The volume
h
h
is given by
h
V = π3y2dx
0
h
= π3
0
r2x2
dx
h2
h
=
FS
2
3x dx
0
h
πr2 x3
c d
h2 3 0
πr2 h3
a − 0b
h2 3
E
= 13πr2h
O
=
h2
PR
O
=
πr2
PA
G
Volume of a sphere
The volume of a sphere of radius r is given by 43πr3. To verify this result, consider
the circle x2 + y2 = r2 with centre at the origin and radius r. If we rotate this circle
by 360° or rotate the top half of the circle y = "r2 − x2 about the x-axis, between
x = −r and x = r, it forms a sphere.
y
r
EC
TE
D
y
r
0
R
O
C
U
N
–r
r x
R
–r
0
r x
–r
r
V = π 3 y2dx
−r
r
= π 3 (r2 − x2) dx by symmetry
−r
r
= 2π3 (r2 − x2) dx
0
= 2π c r2x − 13x3 d
= 2π a
r3
= 43π r3
506 −
1 3
r
3
r
0
− 0b
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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solid volumes formed by rotating curves about the x-axis
When rotating the curve y = f(x) about the x-axis between x = a and x = b, the
volume formed is given by V = π3y2 dx. The integrand must be in terms of constants
and x-values only.
a
b
WorKeD
eXaMPLe
5
The area bounded by the curve y = x2, the x-axis, the origin and the line
x = 2 is rotated about the x-axis to form a solid of revolution. Find the
volume formed.
WritE/draW
FS
tHinK
PR
O
area to rotate.
O
y
6
5
4
3
2
1
1 Sketch the graph and identify the
1 2 3 4 5 x
R
R
EC
TE
D
PA
G
E
–5 –4 –3 –2 –1 0
–1
–2
O
2 Write a definite integral which gives
U
N
C
the volume.
3 Antidifferentiate using the rules.
4 Evaluate the definite integral.
–5 –4 –3 –2 –1 0
–1
–2
–3
–4
–5
–6
1 2 3 4 5 x
V = π3y2 dx
b
a
a = 0, b = 2, y = x2 so y2 = x4
V = π3x4 dx
2
0
V = πc
2
x5
d
5 0
V = π c 32
− 0d
5
V=
5 State the volume.
y
6
5
4
3
2
1
32π
5
The volume formed is
32π
cubic units.
5
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 507
507
08/07/15 2:33 AM
solid volumes formed by rotating curves about the y-axis
y
y
x = f(y)
b
x = f(y)
b
∆y
x
P (x, y)
a
a
x
0
x
0
FS
When the curve x = f( y) is rotated about the y-axis between y = a and y = b, the
volume formed is given by V = π3x2 dy. The integrand must be in terms of constants
and y-values only.
a
WorKeD
eXaMPLe
6
PR
O
O
b
The area bounded by the curve y = x2, the y-axis, the origin and the line
y = 4 is rotated about the y-axis to form a solid of revolution. Find the
volume formed.
WritE/draW
E
tHinK
y
6
5
4
3
2
1
PA
G
1 Sketch the graph and identify the
TE
D
area to rotate.
EC
–5 –4 –3 –2 –1 0
–1
–2
R
y
6
5
4
3
2
1
R
O
C
U
N
2 Write a definite integral which
gives the volume.
1 2 3 4 5 x
–5 –4 –3 –2 –1 0
–1
–2
1 2 3 4 5 x
V = π3x2 dy
b
a
a = 0, b = 4, y = x2
V = π3y dy
4
0
3 Antidifferentiate using the rules.
508
V = πc
y2
2
4
d
0
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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08/07/15 2:33 AM
V = π[8 − 0]
V = 8π
The volume formed is 8π cubic units.
4 Evaluate the definite integral.
5 State the volume.
applications
The volumes of some common geometrical objects can now be found using calculus,
by rotating lines or curves about the x- or y-axes and using the above techniques.
7
A drinking glass has a base diameter of 5 cm, a top diameter of 7 cm and
a height of 11 cm. Find the volume of the glass to the nearest mL.
FS
WorKeD
eXaMPLe
PR
O
O
7
PA
G
E
11
5
TE
D
tHinK
WritE/draW
y
1 Sketch the graph and identify the area
EC
B
R
y = mx + c
R
A
O
0
C
2 Establish the gradient of the line segment AB
U
N
joining the points A
5
a , 0b
2
( , 11)
7
–
2
to rotate. Write the coordinates of the
points A and B.
and B
7
a , 11 b .
2
3 Establish the equation of the line segment
joining the points A and B.
m=
11 − 0
7
2
− 52
( , 0)
5
–
2
x
= 11
y − 0 = 11ax − 52b
y = 11x − 55
2
The glass is formed when the line y = 11x − 55
2
for 52 ≤ x ≤ 72 is rotated about the y-axis.
V = π3x2 dy
b
4 Rotate this line about the y-axis to form the
required volume.
a
a = 0, b = 11
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 509
509
08/07/15 2:33 AM
y = 11x − 55
2
5 Rearrange to make x the subject.
11x = y + 55
2
= 12 (2y + 55)
1
x = 22
(2y + 55)
11
V=
6 Write a definite integral which gives
the volume.
π
(2y + 55) 2 dy
484 3
11
π
V=
c (2y + 55) 3 d
2 × 3 × 484
0
π ( 3
V=
77 − 553)
2904
7 Antidifferentiate using the rules.
V ≈ 99.92π
≈ 314 cm3
9 Since 1 cm3 = 1 mL, state the volume of
PR
O
O
8 Evaluate the definite integral.
FS
0
The volume of the glass is 314 mL.
ExErcisE 10.3 Solids of revolution
The area bounded by the curve y = !x, the x-axis, the origin and the
line x = 4 is rotated about the x-axis to form a solid of revolution. Find the
volume formed.
WE5
TE
D
1
PractisE
PA
G
E
the glass.
2 The area bounded by the curve y = 4 − x2 and the coordinate axis is rotated about
the x-axis to form a solid of revolution. Find the volume formed.
The area bounded by the curve y = !x, the y-axis, the origin and the
line y = 2 is rotated about the x-axis to form a solid of revolution. Find the
volume formed.
EC
WE6
R
3
4 The area bounded by the curve y = 4 − x2 and the coordinate axis is rotated about
A plastic bucket has a base diameter of 20 cm, a top diameter of 26 cm and a
height of 24 cm. Find the volume of the bucket to the nearest litre.
WE7
U
N
C
O
5
R
the y-axis to form a solid of revolution. Find the volume formed.
26
24
20
510
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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08/07/15 2:33 AM
6 A soup bowl has a base radius of 6 cm, a top radius of 8 cm, and a height of
7 cm. The edge of the bowl is a parabola of the form y = ax2 + c. Find the
capacity of the soup bowl.
y
y = ax2 + c
8
7
O
7 a
Find the volume of the solid of revolution formed when the area between the
line y = 3x, the x-axis, the origin and x = 5 is rotated 360° about the x-axis.
b Find the volume of the solid of revolution formed when the area between the
line y = 3x, the y-axis, the origin and y = 5 is rotated 360° about the y-axis.
c A cone is formed by rotating the line segment of 2x + 3y = 6 cut off by the
coordinate axes about:
ithe x-axis
iithe y-axis.
Find the volume in each case.
PA
G
E
PR
O
Consolidate
FS
x
6
8aIf the region bounded by the curve y = 3 sin(2x), the origin, the x-axis and the
EC
TE
D
first intercept it makes with the x-axis is rotated 360° about the x-axis, find the
volume formed.
b If the region bounded by the curve y = 4 cos(3x), the coordinate axes and the
first intercept it makes with the x-axis is rotated 360° about the x-axis, find the
volume formed.
π
c If the region bounded by the curve y = sec(2x), the coordinate axes and x =
8
is rotated 360° about the x-axis, find the volume formed.
x
2
R
d If the region bounded by the curve y = 2e , the coordinate axes and x = 2 is
R
rotated 360° about the x-axis, find the volume formed.
9 a i If the area between the curve y = 3x2 + 4, the coordinate axes and the line
U
N
C
O
x = 2 is rotated 360° about the x-axis, find the volume formed.
iiIf the area between the curve y = 3x2 + 4, y = 4 and y = 10 is rotated about
the y-axis, find the volume formed.
b i For the curve y = x2 − 9, find the area between the curve and the x-axis.
iiIf the area described in i is rotated 360° about the x-axis, find the
volume formed.
iiiIf the area described in i is rotated 360° about the y-axis, find the
volume formed.
y2
x2
10a Find the volume formed by rotating the ellipse
+
= 1 about:
25 16
ithe x-axis
iithe y-axis.
2
x2 y
b i Determine the volume formed if the ellipse
+
= 1, having semi-major
a2 b2
and semi-minor axes a and b respectively, is rotated about the x-axis.
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 511
511
08/07/15 2:33 AM
iiDetermine the volume if the ellipse from part bi is
rotated about the y-axis.
iiiIf a = b, verify that your results from bi and ii give
the volume of a sphere.
c An egg can be regarded as an ellipsoid. The egg has a
total length of 57 mm and its diameter at the centre is
44 mm. Find its volume in cubic millimetres.
11aThe drive shaft of an industrial spinning machine is
x
20
E
PR
O
O
FS
2 metres long and has the form of the curve y = e ,
measured in metres, rotated 360° about the x-axis
between the y-axis.
iFind the volume in cubic metres.
iiWhat is the length of a similar shaft if it encloses a volume of 3.5π m3?
b A piece of plastic tubing has its boundary in the form of the curve
1
y=
.
"4 + 9x2
When this curve is rotated 360° about the x-axis between the y-axis and the
line x = 5, find its volume.
PA
G
3
2
12aA wine glass is formed when the arc OB with the equation y = x is rotated
about the y-axis, as shown. The dimensions of the glass are given in cm.
Find the volume of the glass in mL.
y
TE
D
4
B
0
x
U
N
C
O
R
R
EC
3
–
y = x2
b The diagram shows a wine barrel. The
barrel has a total length of 60 cm, a total
height of 60 cm at the middle and a total
height of 40 cm at the ends.
iIf the upper arc can be represented by
a parabolic boundary, show that its
equation is given by
x2
y = − + 30 for −30 ≤ x ≤ 30.
90
y
30
20
10
–30 –20 –10 0
–10
10
20
30 x
–20
–30
512 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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08/07/15 2:33 AM
FS
iiIf the arc is rotated about the x-axis, find the capacity of the wine barrel to
PA
G
TE
D
8
E
PR
O
O
the nearest litre.
13aA vase has a base radius of 6 cm, its top radius is 8 cm and its height is 20 cm.
Find the volume of the vase in cubic centimetres if the side of the vase is
modelled by:
ia straight line of the form y = ax + c
iia parabola of the form y = ax2 + c
iiia cubic y = ax3 + c
iva quartic of the form y = ax4 + c.
6
R
EC
20
R
b Another vase is modelled by rotating the curve with the equation
x2 ( y − 10)
−
=1
4
12
about the y-axis between the x-axis and y = 20. Find the volume of water
needed to completely fill this vase.
14a For the curve y = e−ax where a > 0:
ifind the area A bounded by the curve, the coordinate axes and x = n
iiif the region in i is rotated 360° about the x-axis, find the volume formed, V.
iiiDetermine lim A and lim V.
U
N
C
O
2
n→∞
n→∞
b The region bounded by the rectangular hyperbola xy = 1, the x-axis and the
lines x = 1 and x = a has an area of A and a volume V when rotated about
the x-axis.
iFind A and V.
iiFind lim A and lim V if they exist.
a→∞
a→∞
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 513
513
08/07/15 2:34 AM
15aIf the region bounded by the curve y = a sin(nx), the origin, the x-axis and
Volumes
Volumes of revolution
R
10.4
EC
TE
D
PA
G
E
PR
O
O
FS
the first intercept it makes with the x-axis is rotated about the x-axis, find the
volume formed.
b If the region bounded by the curve y = a cos(nx), the coordinate axes and
the first intercept it makes with the x-axis is rotated about the x-axis, find the
volume formed.
16aProve that the volume of a right truncated cone of inner and outer radii r1 and
r2 respectively and height h is given by
πhar32 − r31b
.
3(r2 − r1)
b i A hemispherical bowl of radius r contains water to a depth of h. Show that
the volume of water is given by
πh2
(3r − h).
3
r
iiIf the bowl is filled to a depth of , what is the volume of water in the bowl?
2
1 1 1
17 a
The region bounded by the curve + = , the x-axes and x = 0
x y 4
Master
and x = 2 is rotated about the x-axis. Find the volume formed.
b The region bounded by the curve !x + !y = 2
and the coordinate axes is rotated about the x-axis.
Find the volume formed.
18 A fish bowl consists of a portion of a sphere of radius
20 cm. The bowl is filled with water so that the radius
of the water at the top is 16 cm and the base of the
bowl has a radius of 12 cm. If the total height of the
water in the bowl is 28 cm, find the volume of the water
in the bowl.
y
U
N
C
O
R
Volumes around the x-axis
Consider y = g(x) and y = f(x) as two continuous non-intersecting curves on a ≤ x ≤ b
and g(x) ≥ f(x). If the area between the curves is rotated 360° about the x-axis, it forms
a volume of revolution.
y = f(x)
0
514 a
y = g(x)
P(x, y)
b
x
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 514
08/07/15 2:34 AM
However, the volume is not solid and has a hole in it.
Consider a cross-sectional area shaped like a circular
washer, with inner radius r1 = y1 = f(x) and outer
radius r2 = y2 = g(x).
r1
The volume formed is V = π3ar22 − r12 b dx. Note that we
b
r2
a
must express the inner and outer radii in terms of constants
and x-values only.
Find the volume formed when the area bounded by the curve y = 4 − x2
and the line y = 3 is rotated about:
b the y-axis.
PR
O
a the x-axis
FS
8
O
WorKeD
eXaMPLe
tHinK
WritE/draW
a 1 Find the points of intersection between
a Let y1 = 4 − x2 and y2 = 3.
y1 = y2
4 − x2 = 3
x2 = 1
x = ±1
2 Sketch the graph and identify the area to
–2
–1
EC
TE
D
rotate about the x-axis.
PA
G
E
the curve and the line.
R
R
terminals of integration. The volume formed
has a hole in it.
–2
–1
C
O
0
–1
–2
–3
–4
y = 4 – x2
1
2 x
y
y = 4 – x2
4
3
y1
3 2
1
3 Identify the inner and outer radii and the
U
N
y
4
3
2
1
0
–1
–2
–3
–4
1
2 x
y1 = x2 – 4
V = π3ar22 − r12 b dx
b
a
a = −1, b = 1, r2 = y1, r1 = 3
1
4 Write a definite integral for the volume.
V = π 3 ((4 − x2) 2 − 9) dx
−1
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 515
515
08/07/15 2:34 AM
1
V = π 3 (16 − 8x2 + x4 − 9) dx
5 Expand and simplify the integrand.
−1
1
V = 2π3 (7 − 8x2 + x4) dx
6 Use symmetry to write the volume as
a definite integral.
0
1
0
V = 2π c a7 − 83 + 15b − 0 d
b 1 Sketch the graph and identify the area
y
4
3
2
1
b
to rotate about the y-axis. This is a solid
volume of revolution.
–1
y = 4 – x2
0
–1
–2
–3
–4
1
2 x
PA
G
–2
136π
cubic units.
15
PR
O
The volume is
E
9 State the value of the volume.
O
8 Evaluate the definite integral.
FS
V = 2π c 7x − 83x3 + 15x5 d
7 Perform the integration.
V = π3x2 dy
TE
D
b
R
EC
2 Write a definite integral for the volume.
O
R
3 Perform the integration.
C
4 Evaluate the definite integral.
U
N
5 State the value of the volume.
a
a = 3, b = 4, x2 = 4 − y
V = π3 (4 − y) dy
4
3
V = π c 4y −
1 2
y d
2
4
3
V = π c a4 × 4 − 12 × 42b − a4 × 3 − 12 × 32b d
The volume is
π
cubic units.
2
Volumes around the y-axis
In a similar way, if x = g( y) and x = f( y) are two continuous non-intersecting curves
on a ≤ y ≤ b where g(y) ≥ f(y) and the area between the curves is rotated 360° about
the y-axis, it forms a volume of revolution.
However, the volume is not solid and has a hole in it. Consider a typical
cross-sectional area with inner radius r1 = x1 = f( y) and outer radius r2 = x2 = g( y).
516 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 516
08/07/15 2:34 AM
The volume formed is V = π3ar22 − r12 b dy. Note that we must express the inner and
b
a
outer radii in terms of constants and y-values only.
WorKeD
eXaMPLe
9
Find the volume formed when the area bounded by the curve y = x2 − 4,
the x-axis and the line x = 3 is rotated about:
a the y-axis
y=9−4=5
O
intersection between the
curve and the line.
a Let y1 = x2 − 4 and x2 = 3
PR
O
a 1 Find the points of
FS
WritE/draW
y
6
5
4
3
2
1
2 Sketch the graph and
PA
G
identify the area to rotate
about the y-axis.
E
tHinK
b the x-axis.
3 Identify the inner and outer
U
N
C
O
R
R
EC
radii and the terminals of
integration. The volume
formed has a hole in it.
TE
D
–4 –3 –2 –1 0
–1
–2
–3
–4
y
6
5
4
3
2
1
–4 –3 –2 –1 0
–1
–2
–3
–4
1 2 3 4 x
3
r1
1 2 3 4 x
V = π3ar22 − r12 b dy
b
a
a = 0, b = 5, r2 = x2 = 3, r1 = x
V = π3 (9 − x2) dy
5
4 Write a definite integral for
the volume.
0
5
= π3 (9 − ( y + 4)) dy
0
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 517
517
08/07/15 2:34 AM
5
V = π3 (5 − y) dy
5 Simplify the integrand.
0
5
V = π c 5y −
6 Perform the integration.
1 2
y d
2
0
1
2
V = π c 5 × 5 − × 52 − 0 d
b 1 Sketch the graphs and
y
6
5
4
3
2
1
b
identify the area to rotate
about the x-axis. This is a
solid volume of revolution.
–3
–2
y = 4 – x2
V = π3y2 dx
–1
0
–1
–2
–3
–4
–5
–6
1
2
3
4 x
PA
G
–4
y = x2 – 4
FS
volume.
25π
cubic units.
2
O
The volume is
8 State the value of the
PR
O
integral.
E
7 Evaluate the definite
2 Write a definite integral for
the volume.
TE
D
b
a
a = 2, b = 3, y2 = 1 x2 − 4 2 2
R
R
EC
V = π3 (x2 − 4) 2 dx
C
O
3 Expand the integrand.
U
N
4 Perform the integration.
5 Evaluate the definite
integral.
6 State the value of the
3
2
V = π3 (x4 − 8x2 + 16) dx
3
2
V = π c 15x5 − 83x3 + 16x d
3
2
V = π c a15 × 35 − 83 × 33 + 16 × 3b − a15 × 25 − 83 × 23 + 16 × 2b d
The volume is
volume.
113π
cubic units.
15
Composite figures
Sometimes we need to carefully consider the diagram and identify the regions which
are rotated about the axes.
518 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 518
08/07/15 2:34 AM
WorKeD
eXaMPLe
10
Find the volume formed when the area between the curves y = x2 and
y = 8 − x2 is rotated about:
a the x-axis
b the y-axis.
WritE/draW
y1 = y2
x2 = 8 − x2
2x2 = 8
x2 = 4
x = ±2
When x = ±2, y = 4.
y
8
7
6
5
4
3
2
1
2 Sketch the graphs and
y1 = x2
r2
–3
EC
TE
D
–4
PA
G
E
identify the area to rotate
about the x-axis.
FS
between the two curves.
a Let y1 = x2 and y2 = 8 − x2.
R
3 Identify the inner and outer
O
R
radii and the terminals of
integration.
C
4 Write a definite integral for
U
N
the volume.
5 Expand and simplify the
integrand.
–2
–1 0
–1
–2
–3
–4
–5
–6
–7
–8
O
a 1 Find the points of intersection
PR
O
tHinK
r1
1
2
3
4 x
y2 = 8 – x2
V = π3ar22 − r12 b dx
b
a
a = −2, b = 2, r2 = y2, r1 = y1
2
V = π 3((8 − x2) 2 − (x2) 2) dx
−2
2
V = π 3(64 − 16x2 + x4 − x4) dx
−2
2
6 Use symmetry to write
the volume as a definite
integral in simplest form.
V = 2π3 (64 − 16x2) dx
0
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 519
519
08/07/15 2:34 AM
2
V = 2π c 64x −
16 3
x d
3
8 Evaluate the definite integral.
V = 2π c a64 × 2 − 16
× 23 b − 0 d
3
9 State the value of the volume.
The volume is
512π
cubic units.
3
y
8
7
6
5
4
3
2
1
y = x2
FS
the area to rotate about
the y-axis.
y = 8 – x2
1 2 3 4 5 x
–5 –4 –3 –2 –1 0
–1
–2
4
8
two sections. Write a
definite integral for the
total volume.
PA
G
V = π3x12 dy + π3x22 dy
2 The region comprises
O
b
PR
O
b 1 Sketch the graph and identify
0
E
7 Perform the integration.
0
4
4
8
TE
D
= π3y dy + π3 (8 − y) dy
0
4
8
0
4
V = π c 12y2 d + π c 8y − 12y2 d
EC
3 Perform the integration.
4
4 Evaluate the definite integral.
The volume is 16π cubic units.
O
R
R
5 State the value of the volume.
V = π c a12 × 42b − 0 d + π c a8 × 8 − 12 × 82b − a8 × 4 − 12 × 42b d
U
N
C
Exercise 10.4 Volumes
PRactise
520 Find the volume formed when the area bounded by the curve y = 9 − x2 and
the line y = 5 is rotated about:
a the x-axis
b the y-axis.
2 Find the volume formed when the area bounded by the curve y = !x, the y-axis,
and the line y = 2 is rotated about:
a the x-axis
b the y-axis.
3 WE9 Find the volume formed when the area bounded by the curve y = x2 − 9, the
x-axis and the line x = 4 is rotated about:
a the y-axis
b the x-axis.
1
WE8
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 520
08/07/15 2:34 AM
4 Find the volume formed when the area bounded by the curve y = !x, the x-axis
and the line x = 4 is rotated about:
a the y-axis
b the x-axis
5 WE10 Find the volume formed when the area between the curves y = x2 and
y = 18 − x2 is rotated about:
a the x-axis
b the y-axis.
6 When the area between the curves y = x2 − 4x + 4 and y = 4 + 4x − x2 is rotated
about the x-axis, it forms a volume of revolution. Find the volume.
7 Consider the area bounded by the curve y = 16 − x2 and the line y = 12.
a If this area is rotated about the x-axis, it forms a volume of revolution. Find the
FS
Consolidate
U
N
C
O
R
R
EC
TE
D
PA
G
E
PR
O
O
volume formed.
b If this area is rotated about the y-axis, it forms a volume of revolution. Find the
volume formed.
8 Consider the area bounded by the curve y = x2 − 16, the x-axis and the
line x = 5.
a If this area is rotated about the y-axis, it forms a volume of revolution. Find the
volume formed.
b If this area is rotated about the x-axis, it forms a volume of revolution. Find the
volume formed.
9 Let R be the region for the area bounded by the curve y = x3, the x-axis and
the line x = 3.
a If the region R is rotated about the y-axis, it forms a volume of revolution.
Find the volume formed.
b If the region R is rotated about the x-axis, it forms a volume of revolution.
Find the volume formed.
10 The area between the curve y = !2x − 8, the x-axis and the line x = 6 is
rotated about:
a the y-axis
b the x-axis.
Determine the volume of the resulting solid of revolution in each case.
x
11aFind the volume formed when the area bounded by the curve y = 4 cos−1 a b
2
and the lines x = 2 and y = 2π is rotated about the y-axis.
x
3
b Find the volume formed when the area bounded by the curve y = 2 sin−1 a b,
the x-axis and the line x = 3 is rotated about the y-axis.
12aFind the volume formed when the area between the curves y = 2x and y = x2 is
rotated about:
i the x-axis
ii the y-axis.
b Find the volume formed when the area between the curves y = x2 and y = x3 is
rotated about:
i the x-axis
ii the y-axis.
x2
13aIf the area between the two curves y = 2 !x and y =
is rotated about the
4
x-axis, find the volume formed.
x2
b If the area between the two curves y = !ax and y =
where a > 0 is rotated
a
about the y-axis, find the volume formed.
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 521
521
08/07/15 2:34 AM
14 a Find the volume formed when the area in the first quadrant bounded by the
πx
b and the line y = 4x is rotated about the x-axis.
6
Find the volume formed when the area in the first quadrant bounded by the
πx
curve y = 4 sina b and the line y = 4 − x is rotated about the x-axis.
8
Find the volume formed when the area between the curves y = x2 and
y = 32 − x2 is rotated about:
i the x-axis
ii the y-axis.
Find the volume formed when the area in the first quadrant bounded by the
πx
curve y = 8 sina b and the parabola y = 2x2 is rotated about the x-axis.
4
A cylindrical hole of radius 2 cm is cut vertically through the centre of a solid
sphere of cheese of radius 4 cm. Find the volume of the cheese remaining.
a
A cylindrical hole of radius is drilled horizontally through the centre of a
2
sphere of radius a. Show that the volume of the remaining solid is
!3 3
given by
a π.
2
Find the volume of a torus (a doughnutshaped figure) formed when the area
bounded by the circle x2 + (y − 8) 2 = 16
is rotated about the x-axis.
Find the volume of a torus (a doughnutshaped figure) formed when the area
bounded by the circle (x − a) 2 + y2 = r2
where a > r > 0 is rotated about
the y-axis.
Find the volume formed when the
area bounded by the curve y = a2 − x2
a2
and the line y = , where a > 0, is
4
rotated about:
i the x-axis
ii the y-axis.
Find the volume formed when the area bounded by the curve y = x2 − a2, the
x-axis and the line x = 2a, where a > 0, is rotated about:
i the y-axis
ii the x-axis.
Find the volume formed when the area between the curves y = x2 and
y = 2a2 − x2, where a > 0, is rotated about:
i the x-axis
ii the y-axis.
b
MastEr
17 a
O
R
R
18 a
EC
TE
D
b
O
16 a
PR
O
b
E
15 a
PA
G
b
FS
curve y = 12 sina
U
N
C
b
10.5
c
Arc length, numerical integration and
graphs of antiderivatives
Length of a curve
In this section integration will be used to determine the arc length, s, of a plane curve.
Suppose that the curve y = f(x) is a continuous curve on the closed interval a ≤ x ≤ b. The
curve can be thought of as being made up of infinitely many short line segments as shown.
522
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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08/07/15 2:34 AM
y
y = f(x)
S
∆s
x
b
∆x
FS
a
∆y
By Pythagoras’ theorem, the length of a typical small segment Δs is equal to
O
"(Δx) 2 + (Δy) 2. The total length of the curve s from x = a to x = b is obtained
by summing all such segments and taking the limit as Δx → 0.
PR
O
x=b
x=b
Δy 2
s = lim a "(Δx) 2 + (Δy) 2 = lim a 1 + a b Δx
Δx→0
Δx→0
Δx
x=a
x=a Å
11
a
a
PA
G
WorKeD
eXaMPLe
b
E
dy 2
s = 3 1 + a b dx = 3"1 + (f ′(x)) 2 dx
Å
dx
b
Find the length of the curve y =
WritE
1 Find the gradient function
dy
TE
D
tHinK
x3
1
+
from x = 1 to x = 5.
2
6x
x3
1
+
2 6x
x3 1
= + x−1
2 6
dy 3x2 1 −2
=
− x
2
6
dx
2
3x
1
=
− 2
2
6x
y =
by
O
R
R
EC
dx
differentiating and express back with
positive indices.
dy 2
s = 3 1 + a b dx with a = 1 and b = 5
Å
dx
C
b
2 Substitute into the formula and write
U
N
a definite integral which gives the
required length.
a
5
3x2
1 2
s=3 1+a
− 2 b dx
Å
2
6x
1
3x2
3x2
1
1
s = 3 1 + qa b − 2 ×
× 2 + a 2 b r dx
Å
2
2
6x
6x
5
3 Expand using (a −
b) 2
=
a2
− 2ab +
b2.
2
2
1
9x4 1
1
s=3 1+a
− +
b dx
Å
4
2 36x4
5
4 Cancel terms and simplify.
1
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 523
523
08/07/15 2:34 AM
9x4 1
1
s=3 1+a
+ +
b dx
Å
4
2 36x4
5
5 Simplify the integrand.
1
3x2
1
s=3 a
+ 2 b dx
Å 2
6x
5
6 Recognise the integrand as a
perfect square.
2
1
3x2
1
s = 3a
+ 2 b dx
2
6x
5
7 Express the integrand in a form which
can be integrated.
3x2 1 −2
s = 3a
+ x b dx
2
6
1
5
5
s=
9 Evaluate the definite integral.
a
53
2
−6
1
b
× 5
numerical integration
a
13
2
−6
1
b
× 1
PA
G
s = 932
units
15
10 State the final result.
−
E
8 Perform the integration.
PR
O
x3 1
x3
1
s = c − x−1 d = c − d
2 6
2 6x 1
1
O
FS
1
5
12
Express the length of the curve y = x3 from x = 0 to x = 2 as a definite
integral and hence find the length, giving your answer correct to 4 decimal
places.
R
R
WorKeD
eXaMPLe
EC
TE
D
The worked example above is somewhat contrived, because for many simple
curves, the arc length formula results in a definite integral that cannot be evaluated
by techniques of integration. In these situations we must resort to numerical
methods, such as using calculators, which can give numerical approximations to
the definite integrals obtained.
O
tHinK
C
1 Find the gradient function
dx
by
U
N
differentiating.
WritE
dy
2 Substitute into the formulae and write
a definite integral which gives the
required length.
y = x3
dy
= 3x2
dx
dy
s = 3 1 + a b dx with a = 0 and b = 2
Å
dx
b
2
a
s = 3"1 + 9x4 dx
2
0
3 This definite integral must be
evaluated using a calculator. State
the final result.
524
s = 8.6303
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
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using numerical integration in differential equations
For many first-order differential equations, the integral cannot be found by techniques
of integration. In these situations calculators can give a numerical approximation.
dy
Consider the differential equation
= f(x) and y(x0) = y0. We want to find the value
dx
of y1 when x = x1. We obtain y = 3f(t)dy + c, where we have arbitrarily used zero as
x
0
FS
the lower terminal and t as a dummy variable.
Note: x is maintained as the independent variable of the solution function.
We use the given initial condition, y = y0 when x = x0, then substitute
x0
x0
0
O
y0 = 3 f(t)dt + c to find that the constant of integration c = y0 − 3f(t) dt. Substituting
x0
0
0
PR
O
back for c gives y = 3f(t)dt + y0 − 3f(t)dt.
x
0
y = y0 + 3f(t)dt + 3f(t)dt
PA
G
0
x0
E
x
By properties of the definite integral,
0
EC
When x = x1,
TE
D
y = y0 + 3f(t)dt + 3f(t)dt
U
N
Given that
0
0
x
x0
= y0 + 3f(t)dt
x0
x1
y1 = y0 + 3f(t)dt.
R
R
O
13
C
WorKeD
eXaMPLe
x
x0
dy
dx
= ex , y (1) = 2:
2
a find a definite integral for y in terms of x
b determine the value of y correct to 4 decimal places when x = 1.5.
tHinK
WritE
a 1 Antidifferentiate the differential equation.
a
dy
2
= ex
dx x
y = 3et dt + c
2
0
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 525
525
08/07/15 2:34 AM
Substitute x = 1 when y = 2:
2 Use the given initial conditions to find
2 = 3et dt + c
1
the value of the constant of integration.
2
0
c = 2 − 3et dt
1
2
0
y = 3et dt + 2 − 3et dt
1
x
2
0
0
0
y = 2 + 3et dt + 3et dt
x
2
2
1
y = 2 + 3et dt
x
2
integral involving x.
1
b 1 Find the value of y at the required x-value.
PR
O
0
4 State the solution for y as a definite
FS
simplify using the properties of definite
integrals.
2
O
3 Substitute back for the constant and
b Substitute x = 1.5:
y = 2 + 3 et dt
E
1.5
PA
G
2
1
2 This definite integral must be evaluated
y = 4.6005
TE
D
using a calculator. State the final result.
U
N
C
O
R
R
EC
approximating volumes
For many curves, the volume obtained also results in definite integrals that cannot
be evaluated by any techniques of integration. Other volumes can only be evaluated
by techniques that that we have not as yet covered. In either of these situations we
must again resort to numerical methods, such as using calculators, to find numerical
approximations to the definite integrals obtained.
When a question asks for an answer correct to a specified number of decimal
places, a calculator can be used to obtain a decimal approximation to the
definite integral.
WorKeD
eXaMPLe
14
a Find the exact volume formed when the area bounded by the curve
π
y = 2 sin (3x) , the x-axis, the origin and the line x = is rotated about
6
the x-axis.
b Set up a definite integral for the volume formed when the area
bounded by the curve y = 2 sin (3x) , the x-axis, the origin and the line
π
x = is rotated about the y-axis. Hence, find the volume correct to
6
4 decimal places.
526
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c10FurtherApplicationsOfIntegration.indd 526
08/07/15 2:34 AM
THINK
WRITE/DRAW
a 1 Sketch the graph and identify the
y
3
a
area to rotate about the x-axis.
2
1
– –6π
0
π
–
6
–1
π
–
3
π
–
2
2π
––
3
–2
O
–3
volume.
PR
O
y
3
2 Write a definite integral for the
2
0
PA
G
– –6π
E
1
– –3π
x
FS
– –3π
–1
π
–
6
π
–
3
x
TE
D
–2
–3
O
R
R
EC
V = π3y2dx
π
a = 0, b = , y = 2 sin(3x)
6
π
6
V = π34 sin2 (3x)dx
π
6
C
U
N
a
0
3 Use the double angle
formula sin2 (A) =
b
1
(1
2
− cos(2A)).
4 Perform the integration.
5 Evaluate the definite integral.
6 State the value of the volume.
V = 2π3 (1 − cos(6x))dx
0
V = 2π c x −
π
6
1
(6x) d
sin
6
0
π 1
1
V = 2π c a − sin(π) b − a0 − sin(0) b d
6 6
6
V=
π2
units3
3
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 527
527
08/07/15 2:34 AM
π
6
π
2
b When x = , y = 2 sin a b = 2.
b 1 Sketch the graph and identify the
area to rotate about the y-axis.
y
3
r1 = x1
r2 = –π6
2
1
– –6π
0
π
–
6
–1
π
–
3
x
FS
– –3π
O
–2
V = π3ar22 − r12 b dy
b
2 Identify the inner and outer radii
and the terminals of integration.
a
PR
O
–3
TE
D
PA
G
E
π
a = 0, b = 2, r2 = x2 = , r1 = x1
6
y = 2 sin(3x)
y
= sin(3x)
2
y
3x = sin−1 a b
2
EC
x1 =
3 Write a definite integral for
R
the volume.
4 This definite integral cannot
y
π2 1
− asin −1 a b b r dy
V = π3 q
2
36 9
2
2
0
V = 1.3963 units3
U
N
C
O
R
be evaluated by integration
techniques. Find a numerical
value for the definite integral
using a calculator, and state
the final result.
1 −1 y
sin a b
3
2
Graphs of antiderivatives of functions
Given a function f(x), we can sketch the graph of the antiderivative F(x) = 3f(x)dx
by noting key features and considering F′(x) = f(x). The table below shows the
relationships between the graphs. Note that the graph of the antiderivative cannot be
completely determined as it includes a constant of integration, which is a vertical
translation of the graph of F(x) parallel to the y-axis.
528 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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Graph of antiderivative F
F(x) has a negative gradient, or is
decreasing for x ∈ (a, b)
positive, f(x) > 0 for x ∈ (a, b)
F(x) has a positive gradient, or is
increasing for x ∈ (a, b)
f(x) cuts the x-axis at x = a from
negative to positive
F(x) has a local minimum at x = a
f(x) cuts the x-axis at x = a from
positive to negative
F(x) has a local maximum at x = a
f(x) touches the x-axis at x = a
F(x) has stationary point of
inflexion at x = a
f(x) has a turning point at x = a
F(x) has a point of inflexion at
x = a (non-stationary unless
f(x) = 0)
PR
O
O
FS
Graph of function f
negative, f(x) < 0 for x ∈ (a, b)
In particular
PA
G
E
• If f(x) is a linear function, then the graph of the antiderivative F(x) will be a
quadratic function.
• If f(x) is a quadratic function, then the graph of the antiderivative F(x) will be a
cubic function.
• If f(x) is a cubic function, then the graph of the antiderivative F(x) will be a
quartic function.
a Given the graph, sketch a possible graph of the antiderivative.
TE
D
15
y
4
3
2
1
–5 –4 –3 –2 –1 0
–1
–2
–3
–4
1 2 3 4 5 6 x
U
N
C
O
R
R
EC
WorKeD
eXaMPLe
b The graph of the gradient function is shown. Sketch a possible graph of
the function.
y
4
3
2
1
–4 –3 –2 –1 0
–1
–2
–3
–4
1 2 3 4 x
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 529
529
08/07/15 2:34 AM
THINK
WRITE/DRAW
a 1 The given graph crosses the x-axis at
x = 3, so the graph of the antiderivative
has a stationary point at x = 3.
y
8
6
4
2
a
2 At x = 3 the given graph changes from
2 4 6 8 x
–8 –6 –4 –2 0
–2
–4
–6
–8
positive to negative as x increases,
so the stationary point is a maximum
turning point.
3 No further information is provided, so
2 At x = −1 the gradient changes from
O
PR
O
1 2 3 4 5 x
TE
D
positive to negative as x increases, so
the stationary point is a maximum turning
point. At x = 3 the gradient changes
from negative to positive as x increases,
so the stationary point is a minimum
turning point.
–5 –4 –3 –2 –1 0
–2
–4
–6
–8
E
x = −1 and x = 3, so the graph of the
antiderivative has a stationary point at
these values.
y
8
6
4
2
b
PA
G
b 1 The given graph crosses the x-axis at
FS
we cannot determine the y-value of the
turning point or any values of the axis
intercepts. The graph of the antiderivative
could be translated parallel to the y-axis.
3 The gradient function has a turning point
EC
at x = 1 so the graph of the antiderivative
has a point of inflexion at x = 1.
4 No further information is provided.
U
N
C
O
R
R
We cannot determine the y-values
of the stationary points or the point
of inflexion, or any values of the
axis intercepts. The graph could be
translated parallel to the y-axis.
Exercise 10.5 Arc length, numerical integration and graphs
of antiderivatives
PRactise
530 x3 1
+ from x = 1 to x = 6.
12 x
3x4 + 4
2 Find the length of the curve y =
from x = 2 to x = 8.
12x
3 WE12 Express the length of the curve y = x2 from x = 0 to x = 3 as a
definite integral and hence find the length, giving your answer correct to
4 decimal places.
1
WE11
Find the length of the curve y =
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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2
from x = 1 to x = 4 as a definite integral
x2
and hence find the length, giving your answer correct to 4 decimal places.
dy
5 WE13 Given that
= sin(x2), y(0.5) = 3:
dx
a find a definite integral for y in terms of x
b determine the value of y correct to 4 decimal places when x = 1.
dy
π
1
1
6 Given the differential equation
= tana b, ya b = , find a definite integral
x
dx
8
2
for y in terms of x.
WE14 a Find the exact volume formed when the area bounded by the
π
curve y = 3 tan(2x), the x-axis, the origin and the line x = is rotated about
8
the x-axis.
O
7 FS
4 Express the length of the curve y =
PR
O
b Set up a definite integral for the volume formed when the area bounded by the
curve y = 3 tan(2x), the x-axis, the origin and the line x =
y-axis. Hence, find the volume correct to 4 decimal places.
π
is rotated about the
8
8aSet up a definite integral for the volume formed when the area bounded by the
a Given the graph, sketch a possible graph of the antiderivative.
TE
D
WE15 U
N
C
O
R
R
EC
9
PA
G
E
curve y = 3 sin−1 (2x), the x-axis, the origin and the line x = 12 is rotated about
the x-axis. Hence, find the volume correct to 4 decimal places.
b Find the exact volume formed when the area bounded by the curve
y = 3 sin−1 (2x), the x-axis, the origin and the line x = 12 is rotated about
the y-axis.
y
4
3
2
1
–5 –4 –3 –2 –1 0
–1
–2
–3
–4
1 2 3 4 5 x
b The graph of the gradient function is shown. Sketch a possible graph of
the function.
y
8
6
4
2
–4 –3 –2 –1 0
–2
–4
–6
–8
1 2 3 4 x
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 531
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08/07/15 2:34 AM
10 The graph of y = f ′(x) is shown. For the graph of y = f(x), state the x-values of
the stationary points and their nature.
y
25
20
15
10
5
O
11 a Find the length of y = 3x + 5 from x = 1 to x = 6.
x3
1
+
from x = 1 to x = 2.
6 2x
x4 + 48
c Find the length of the curve y =
from x = 2 to x = 4.
24x
PR
O
Consolidate
1 2 3 4 5 x
FS
–5 –4 –3 –2 –1 0
–5
–10
–15
–20
–25
b Find the length of the curve y =
1
PA
G
E
d Find the length of the curve y = 2 (ex + e−x) from x = 0 to x = 1.
3
4"2 2
12a Determine the length of the curve y =
x − 1 from x = 0 to x = 1.
3
b For the curve 27y2 = 4(x − 2) 3, find the length of the curve
TE
D
from x = 3 to x = 8.
2
c Find the length of the curve y = 3 "(x − 1) 3 from x = 1 to x = 9.
2
d Find the length of the curve y = 3 "(2x − 1) 3 from x =
5
2
to x = 92.
EC
13a Find the length of the curve y = "1 − x2 from x = 0 to x = 1.
b Find the length of the curve y = "9 − x2 from x = 0 to x = 3.
U
N
C
O
R
R
What length does this represent?
14 Set up definite integrals for the lengths of the following curves, and
hence determine the arc length in each case. Give your answers correct
to 4 decimal places.
a y = 3x2 + 5 from x = 1 to x = 6.
π
b y = 4 cos(2x) from x = 0 to x =
4
c y = 3e−2x from x = 0 to x = 1
d y = loge (2x + 1) from x = 0 to x = 3
2
x2 y
+ = 1.
15aSet up a definite integral for the total length of the ellipse
9
4
Find this length, giving your answer correct to 4 decimal places.
x
b Set up a definite integral for the total length of the curve y = 3 cos−1 a b.
2
Find this length correct to 4 decimal places.
16aFind the exact volume formed when the area bounded by the curve
y = 3 cos(2x), the coordinate axes and x =
532 π
is rotated about the x-axis.
4
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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08/07/15 2:34 AM
b Set up a definite integral for the volume formed when the area bounded by the
π
is rotated about the y-axis.
4
Determine this volume correct to 4 decimal places
curve y = 3 cos(2x), the coordinate axes and x =
PR
O
O
FS
1
dy
x
17a Given that
= e , y(1) = 3:
dx
ifind a definite integral for y in terms of x
iidetermine the value of y when x = 2.
dy
b Given that
= sin−1 (x2), y(0.1) = 1:
dx
ifind a definite integral for y in terms of x
iidetermine the value of y when x = 0.5.
dy
1
1
c Given that
, y(1) = :
=
dx "x3 + 8
3
ifind a definite integral for y in terms of x
iidetermine the value of y when x = 2.
graph of the original function.
a
E
18 The following graphs are of gradient functions. In each case sketch a possible
b
1 2 3 4 x
–6 –5 –4 –3 –2 –1 0
–1
–2
–3
–4
–5
EC
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–4 –3 –2 –1 0
–1
–2
–3
–4
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O
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c
y
5
4
3
2
1
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y
4
3
2
1
d
y
4
2
–5 –4 –3 –2 –1 0
–2
–4
–6
–8
–10
1 2 3 4 5 x
1 2 3 4 5 6 x
y
10
8
6
4
2
–5 –4 –3 –2 –1 0
–2
–4
1 2 3 4 5 x
19 Let A be the area bounded by the graph of y = x2 − 6x and the x-axis.
a Find the value of A.
b If the area A is rotated about the x-axis, find the volume formed.
c Find the length of the curve y = x2 − 6x from x = 0 to x = 6, giving your
answer correct to 4 decimal places.
d If the area A is rotated about the y-axis, find the volume formed.
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 533
533
08/07/15 2:34 AM
x
4
20 Let A 1 be the area bounded by the graph of y = 2 sin−1 a b, the x-axis, the origin
x
and the line x = 4. Let A 2 be the area bounded by the graph of y = 2 sin−1 a b,
4
the y-axis, the origin and the line y = π.
x
4
a Differentiate x sin−1 a b and hence find the value of A1.
b Find the value of A2.
c If the area A1 is rotated about the x-axis, find the volume formed, giving your
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answer correct to 4 decimal places.
d If the area A2 is rotated about the y-axis, find the exact volume formed.
e If the area A1 is rotated about the y-axis, find the exact volume formed.
f If the area A2 is rotated about the x-axis, find the volume formed, giving your
answer correct to 4 decimal places.
21 a i W
rite a definite integral which gives the length of the curve y = xn
Master
from x = a to x = b.
iiHence, find correct to 4 decimal places the length of the curve y = !x
from x = 1 to x = 9.
b i W
rite a definite integral which gives the length of the curve y = sin(kx)
from x = a to x = b.
iiHence, find correct to 4 decimal places the length of the curve y = sin(2x)
from x = 0 to x = π.
c i W
rite a definite integral which gives the length of the curve y = ekx
from x = a to x = b.
iiHence, find correct to 4 decimal places the length of the curve y = e2x
from x = 0 to x = 1.
22aFor the line y = mx + c, verify that the arc length formula gives the distance
(p − 1)(12a2p(p2 + p + 1) + 1)
12ap
O
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along the line between the points x = a and x = b.
b If a and p are positive real constants, show that the length of the curve
1
y = ax3 +
from x = 1 x = p is given by
12ax
.
U
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c Prove that the circumference of a circle of radius r is 2πr.
d Show that the total length of the ellipse
definite integral
2
x2 y
+
= 1 is given by the
a2 b2
a4 + x2 (b2 − a2)
4
s= 3
dx.
a Å
a2 − x2
a
0
10.6
Water flow
Torricelli’s theorem
Evangelista Torricelli (1608–1647) was an Italian scientist interested in mathematics
and physics.
534 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 534
08/07/15 2:34 AM
He invented the barometer to measure atmospheric pressure and was also one of
the first to correctly describe what causes the wind. Hhe also designed telescopes
and microscopes. Modern weather forecasting owes much to the work of Torricelli.
His main achievement is the theorem named after him, Torricelli’s theorem, which
describes the relationship between fluid leaving a container through a small hole in
the container and the height of the fluid in the container. Basically, the theorem states
that the rate at which the volume of the fluid leaves the container is proportional to
the square root of the height of the fluid in the tank. This theorem applies for all types
of containers.
16
A vase has a circular base and top with radii of 4 and 9 cm respectively,
and a height of 16 cm. The origin, O, is at the centre of the base of the
vase. The vase is formed when the curve y = a !x + b is rotated about
the y-axis. Initially the vase is filled with water, but the water leaks out
at a rate equal to 4 !h cm3/min, where h cm is the height of the water
remaining in the vase after t minutes. Set up the differential equation
for h and t, and determine how much time it takes for the vase to become
empty.
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eXaMPLe
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Problem solving
In solving problems involving fluid flow, we need to use the techniques of
finding volumes and use related rate problems to set up and solve differential
equations. Sometimes we may need to use numerical methods to evaluate a
definite integral.
tHinK
WritE/draW
y
1 Set up simultaneous equations which can be
9
16
h
U
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C
O
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R
EC
solved for a and b. The height of the vase is
16 and the height of the water in the vase is
h, so 0 ≤ h ≤ 16. Note all dimensions
are in cm.
2 Determine the values of a and b.
3 Determine the volume of the vase.
O
4
x
The curve y = a !x + b passes through the
points (4, 0) and (9, 16). Substituting:
(4, 0) ⇒ (1) 0 = 2a + b
(9, 16) ⇒ (2) 16 = 3a + b
(2) − (1) ⇒ a = 16, so b = −32.
The vase is formed when y = 16 !x − 32 for
4 ≤ x ≤ 9 is rotated about the y-axis.
When a curve is rotated about the y-axis,
the volume is V = π3x2dy.
h
0
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 535
535
08/07/15 2:34 AM
y = 16 !x − 32
4 Rearrange the equation to make x the subject.
16 !x = y + 32
1
"x = 16
( y + 32)
1
( y + 32) 2
x = 256
x2 =
( y + 32) 4
65 536
(y + 32) 4
V = π3
dy
65536
5 Determine a definite integral for the
volume of water when the vase is filled to
a height of h cm.
time, t. The rate is negative as it is a
decreasing rate.
O
0
Since the water leaks out at rate proportional to
the square root of the remaining height of the
dV
water,
= −4 !h.
dt
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6 Determine the given rates in terms of
FS
h
(y + 32) 4
V = π3
dy
65 536
7 Note the result used is from the numerical
techniques described earlier.
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0
E
h
dV π(h + 32) 4
=
dh
65 536
dh dh dV
=
dt dV dt
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8 Use related rates and a chain rule.
9 Set up the differential equation for the
EC
height h at time t.
10 To solve this type of differential equation,
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R
invert both sides.
O
11 Set up a definite integral for the time for
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the vase to empty.
12 Use a calculator to numerically evaluate
the definite integral.
13 State the result.
536 Substitute for the rates and use
dh −65 536 × 4 !h
=
dt
π(h + 32) 4
dV
dh
= 1/ :
dV
dh
4
dt −π(h + 32)
=
dh
262 144 !h
t=3
0
−π(h + 32) 4
dh
262 144"h
Note the order of the terminals from h = 16
to h = 0
16
t = 205.59; note that the time is positive.
The tank is empty after a total time of
205.59 minutes.
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 536
08/07/15 2:34 AM
WorKeD
eXaMPLe
17
2
y
x2
−
= 1 for 4 ≤ x ≤ 6,
16 500
y ≥ 0, is rotated about the y-axis to form a volume of revolution. The x- and
y-coordinates are measured in centimetres. The vase has a small crack in the
base, and the water leaks out at rate proportional to the square root of the
remaining height of the water. Initially the vase was full, and after 10 minutes
the height of the water in the vase is 16 cm. Find how much longer it will be
before the vase is empty.
Another vase is formed when part of the curve
WritE/draW
FS
tHinK
V = π3x2dy
1 The vase is formed when the given curve is
rotated about the y-axis.
O
b
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a
Find the value of y when x = 6.
y2
62
=
−1
500 16
2 Determine the height of the vase.
E
= 54
3 Sketch the region of the hyperbola which
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G
y2 = 625
y = 25 as y > 0
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forms the vase. The height of the vase is
25 cm and the height of the water in the vase
is h, so 0 ≤ h ≤ 25.
U
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R
4 Transpose the equation to make x2 the subject.
y
30
25
20
15
10
5
25
–6 –4
–2
O
–5
–10
6
h
2
4
4
6 x
y2
x2
−
=1
16 500
2
y
x2
=1+
16
500
16(500 + y2)
x2 =
500
4(500 + y2)
x2 =
125
4π
V=
(500 + y2)dy
1253
h
5 Find a definite integral for the volume of
water when the vase is filled to a height of
h cm, where 0 ≤ h ≤ 25.
6 Note the result used is from the numerical
techniques described earlier.
0
2
dV 4π(500 + h )
=
dh
125
topic 10 Further aPPLICatIons oF IntegratIon
c10FurtherApplicationsOfIntegration.indd 537
537
08/07/15 2:34 AM
7 Determine the given rates in terms of time t
Since the water leaks out at rate
­proportional to the square root of the
dV
­remaining height of the water,
= −k!h,
dt
where k is a positive constant.
in minutes.
dh dh dV
=
dt dV dt
8 Use related rates and a chain rule.
9 Set up the differential equation for the
FS
Substitute for the related rates, using
dV
dh
= 1/
dV
dh
height h at time t.
O
−125k!h
dh
=
dt
4π(h2 + 500)
2
4π
dt −A(h + 500)
=
where A =
dh
125 k
!h
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O
10 Incorporate the constants, into one constant.
To solve this type 2 differential equation,
invert both sides.
3
1
−
dt
= −Aah2 + 500h 2 b
dh
E
(1) t = −A3 ah + 500h
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11 Integrate with respect to h.
t=
12 Perform the integration.
5
2 2
−A c 5h
3
2
1
2
−
1
2
b dh
+ 1000h d + c, where c is the
­constant of integration.
R
R
14 Simplify the relationship.
EC
the values of the two unknowns.
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13 Two sets of conditions are required to find
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unknowns.
O
15 Find another relationship between the
16 Simplify the relationship and solve for the
constant of integration.
Initially, when t = 0, h = 25, since the vase
was full. Substitute t = 0 and h = 25:
0=
5
2
−A c 5 (25) 2
1
2
+ 1000 × 25 d + c
0 = −A c 25 × 3125 + 1000 × 5 d + c
c − 6250A = 0
c
6250
Since after 10 minutes the height of the ­
water in the vase is 8 cm, substitute
t = 10 and h = 16 into (1):
A=
10 =
5
2
−A c 5 (16) 2
1
+ 1000 × 162 d + c
Substitute for A:
10 = −
c 2
c × 1024 + 1000 × 4 d + c
6250 5
1 2
10 = c a1 − 6250
c × 1024 + 4000 d b
5
250
c = 156
4601
c = 33.96
538 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 538
08/07/15 2:34 AM
Since t =
17 Determine when the vase will be empty.
5
2 2
−A c 5h
1
2
+ 1000h d + c, the vase is
empty when h = 0, that is at time when t = c.
18 State the final result.
The vase is empty after a total time
of 33.96 minutes, so it takes another
23.96 ­minutes to empty.
1
A vase has a circular base and top with radii of 4 and 9 cm respectively,
and a height of 16 cm. The origin, O, is at the centre of the base of the vase. The
vase is formed when the line y = ax + b is rotated about the y-axis. Initially the
vase is filled with water, but the water leaks out at a rate equal to 2 !h cm3/min,
where h cm is the height of the water remaining in the vase after t minutes. Set up
the differential equation for h and t, and determine how long it will be before the
vase is empty.
WE16
PR
O
O
PRactise
FS
Exercise 10.6 Water flow
2 A vase has a circular base and top with radii of 4 and 9 cm respectively and a
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height of 16 cm. The origin, O, is at the centre of the base of the vase. The vase is
formed when the curve y = ax2 + b is rotated about the y-axis. Initially the vase
is filled with water, but the water leaks out at a rate equal to 2 !h cm3/min, where
h cm is the height of the water remaining in the vase after t minutes. Set up the
differential equation for h and t, and determine how long before the vase is empty.
65y2
x2
−
= 1 for 4 ≤ x ≤ 9,
16 4096
y ≥ 0, is rotated about the y-axis to form a volume of revolution. The x- and
y-coordinates are measured in centimetres. The vase has a small crack in the base,
and the water leaks out at rate proportional to the square root of the remaining
height of the water. Initially the vase was full, and after 10 minutes the height
of the water in the vase is 9 cm. Find how much time it will take for the vase to
become empty.
A vase is formed when part of the curve
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3
U
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O
4 A vase has a base radius of 4 cm, a top radius is 9 cm, and a height of 16 cm.
Consolidate
The origin, O, is at the centre of the base of the vase. Initially the vase is filled
with water, but the water leaks out at a rate proportional to the square root of the
remaining height of the water. The side of the vase is modelled by a quadratic
y = ax2 + b. Initially the vase was full, and after 10 minutes the height of
the water in the vase is 9 cm. Find how much time it will take for the vase to
become empty.
5 a
A cylindrical coffee pot has a base radius of 10 cm and a height of 49 cm and
is initially filled with hot coffee. Coffee is removed from the pot at rate equal
to 2 !h cm3/sec, where h cm is the height of the coffee remaining in the coffee
pot after t seconds. Set up the differential equation for h and t, and determine
how long it will be before the coffee pot is empty.
b A small cylindrical teapot with a base radius of 5 cm and a height of 16 cm
is initially filled with hot water. The hot water is removed from the teapot at
a rate proportional to !h cm3/min, where h cm is the height of the hot water
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 539
539
08/07/15 2:34 AM
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7aThe volume of a hemispherical bowl is given by
O
FS
remaining in the teapot after t minutes. Set up the differential equation for
h and t. If after 10 minutes the height of the hot water is 9 cm, what further
time elapses before the teapot is empty?
6a
A rectangular bathtub has a length of 1.5 metres and is 0.6 metres wide. It is
filled with water to a height of 1 metre. When the plug is pulled, the water
flows out of the bath a rate equal to 2 !h m3/min, where h is the height of the
water in metres in the bathtub at a time t minutes after the plug is pulled. How
long will it take for the bathtub to empty?
b A cylindrical hot water tank with a capacity of 160 litres is 169 cm tall and is
filled with hot water. Hot water starts leaking out through a crack in the bottom
of the tank at a rate equal to k!h cm3/min, where h cm is the depth of water
remaining in the tank after t minutes. If the tank is empty after 90 minutes, find
the value of k.
R
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E
πh2
(30 − h) cm3, where h cm is the depth of the
3
h
water in the bowl.
Initially the bowl has water to a depth of 9 cm.
The water starts leaking out through a small hole in
the bowl at a rate equal to k!h cm3/min. After
1314 minutes the bowl is empty. Find the value of k.
b A drinking trough has a length of 2 m.
Its cross-sectional face is in the shape
90
of a trapezium with a height of 25 cm
and with lengths 40 and 90 cm. Both
25
200
h
45°
sloping edges are at an angle of 45° to
45°
the vertical, as shown in the diagram.
40
The trough contains water to a height
of h cm. The water leaks out through a
crack in the base of the trough at a rate proportional to !h cm3/min. Initially
the trough is full, and after 20 minutes the height of the water in the trough is
16 cm. How long will it be before the trough is empty?
U
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8aA plastic bucket has a base diameter of 20 cm,
540 a top diameter of 26 cm and a height of 24 cm.
The side of the bucket is straight. Initially the
bucket is filled with water to a height of 16 cm.
However, there is a small hole in the bucket,
and the water leaks out of the bucket at a rate
proportional to !h cm3/min, where h cm is
the height of the water in the bucket. When the
height of the water in the bucket is 16 cm, the
height is decreasing at a rate of 0.1 cm/min.
Determine how long it will be before the
bucket is empty.
b A plastic coffee cup has a base diameter of
5 cm, a top diameter of 8 cm and a height of
9 cm. Initially the cup is filled with coffee.
8
9
5
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 540
08/07/15 2:34 AM
However, the coffee leaks out of the cup at a rate equal to k!h cm3/min, where
h cm is the height of the coffee in the cup. If the cup is empty after 3 minutes,
determine the value of k.
9a
A conical vessel with its vertex downwards has a height of 20 cm and a radius
O
FS
of 10 cm. Initially it contains water to a depth of 16 cm. Water starts flowing
out through a hole in the vertex at a rate proportional to the square root of the
remaining depth of the water in the vessel. If after 10 minutes, the depth is
9 cm, how much longer will it be before the vessel is empty?
b A conical funnel with its vertex downwards has a height of 25 cm and a radius
of 20 cm. Initially the funnel is filled with oil. The oil flows out through a hole
in the vertex at a rate equal to k!h cm3/s, where h cm is the height of the oil in
the funnel. If the funnel is empty after 40 seconds, determine the value of k.
fountain contains V litres of water, where
V = 500ah2 −
h4
b.
4
PR
O
10 When filled to a depth of h metres, a
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Initially the fountain is empty. Water
is pumped into the fountain at a rate of
300 litres per hour and spills out at a rate
equal to 2 !h litres per hour.
a Find the rate in metres per hour at which
the water level is rising when the depth is
0.5 metres.
b The fountain is considered full when the height of the water in the fountain is
1 metre. How long does it take to fill the fountain?
c When the fountain is full, water is no longer pumped into the fountain, but
water still spills out at the same rate. How long will it be before the fountain is
empty again?
y
11aA wine glass is formed when the arc OB
4
3
2
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R
with the equation y = x is rotated about
the y-axis. The dimensions of the glass are
given in cm.
The wine leaks out through a crack in the
base of the glass at a rate proportional to
!h cm3/min. Initially the glass is full, and
after 3 minutes the height of the wine in
the glass is 1 cm. What further time elapses
before the wine glass is empty?
b A large beer glass is formed when a portion
B
3
–
y = x2
x
0
4
3
of the curve with the equation y = x is rotated about the y-axis between the
origin and y = 16. The dimensions of this glass are given in cm. Beer leaks out
through a crack in the base of this glass at a rate proportional to !h cm3/min.
Initially the glass is full, and after 3 minutes the height of the beer in this glass
is 12 cm. How much longer will it be before this glass is empty?
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 541
541
08/07/15 2:34 AM
12aAn ornamental vase has a circular top and base, both with radii of 4 cm, and a
PR
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FS
height of 16 cm. The origin, O, is at the centre of the base of the vase, and the
vase is formed when the hyperbola
2
25x2 (y − 8)
−
=1
144
36
is rotated about the y-axis, with dimensions in cm. Initially the vase is filled
with water, but the water leaks out at a rate equal to 2 !h cm3/s, where h cm
is the height of the water in the vase. Determine the time taken for the vase to
empty and the capacity of the vase.
b A different ornamental vase has a circular top and base, both with radii of
3.6 cm, and a height of 16 cm. The origin, O, is at the centre of the base of
the vase, and the vase is formed when the ellipse
2
x2 (y − 8)
+
=1
36
100
is rotated about the y-axis, with dimensions in cm. Initially this vase is filled
with water, but the water leaks out at a rate equal to 2 !h cm3/s, where h cm
is the height of the water in the vase. Determine the time taken for this vase to
empty and the capacity of the vase.
E
13aA cylindrical vessel is initially full of water. Water starts flowing out through a
EC
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hole in the bottom of the vessel at a rate proportional to the square root of the
remaining depth of the water. After a time of T, the depth of the water is half its
2T
initial height. Show that the vessel is empty after a time of
.
2 − !2
c A conical tank is initially full of water. Water starts flowing out through a hole
in the bottom of the tank at a rate proportional to the square root of the
remaining depth of the water. After a time of T, the depth of the water is half its
T
initial height. Show that the tank is empty after a time of
.
1 − "2−5
y
14 The diagram shows a vase. The base and the top
4
U
N
C
O
R
R
are circular with radii of 9 and 4 cm respectively,
and the height is 16 cm. The origin, O, is at the
centre of the base of the vase, with the coordinate
axes as shown.
Initially the vase is filled with water, but the water
leaks out at a rate equal to 2 !h cm3/s, where h cm
is the height of the water in the vase. Determine the
time taken for the vase to empty and the capacity of
the vase if the side of the vase is modelled by:
a
a a hyperbola, y = + b
x
a
b a truncus, y =
+ b.
x2
Master
542 16
h
O
x
9
15 A vase has a base radius of 4 cm, a top radius of 9 cm and a height of 16 cm. The
origin, O, is at the centre of the base of the vase, and the vase can be represented
by a curve rotated about the y-axis. Initially the vase is filled with water, but the
water leaks out at a rate equal to 2 !h cm3/s, where h cm is the height of the
water in the vase.
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 542
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y
9
16
7
O
x
4
U
N
C
O
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R
EC
TE
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PA
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E
PR
O
O
FS
Determine the time taken for the vase to empty and the capacity of the vase if the
side of the vase is modelled by:
a a cubic of the form y = ax3 + b
b a quartic of the form y = ax4 + b.
16 A hot water tank has a capacity of 160 litres and initially contains 100 litres of
πt
water. Water flows into the tank at a rate of 12 !t sin2 a b litres per hour over the
4
time 0 ≤ t ≤ T hours, where 4 < T < 8. During the time interval 0 ≤ t ≤ 4, water
flows out of the tank at a rate of 5 !t litres per hour.
a After 3 hours, is the water level in the tank increasing or decreasing?
b After 4 hours, is the water level in the tank increasing or decreasing?
c At what time is the water level 100 litres?
d After 4 hours, find the volume of water in the tank.
e After 4 hours, no water flows out of the tank. However, the inflow continues at
the same rate until the tank is full. Find the time, T, required to fill the tank.
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 543
543
08/07/15 2:34 AM
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U
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R
R
EC
TE
D
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544
Maths Quest 12 sPeCIaLIst MatheMatICs VCe units 3 and 4
c10FurtherApplicationsOfIntegration.indd 544
08/07/15 2:34 AM
10 Answers
Exercise 10.2
x
a
16a sin−1 a b +
1
1 27 "(9x 2 + 16) 3 + c
1
2 −15 cos5 (3x) + c
x
a
b cos−1 a b −
3 π
π
24
1
6 −4 (2x + 1)e−2x + c
1
7 x sin −1 (5x) + 5 "1 − 25x2 + c
1
8 16 (π − loge (4))
x
2
9a 2x sina b + 4 cosa b + c
x
1
sin(2x) − cos(2x) + c
4
2
−
x
2
+c
1
b x cos−1 (4x) − 4"1 − 16x2 + c
b −6x sin(6x) + cos(6x),
1
72
x
3
C
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x
c e + e , 9 units2
3
π
π
π
12a
b
c
24
18
6
2x
1
−1
2
13a cos (2x) −
, units
"1 − 4x2 2
3x
1
b sin−1 (3x) +
, (π − 2) units2
"1 − 9x2 6
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N
14a eax (ax + 1) ,
4x
, π − loge (4) units2
x2 + 16
1 ax
e (ax − 1) + c
a2
b ax cos (ax) + sin (ax) ,
1
(ax sin (ax) + cos (ax)) + c
a2
c cos (ax) − ax sin (ax) ,
1
(sin (ax) − ax cos (ax)) + c
a2
15a
1
(π2
256
+ 4π − 8)
b
1
(4π2
1296
c
5
2
− 6π !3 + 27)
loge (5) − 2
x3
(3 loge (4x) − 1) + c
9
xn+1
((n + 1) loge (4x) − 1) + c, n ≠ −1
(n + 1) 2
If n = −1, the result is 1 ( loge (4x)) 2 + c.
b −
2
e−3x
(2 cos (2x) + 3 sin (2x)) + c
13
c (a2 + b2) eax cos (bx) ,
TE
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x
5
loge (x2 + 25) + c
5
2
1
11a 4x cos(4x) + sin(4x), 32 (π − 2)
c x tan−1 a b −
x
4
c
e2x
(2 cos (3x) + 3 sin (2x)) + c
18a 13e2x cos (3x) ,
13
x
3
c tan−1 a b +
x2
(2 loge (4x) − 1) + c
4
d
10a x sin−1 a b + "9 − x2 + c
x
3
b
E
c −(2x + 4)e
a2
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b
x
, x cos−1 a b − "a2 − x2 + c
a
x
ax
a
x tan−1 a b − loge (x2 + a2) + c
2
a
2
+x
(
(
)
)
17a x loge 4x − 1 + c
c tan−1 a b +
1
5 3x sin(3x) + 9 cos(3x) + c
x
2
−
x2
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x
a
x
"a2
x
, x sin−1 a b + "a2 − x2 + c
a
O
1
−
x2
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4
x
"a2
a2
eax
(a cos (bx) + b sin (bx)) + c
+ b2
d (a2 + b2) eax sin (bx) ,
3
a1
13
+e
−
2π
3
x
ax
1 2
(x + a2) tan−1 a b −
+c
a
2
2
b Check with your teacher.
20a sec (x) , loge (1 + !2) units2
b −a cosec (ax) , loge ( !3) units2
19a
c
1
4
b units2
loge ( !3 + 2) units2
Exercise 10.3
All answers are in cubic units unless
otherwise stated.
1 8π
512π
15
32π
3
5
4 8π
5 10 litres
6 350π cm3
2
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 545
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128π
5
5a 1296π
4a
4π2
3
d 4π (e2 − 1)
3
9a i 153 5π ii 6π
486π
5
416π
10a
15
9a
1
1
b i 36 ii 259 5 π iii 40 2 π
2
10a i 106 3π ii 133 13π
4
4
bi3πab2 ii 3 πa2b
c 18 392π mL3
b 136 L
1
b
a
iii Does not exist, π
ii V = πa1 −
O
5πr3
24
b
64π
15
2a 8π
3a
546 18a i
49π
2
b 2π2ar2
11!3πa5
9πa4
ii 40
32
b i
9πa2
38πa5
ii 2
15
c i
16πa5
ii πa4
3
Exercise 10.5
75
1 4 units
1009
8
units
3 9.7471 units
All answers are in cubic units unless otherwise
stated.
704π
5
9π2
2
b Check with your teacher.
2
Exercise 10.4
1a
16a 32π !3 cm3
R
b
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C
17a 32π (3 − 4 loge (2))
2187π
7
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15a
R
27 776π
3
14a
EC
π2a2
π2a2
b
2n
4n
16a Check with your teacher.
b i Check with your teacher.
18
13a
b
64π
8π
ii 15
3
2π
π
i
ii 35
10
96π
3πa3
b
5
10
32π
72π
b
3
16 384π
i
ii 256π
3
192π
5
17a 256π
15a
ii b
TE
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cm3
iii 3183.03 cm3 iv 3223.69 cm3
2720π
b
9
1
14a i A = (1 − e−na)
a
π
(1 − e−2na)
ii V =
2a
1 π
iii ,
a 2a
b i A = loge (a)
13a i3099.70
cm3 ii 3141.59
383π
15
b
12a i
b 0.24π m3
b
b 4π
11a 4π2
11a i 6.96 m3 ii 3.0 m
12a 172.34 mL
b 8π
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b
b 81π
512π
6
3
4352π
7a
15
81π
8a
2
E
9π2
4
π
c
2
8a
b 8π
FS
125π
b
27
ci
4π
ii 6π
O
7a 375π
4 3.9140 units
5a 3sin(t2) dt + 3
x
1
2
b 8π
32π
5
76π
b
5
b
b 3.2688
6 3tan a b dt +
x
π
8
1
t
1
2
Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
c10FurtherApplicationsOfIntegration.indd 546
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9π (4 − π)
cubic units
8
b 0.8755 cubic units
8a 6.6077 cubic units
x
17a i 3e dt + 3 ii 5.020
7a
1
2
1
x
b i 3 sin−1 (t2) dt + 1 ii 1.042
3π2
cubic units
16
9a x = 2 is a minimum turning point.
b
0.1
x
c i 3
y
y = (x – 2)2 + 5
1
1
"t3
1
dt + ii 0.628
3
+8
18a A straight line with a gradient of 2
2
FS
0
–2
6 x
4
O
–4
b x = −3 is a minimum turning point,
x = 1 is a maximum turning point,
x = −1 is an inflexion point.
y
–4
–2
2
4
6 x
E
–6
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–6
y
x
1
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–1 0
–3
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b x = −3 is a minimum turning point.
EC
10 x = ±4 are minimum turning points,
units
12a
13
6
units
52
units
3
C
c
d
R
17
6
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c
R
x = 0 is a maximum turning point.
17
11a 5!10 units
b 12 units
1
1
ae − b units
e
2
b
38 !3
9
d
49
6
units
U
N
c 2.8323 units
15a 15.8654 units
16a
9π2
8
–7 –6 –5 –4 –3 –2 –1 0
c A maximum turning point at x = −2, a minimum
turning point at x = 4 and a point of inflexion at x = 1
y
–10–8 –6 –4 –2 0
2 4 6 8 10 x
d A stationary point of inflexion at x = 2
y
d 3.6837 units
b 10.3978
0
cubic units
1 2 3 x
units
π
units
2
3π
b
, one-quarter of the circumference of a circle
2
of radius 3
14a 105.1490 units
b 4.1238 units
13a
y
2
x
b 2.6898 cubic units
Topic 10 Further applications of integration c10FurtherApplicationsOfIntegration.indd 547
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1296π
cubic units
5
d 216π cubic units
b
1 431.4 s
2 473.3 s
3 10.8 min
x
4
20a sin−1 a b + "16 − x2, 4π − 8
4 12.8 min
b 8 square units
5a 36.65 min
b 20 min
c 23.4941 cubic units
6a 54 s
b 273.5
π
7a
5
8a 253.63 s
8π2
cubic units
e
units
f 100.5310 cubic units
8π2cubic
21a i 3
a
∣x∣
b
9a 3.11 min
b
"n2x2n + x2
b 50.25 min
dx ii 8.2681 units
b
b i 3"1 + k2cos2 (kx) dx ii 5.2704 units
92π
5
b 20π
10a 0.68 m/h
b 1.26 h
11a 2.07 min
b 3.86 min
12a 132.4 s, 461.1 cm3
13Check with your teacher.
c 261.9 h
b 317.3 sec, 1423.5 cm3
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d
FS
c 19.4942 units
Exercise 10.6
O
19a 36 square units
a
14a 609.1 s, 1809.6 cm3
b 560.7 s, 1625.5 cm3
b
15a 515.3 s, 2631.6 cm3
b 555.6 s, 2802.8 cm3
c i 3"1 + k2e2kx dx ii 6.4947 units
16aIncreasing
d 106.65 L
E
a
c 0.89, 3.11 h
e 7.04 h
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EC
TE
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G
22Check with your teacher.
b Decreasing
548 Maths Quest 12 SPECIALIST MATHEMATICS VCE Units 3 and 4
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08/07/15 2:35 AM
FS
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PR
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E
PA
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EC
R
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C
U
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