4
More Applications of Definite Integrals: Volumes,
arclength and other matters
Volumes of surfaces of revolution
4.1
Find the volume of a cone whose height h is equal to its base radius r, by using the disc
method. We will place the cone on its side, as shown in the Figure 1, and let x represent
position along its axis.
(a) Using the diagram shown below (Figure 1), explain what kind of a curve in the xy plane
we would use to generate the surface of the cone as a surface of revolution.
y
generating curve
r
x
h
Figure 1: For problem 4.1
(b) Using the proportions given in the problem, specify the exact function y = f (x) that we
need to describe this “curve”.
(c) Now find the volume enclosed by this surface of revolution for 0 ≤ x ≤ 1.
(d) Show that, in this particular case, we would have gotten the same geometric object, and
also the same enclosed volume, if we had rotated the “curve” about the y axis.
(e) Suppose you are now given a new line, y = ax and told to compute the two volumes as
in (c) and (d) above. By what factor would they differ?
Solution
(a) (b) The curve that generates the cone surface of revolution is the straight line y =
(since r = h).
r
h
x=x
(c) “Disk method” - rotate about the x−axis. Each disk has width dx and radius y. Volume
of each disk is (area of base) × (width) = πy 2 dx. Because the curve used to generate
this surface of revolution is y = f (x) = x, the volume of a disk is π[f (x)]2 dx = πx2 dx.
1
Summing up the volumes enclosed by each of the disks, we find the total volume enclosed
by this surface of revolution for 0 < x < 1 :
V =
Z
1
π[f (x)]2 dx =
0
Z
1
πx2 dx = π/3 cubic units
0
(d) Rotate about the y−axis: since r = h, we obtain the same geometric object. In this case,
each disk has width dy and radius x. The volume of each disk is therefore π[f (y)]2 dy =
πy 2 dy, since f (y) = y.
Volume enclosed by surface of revolution is then for 0 < y < 1 and is
Z
1
2
π[f (y)] dy =
0
Z
1
πy 2 dy = π/3 cubic units
0
(same as in part (c)).
(e) The two volumes are: a2 π/3 and a−2 π/3 respectively. Thus, they differ by a factor a4 .
4.2
Find the volume of the cone generated by revolving the curve y = f (x) = 1−x (for 0 < x < 1)
about the y axis. Use the disk method, with disks stacked up along the y axis.
Solution
Figure 2:
The disks have thickness ∆y → dy and radii in the x direction so that x = f (y) = 1 − y
is the function to use. From the interval 0 < x < 1, we find the interval on the y-axis is
0 < y < 1. Thus,
Z
Z
1
V =π
1
[f (y)]2 dy = π
0
0
2
[1 − y]2 dy
V =π
Z
1
0
!
y 2 y 3 1
[1 − 2y + y ] dy = π y − 2 +
2
3 0
2
1
π
V = π(1 − 1 + ) = .
3
3
4.3
Find the volume of the “bowl” obtained by rotating the curve y = 4x2 about the y axis for
0 ≤ x ≤ 1.
Solution
We are rotating about the y axis, so, using the ”disk” method, disk
q width is dy, the radius
of each disk is in the x direction, i.e. it is given by x = f (y) = y/4. The volume will be
given by
Z ymax
π[f (y)]2 dy.
V =
ymin
To find ymax and ymin we use the interval 0 ≤ x ≤ 1. But y(0) = 0 and y(1) = 4. Therefore
the interval on the y-axis is 0 ≤ y ≤ 4.
Z
4
0
πZ4
π[f (y)] dy =
y dy = 2π.
4 0
2
4.4
On his wedding day, Kepler wanted to calculate the amount of wine contained inside a wine
barrel whose shape is shown below in Figure 3. Use the disk method to compute this volume.
You may assume that the function that generates the shape of the barrel (as a surface of
revolution) is y = f (x) = R − px2 , for −1 < x < 1 where R is the radius of the widest part
of the barrel. (R and p are both positive constants.)
y
y = R − px 2
1
x
Figure 3: For problem 4.4
Solution
3
radius
r=R−px^2
dx
Figure 4: For problem 4.4 solution
−1 ≤ x ≤ 1, R > 0, p > 0.
Disk method: See Figure 4.
The radius of each disk is y = R − px2 , and the thickness is ∆x → dx. The volume of each
disk is therefore ∆V = π(R − px2 )2 ∆x
Because f (x) = R − px2 is an even function, the volume for −1 < x < 1 is twice that for
0 < x < 1.
Volume
= 2π
= 2π
Z
Z
1
0
1
0
(R − px2 )2 dx
(R2 − 2Rpx2 + p2 x4 ) dx
!
5 1
x3
2 x
= 2π R x − 2Rp
+p
3
5 0
!
2Rp p2
2
= 2π R −
+
3
5
2
2RP
p2
So the amount of wine in the barrel is 2π R −
+
3
5
2
!
cubic units.
4.5
Consider the curve
y = f (x) = 1 − x2
0<x<1
rotated about the y axis. Recall that this will form a shape called a paraboloid. Use the
cylindrical shell method to calculate the volume of this shape.
Solution
The cylindrical shells will encase one another with their radii in the direction of the x axis,
their heights along the y axis, and their thickness ∆x → dx along the x axis. The radius of
4
a given shell is x, the height is y = f (x) = 1 − x2 . The volume of one shell is
Vshell = 2πrhτ = 2πxf (x) dx
The volume of the entire shape for 0 < x < 1 is
V = 2π
V = 2π
4.6
Z
1
0
Z
1
xf (x) dx = 2π
0
Z
1
0
x(1 − x2 ) dx.
!
π
x2 x4 1
1 1
=
−
−
(x − x ) dx = 2π
= 2π
2
4 0
2 4
2
3
√
Consider the curve y = f (x) = 1 − x2 (which is a part of a circle of radius 1) over the
interval 0 < x < 1. Suppose this curve is rotated about the y axis to generate the top half of
a sphere. Set up an integral which computes the volume of this hemisphere using the shell
method, and compute the volume. (You will need to make a substitution to simplify the
integral. To do this question, some techniques of anti-differentiation are required.)
Solution
√
The generating curve is y = f (x) = 1 − x2 . We can consider a typical ”shell” of thickness
∆x whose height is determined by√the value of y on the curve. Then the radius of the shell
is r = x, the height is y = f (x) = 1 − x2 , and the volume is
√
Vshell = 2πx 1 − x2 ∆x.
The integral (sum of all shell volumes as ∆x → 0) is
V = 2π
Z
1
0
√
x 1 − x2 dx.
To compute the integral, let u = 1 − x2 . Then du = −2dx so
V = −2π
Z
u
1/2
u3/2
du
= −π
2
3/2
1
2
2
V = −π (1 − x2 )3/2 = π.
3
3
0
Since the radius of the generating circle is 1 and we have half of a sphere, the volume is
(1/2)(4/3)πr 2 = (2/3)π.
5
4.7
Find the volume of the solid obtained by rotating the region bounded by the given curves
f (x) and g(x) about the specified line.
√
(a) f (x) = x − 1, g(x) = 0, from x = 2 to x = 5, about the x-axis.
√
(b) f (x) = x, g(x) = x/2, about the y-axis.
(c) f (x) = 1/x, g(x) = x3 , from x = 1/10 to x = 1, about the x-axis.
Solution
(a) The volume of the solid is equal to
Z
5
2
π(x − 1)dx = π(x
2
5
/2 − x)
= 7.5π.
2
(b) To calculate the volume of the solid, we have to find the x-coordinates of the 2 curves
as a function of the y-coordinates, and we have to determine the boundaries on the y-axis.
For the first task, we simply take the inverse functions: x = y 2 and x = 2y. Thus, the
solid is obtained by rotating the region between the curve x = 2y and the curve x = y 2
around the y-axis (see Figure 5). This region is bounded by the two points (x, y) = (0, 0)
and (x, y) = (4, 2) (the latter point can be found by solving f (x) = g(x)). Therefore, we
have to integrate from 0 to 2 on the y-axis. From a sketch, we see that the volume is a
paraboloid V1 with a conical hollow core V2 . The volume can be computed by subtraction
V = V1 − V2 . We get the volume of the solid as
V =
Z
2
0
V =
π(2y)2 dy −
Z
2
0
Z
2
π(y 2 )2 dy
0
π[(2y)2 − (y 2 )2 ]dy
4y 3 y 5 2 64π
V = π(
− ) =
3
5 0
15
(c) Since f (x) ≥ g(x) in the region considered, the volume of the solid is
V =
Z
1
2
2
(πf (x) − πg(x) )dx = π
1
10
1
V = π(−1/x − x7 /7)
1
10
Z
1
1
10
(x−2 − x6 )dx
= π(62 + 1/107 )/7 ≈ 8.857π.
6
F(x)
X
Figure 5: For problem 4.7 solution
4.8
Let R= the region contained between y = sin(x) and y = cos(x) for 0 ≤ x ≤ π/2. Write
down the expression for the volume obtained by rotating R about (a) x-axis (b)the line y=-1.
Solution
(a) V =
=2
R
π
4
0
R
π
4
0
π
4
0
π(cos2 (x)−sin2 (x)) dx +
π cos(2x) dx
(b)V =
2
R
R
π
4
0
R
π
2
π
4
π(sin2 (x)−cos2 (x)) dx = 2
π[(1 + cos(x))2 − (1 + sin(x))2 ] dx +
π[(1 + cos(x))2 − (1 + sin(x))2 ] dx
R
π
2
π
4
R
π
4
0
π(cos2 (x)−sin2 (x)) dx
π[(1 + sin(x))2 − (1 + cos(x))2 ] dx
4.9
Find the volume of a solid torus (donut shaped region) with radii r and R as shown in
Figure 6. (Hint: There are several ways to do this. You can consider this as a surface of
revolution and slice it up into little disks with holes (“washers”) as shown.)
torus
R
r
"washer"
(disk with a circular hole)
Figure 6: For problem 4.9
Solution
7
y
line
x=R
x
R
Figure 7: For problem 4.9 solution
cylindrical shell
x
y=(r^2−(x−R)^2)^(1/2)
rotating about
the y−axis
dx
dx
Figure 8: For problem 4.9 solution: shell
Consider the upper semi-circle (shaded region in Figure 7) rotated about the y−axis. The
volume of the torus is then 2 times the resulting volume of revolution.
The q
circle centered at (R, 0) with radius r is (x − R)2 + y 2 = r 2 , so the upper semi-circle is
y = r 2 − (x − R)2 for R − r ≤ x ≤ R + r.
One Way:
See Figure 8.
q
Volume of one shell = (width)(height)(circumference) = dx r 2 − (x − R)2 2πx.
Volume of torus
= 2
Z
= 4π
R+r
R−r
Z
R+r
R−r
q
2πx r 2 − (x − R)2 dx
s
x−R
xr 1 −
r
2
dx
Let sin u = x−R
⇒ x = R + r sin u, dx = r cos u du. Notice that the limits of integration
r
now change to −π/2 ≤ u ≤ π/2.
Volume
= 4π
Z
π/2
−π/2
[r(R + r sin u) cos u · r cos u] du
8
x = R− (r^2−y^2)^(1/2)
"washer"
x = R+ (r^2−y^2)^(1/2)
dy
rotating about
the y−axis
dy
for 0<y<r
Figure 9: For problem 4.9 solution: washer
= 4πr
2
2
Z
π/2
= 4πr R
=
=
=
=
cos2 u(R + r sin u) du
−π/2
Z
π/2
2
cos u du + 4πr
−π/2
3
Z
π/2
cos2 u sin u du
−π/2
1 + cos 2u
4πr 2 R
du + 0
2
−π/2
sin 2u π/2
2
2πr R u +
2
u=−π/2
π
π
− −
2πr 2 R
2
2
2 2
2π r R cubic units
Z
π/2
Another Way:
See Figure 9.
For this method we use the disk method, rotating about the y-axis.
2
2
Volume of one washer = (volume of solid disk) − (volume of hole) = πrdisk
h − πrhole
h.
The radius of the solid disk is equal to the equation for the √
right-hand side of the circle in
Figure 7. This equation, as shown in Figure 9, is rdisk = R + r 2 − y 2 . Similarly, the radius
of the hole is the defined by the equation that describes the innermost
√ 2 part2 of the circle, the
left-hand side as seen in Figure 7 and Figure 9. Thus rhole = R − r − y . Combining this
with the height of each washer (dy), we get an equation for the volume of one washer:
Vwasher = π(R +
q
r 2 − y 2 )2 dy − π(R −
q
r 2 − y 2 )2 dy.
To find the volume of the torus, we would integrate this over −r ≤ y ≤ r. But because the
function is symmetric about the x-axis, we instead integrate over 0 ≤ y ≤ r, and double the
result.
Volume of torus
= 2
Z
r
0
= 2π
Z
π (R +
r
0
q
q
r 2 − y 2 )2 − (R −
4R r 2 − y 2 dy
9
q
r 2 − y 2 )2 dy
= 8πR
Z
r
0
q
r 2 − y 2 dy
!
1
πr 2
(which is of the area of a circle of radius r)
= 8πR
4
4
2
2
= 2π Rr cubic units
4.10
In this problem you are asked to find the volume of a height h pyramid with a square base
of width w. (This is related to the Cheops pyramid problem, but we will use calculus.)
Let the variable z stand for distance down the axis of the pyramid with z = 0 at the top,
and consider “slicing” the pyramid along this axis (into horizontal slices). This will produce
square “slices” (having area A(z) and some width ∆z). Calculate the volume of the pyramid
as an integral by figuring out how A(z) depends on z and integrating this function.
z
h
0
w
Figure 10: For problem 4.10
Solution
Let the side length of the square cross-section at height z be x. Then by similar triangles,
x/z = w/h, or x = (w/h)z. Now the area of the square slice at height z is A(z) = x2 =
(w/h)2 z 2 and its thickness is ∆z. Thus
V =
Z
h
0
(w/h)2 z 2 dz = (w/h)2 [z 3 /3]|h0 = w 2 h/3.
10
Length of a curve and arclength
4.11
Set up the integral that represents the length of the following curves: Do not attempt to
calculate the integral in any of these cases
(a)
y = f (x) = sin(x) 0 < x < 2π.
(b)
y = f (x) =
√
x 0 < x < 1.
(c)
y = f (x) = xn
− 1 < x < 1.
Solution
(a)
y = f (x) = sin(x) 0 < x < 2π.
dy/dx = cos(x) so
L=
Z
2π
0
q
(b)
y = f (x) =
1 + cos2 (x) dx
√
x 0 < x < 1.
dy/dx = 12 x−1/2 , and (dy/dx)2 = ( 21 )2 x−1 =
1
4x
1
s
L=
Z
0
so
1+
1
dx
4x
(c)
y = f (x) = xn
dy/dx = nxn−1 so
L=
Z
1
√
− 1 < x < 1.
1 + n2 x2n−2 dx
−1
11
4.12
Compute the length of the line y = 2x + 1 for −1 < x < 1 using the arc-length formula.
Check your work by using the simple distance formula (or Pythagorean theorem).
Solution
dy/dx = 2 so
L=
Z
√ Z
L= 5
1
q
1+
−1
1
−1
(dy/dx)2 dx
=
Z
1
√
1 + 4 dx
−1
√
√
√ 1
dx = 5x = 5(1 − (−1)) = 2 5.
−1
We can compare this to the simple distance formula: The endpoints of the line are (−1, −1)
and (1, 3) so the length of the line is
q
√
√
√
d = (1 − (−1))2 + (3 − (−1))2 = 22 + 42 = 20 = 2 5.
4.13
Find the length of the curve
y = f (x) = x3/2
0<x<1
(To do this question, some techniques of anti-differentiation are required.)
Solution
We use the arclength integral for the curve y = f (x) = x3/2 . To do so, we first find
3
f 0 (x) = x1/2 .
2
Then the length we want is
L=
Z
1
0
q
1 + (f 0 (x))2 dx =
Z
1
0
s
3
1 + ( x1/2 )2 dx,
2
and, simplifying,
s
9
1 + x dx.
4
0
Substitute u = 1 + (9/4)x. Then du = (9/4)dx, or dx = (4/9)du. Plugging into the integral
leads to
Z
4 u3/2
4 (1 + (9/4)x)3/2 1
4 x=1 √
u du =
=
L=
.
9 x=0
9 3/2
9
3/2
0
We compute
13 √
8 8
(1 + (9/4))3/2 − 1 =
≈ 1.44
L=
13 −
27
27
27
L=
Z
1
12
4.14
To do this question, some techniques of anti-differentiation are required. (a) Use the chain
rule to show that F 0 (x) = sec(x) is the derivative of the function
F (x) = ln(sec(x) + tan(x))
So, what is the anti-derivative of f (x) = sec(x)?
(b) Use integration by parts to show that
Z
1
sec3 (x)dx = (sec(x) tan(x) + ln(sec(x) + tan(x)) + C.
2
You will want to recall the trigonometric identity
1 + tan2 (x) = sec2 (x).
(c) Set up an integral to compute the arclength of the curve y = x2 /2 for 0 < x < 1.
(d) Use the substitution x = tan(u) to reduce this arclength integral to an integral of the
form
Z
sec3 (u) du.
Solution
(a) Differentiating leads to
F 0 (x) =
1
· (sec(x) tan(x) + sec2 (x))
sec(x) + tan(x)
where we have used the chain rule and derivatives of sec(x) and tan(x). Now cancelling the
common factor leads to F 0 (x) = sec(x). From these results, we see that
Z
sec(x) dx = ln(sec(x) + tan(x)).
(b) Let u = sec(x) ⇒ du = sec(x) tan(x)dx. Let dv = sec2 (x)dx ⇒ v = tan(x). Then
I=
Z
sec3 (x)dx = sec(x) tan(x) −
Z
sec(x) tan2 (x) dx
Using tan2 (x) = sec2 (x) − 1 we get
I = sec(x) tan(x) −
Z
sec(x)(sec2 (x) − 1) dx
13
Multiplying through gives
Z
Z
Z
3
sec (x)dx = sec(x) tan(x) −
3
sec (x)dx = sec(x) tan(x) −
Z
Z
(sec3 (x) − sec(x)) dx.
3
sec (x) dx +
Z
sec3 (x)dx = sec(x) tan(x) + ln(sec(x) + tan(x)) −
We can move the
R
sec(x) dx.
Z
sec3 (x) dx.
sec3 (x) dx to the other side and divide by 2 to get
Z
1
sec3 (x)dx = (sec(x) tan(x) + ln(sec(x) + tan(x))
2
(c) y = f (x) = x2 /2 so f 0 (x) = x. The length of the curve is then
L=
Z
1
√
1 + x2 dx.
0
(d) Let x = tan(u). Then dx = sec2 (u)du. We also use the fact that 1 + tan2 (u) = sec2 (u)
so that the integral becomes
L=
Z
1
0
√
1+
x2
dx =
Z
2
sec (u)
q
sec2 (u)
du =
Z
sec3 (u)du.
4.15
Second Spreadsheet assignment: You are told that the derivative of a certain function is
f 0 (x) = 1 − 2 sin2 (x/3)
and that f (0) = 0. Use the spreadsheet to create one plot that contains all of the following
graphs: (1) The graph of the function y = f (x) (whose derivative is given to you). This
should
q be plotted over the interval from 0 to 9. (2) The graph of the ”element of arclength”
dl = 1 + (f 0 (x))2 dx showing how this varies across the same interval. (3) The graph of the
cumulative length of the curve L(x). Briefly indicate what you did to find the function in
part (1). (You might consider how the spreadsheet would help you calculate the values of
the desired function, y = f (x), rather than trying to find an expression for it.)
Solution
Here is how we could organize the spreadsheet: We tabulate x values (in increments of
∆x = 0.1 in the a column, and values of the given f 0 (x) in the b column. We then compute
f (x) in the c column by the linear approximation formula
f (x + ∆x) = f (x) + f 0 (x)∆x.
14
12.0
Arc length
L(x) - total length
f(x)
dl (element of arc-length)
|
V
the given f’(x) (optional)
-2.0
0.0
9.0
Figure 11:
Thus, we start with f (0) = 0 in cell c0, and set cell c1 to ”c0+b0*0.1”, computing the value
of f at this place, and so on. (Drag the entry down to generate the whole column.) In
column d we calculate
q
dl = 1 + (f 0 (x))2 dx
by setting cell d0 to ”sqrt(1 + b02 ) ∗ 0.1”, and dragging the entry down. We add up these
values in column e by setting e0 = 0, e1 = e0 + d0, and so on, to get the total length L(x)
as it accumulates along the curve.
The resulting plot is shown in Figure 11. We also show an expanded version (which
makes dl clearer to see) in Figure 12.
Other applications
4.16
work 1
A spring has a natural length of 16 cm. When it is stretched x cm beyond that, Hooke’s Law
states that the spring pulls back with a restoring force F = kx dyne, where the constant k
is called the spring constant, and represents the stiffness of the spring. For the given spring,
8 dyne of force are required to hold it stretched by 2 cm. How much work (dyne-cm) is done
in stretching this spring from its natural length to a length 24 cm?
Solution
15
0.5
|
V
0.0
0.0
9.0
Figure 12:
Since F = 8 when x = 2 we find that 8 = 2k so that k = 4 and F = 4x is the force. When
the spring is at its natural length, 16 cm, no work is done. To stretch the spring by ∆x
beyond this would require work
W = F ∆x = 4x∆x.
To sum up all the work involved in stretching from 16 to 24 cm, i.e. for the increase in length
x = 0 to x = 8, we would calculate
W =
4.17
Z
F dx =
Z
8
4xdx = 2x
0
work 2
8
2
0
= 128 dynes − cm.
Calculate the work done in pumping water out of a parabolic container. Assume that the
container is a surface of revolution generated by rotating the curve y = x2 about the y axis,
that the height of the water in the container is 10 units, that the density of water is 1gm/cm 3
and that the force due to gravity is F = mg where m is mass and g = 9.8m/s2 .
Solution
The paraboloid can be thought of as a bowl full of water. The total work done is equivalent
to the sum of work done to pump each bit of the water up to the height 10, or equivalently,
16
a sum over “disks” of water, each at some height y. The work done is Work = (force)
(distance). The distance from a disk at height y to height 10 is 10 − y. Therefore, Work =
Force × (10 − y). The force is F = mg where g = 9.8m/s2 = 980cm/s2 . The mass of a disk
at height y is
my = (1 gm/cm3 )(πr 2 ∆y cm3 )
where the radius of the disk is the x coordinate of a point on the parabola y = x2 , giving
√
r = x = y. Thus, the work done to move this disk to height 10 cm is
√
wy = my g(10 − y) = π( y)2 ∆y(980)(10 − y) = 980π(10 − y)y∆y,
and the total work, which is a sum over all disks, is the integration of wy over 0 < y < 10:
W = 980π
Z
10
0
(10 − y)y dy = 980π
!
Z
10
0
(10y − y 2 ) dy
y 2 y 3 10
1 1
W = 980π 10 −
−
|0 = 980π · 103
= 5.131 × 105 ergs.
2
3
2 3
(One erg is 1 gm cm2 s−2 or 1 dyne-cm).
4.18
Energy
In a harmonic oscillator (such as a spring) the energy is partly kinetic (associated with
motion), and partly potential (stored in the stretch of the spring). Energy conservation
implies that the sum of the two forms (E) is constant so that
1 2 1 2
mv + kx = E
2
2
where m is the mass on the spring, k the spring constant. (To do this question, some
techniques of anti-differentiation are required.)
(a) Use the fact that the velocity is v = dx/dt to show that this energy conservation equation
implies that
!1/2
2
k 2
dx
=
E− x
.
dt
m
m
(b) Rewrite this in integral form by dividing both sides of the equation by the term in the
square-root to obtain
Z
Z
dx
dt.
1/2 =
2
k 2
E
−
x
m
m
(c) Integrate both sides. You will need to use substitution, and the result will be an inverse
trig function for x.
17
(d) Find the displacement x as a function of time t. Show that the motion of the spring is
periodic.
Solution
dx
1
(a) The velocity v = dx/dt so that m
2
dt
!
2 1/2
dx
2E − kx
Simplifying gives
=
.
dt
m
(b)
Z
dx
2E−kx2
m
=
1/2
Z
!2
dx
1
+ kx2 = E, so that m
2
dt
!2
= 2E − kx2 .
dt is obtained by “cross multiplying” and integrating both sides, and
is called separation of variables.
(c) We simplify to
dx
Z
q q
k
m
r
Let a2 = 2E/k. Then the above is
trig-type integral, and we get
(d)
r
m
k
Z
√
Thus x = a sin 
frequency ω =
s

= t + C.
dx
= t + C. The integral on the left is a
a2 − x2
m −1
sin (x/a) = t + C.
k
m −1
sin (x/a) = t + C, so sin−1 (x/a) =
k
s
4.19
r
(2E/k) − x2
k
(t + C) =
m
s
k
and amplitude
m
s
s
s

k
k
(t + C), i.e (x/a) = sin 
(t + C).
m
m

2E
k
sin 
(t + C). This is a periodic function with
k
m
s
2E
.
k
Shear
A vertical shear is a transformation of the plane which takes the vertical line x = a and
pushes it upward by a distance sa (the constant s is called the strength of the shear). For
instance, the rectangle shown below in Figure 13 is transformed into a trapezoid.
(a) Set up a definite integral which shows that the area of the trapezoid is the same as the
area of the original rectangle. We say that the shear has preserved the area of the rectangle.
(b) Now consider a more general shape:
Show that the area before the shear is the same as the area after the shear.
Solution
(a) See Figure 15.
18
y
Before shear
After shear
y
x=a
x=a
x
x
Figure 13: For problem 4.19
Before Shear
After shear
Figure 14: For problem 4.19(b)
Area of rectangle = ab.
Area of trapezoid =
Ra
0
[(sx + b) − sx] dx =
So shear is area-preserving for a rectangle.
Ra
0
b dx =
a
bx
= ab.
0
(b) See Figure 16.
Consider the circle as made up of rectangle strips, each of which is undergoing a shear of
strength s. From part (a) we know that this shear is area-preserving. Therefore, the area of
the circle (before shear) is the same as the area of the ellipse (after shear).
4.20
Surface Area
Calculate
the surface area of a cone shaped surface obtained by rotating the curve y =
√
x on the interval [0, 2] around the x-axis. To do this question, some techniques of antidifferencation are required.
Solution
Using the formula for surface area
S=
Z2
0
q
2πy(x) 1 + (y 0 (x))2 dx.
19
y
y
Before shear
After shear
line x = a
y = sx + b
y=b
b
b
y = sx
sa
x=a
a
x
x
Figure 15: For problem 4.19 (a) solution
a
a
a
−a
a
−a
−a
−a
Circle (before shear)
Ellipse (after shear)
Figure 16: For problem 4.19 (b) solution
Now, y =
S=
Z2
0
√
√
x and y 0 (x) = 1/(2 x). Thus
s
2
Z2 √
√
1
π
π
πZ
4x + 1dx =
2π x 1 + dx = π
(4x+1)−1/2 d(4x+1) = (4x+1)1/2 |20 = [3−1] = π.
4x
4
2
2
0
0
20