### Extension 2 past pap..

```4 UNIT MATHEMATICS – INTEGRATION – HSC
Integration
4U97-1a)!
5

Evaluate
0
2
dx .¤
x+4
« 4 »
4U97-1b)!

4
sin
 cos4 d .¤
Evaluate
0
«


1
2 2 1 »
3
4U97-1c)!
Find
x
2
1
dx .¤
 2x + 3
«
x +1
1
tan -1
C »
2
2
4U97-1d)!
Find
4t - 6
 (t + 1)(2t
4U97-1e)!
2
 3)
dt .¤
« Ln
c (2t 2  3)
»
( t + 1) 2
«
3
  Ln2 »
3

3
Evaluate
 x sec2x dx .¤
0
4U96-1a)!
3
Evaluate
4
 (2  x)
2
dx .¤
1
«
4U96-1b)!
Find
8
»
15
 sec  tan  d .¤
2
1
2
« tan2 + c »
4U96-1c)!
Find
5t 2  3
 t(t 2  1) dt .¤
« 3 Ln t+ Ln (t2 + 1) + c »
4U96-1d)!
Using integration by parts, or otherwise, find
¤©BOARD OF STUDIES NSW 1984 - 1997
 x tan1 x dx .¤
4 UNIT MATHEMATICS – INTEGRATION – HSC
«
1 2 -1
(x tan x + tan-1x – x) + c »
2
4U96-1e)!
Using the substitution x  2sin , or otherwise, calculate
3
x2

4  x2
1
dx .¤
«  –
4U96-3b)!
i.

2
Show that  (sin x)2 k cos xdx 
0
1
, where k is a positive integer.
2k  1

2
ii.
By writing (cosx)  (1 sin x) , show that  (cos x)
2n
3 »
2
n
2 n 1
0
(1) k  n
dx  
 .
k  0 2k  1  k
n

2
iii.
Hence, or otherwise, evaluate
 cos5 xdx .¤
0
« i) Proof ii) Proof iii)
4U96-7b)!
i.
ii.
8
»
15
Let f(x) = Lnx - ax + b, for x > 0, where a and b are real numbers and a > 0. Show that
y = f(x) has a single turning point which is a maximum.
The graphs of y = Lnx and y = ax - b intersect at points A and B. Using the result of part (i),
or otherwise, show that the chord AB lies below the curve y = Lnx.
k

Lnx dx = k Lnk - k + 1.
iii.
Using integration by parts, or otherwise, show that
iv.
Use the trapezoidal rule on the intervals with integer endpoints 1, 2, 3,..., k to show that
1
k
 Lnx dx is approximately equal to 2 Lnk  Ln(k  1)! .
1
1
v.
 k
 e
k
Hence deduce that k !  e k   .¤
« Proof »
4U95-1a)!
Find 
dx
.¤
x(lnx)2
« 
1
c »
ln x
4U95-1b)!
Find  x ex dx .¤
« ex - xex + c »
4U95-1c)!
Show that

4
1
6t  23
dt  ln 70 .¤
(2t  1)(t  6)
« Proof »
4U95-1d)!
¤©BOARD OF STUDIES NSW 1984 - 1997
4 UNIT MATHEMATICS – INTEGRATION – HSC
Find
d
(x sin 1x) , and hence find  sin 1x dx .¤
dx
«
4U95-1e)!
x
2
Using the substitution t = tan , or otherwise, calculate
x
 sin 1x , x sin 1x  1  x 2  c »
2
1 x

2
0

dx
.¤
5  3sin x  4cos x
«
4U95-4c)!

sin(2m  1)x sin(2m  1)x

 2 cos(2mx) .
, then
sin x
sin x
2
i.
Show that, if 0 < x <
ii.
Show that, for any positive integer m,
iii.
iv.
v.
1
»
6

2
0

cos(2mx)dx  0 .

2
0
sin(2m  1)x
Deduce that, if m is any positive integer, 
dx 
sin x

sin(2m  1)x
.
Show that, if m = 1 then,  2
dx

0
sinx
2

sin 5x

Hence show that  2
dx  .¤
0 sin x
2

2
0

sin(2m  1)x
dx .
sin x
« Proof »
4U95-7a)!
Let I n 

2
0

 sin x n dx , where n is an integer, n  0.
i.
Using integration by parts, show that, for n  2, In  
ii.
Deduce that I 2n 
I 2n +1
iii.
iv.
2n  1 2n  3
3 1 

...   and
2n
2n  2
4 2 2
2n
2n  2
4 2


...   1 .
2n + 1 2n  1
5 3
n  1 In  2
n 
Explain why Ik > Ik + 1.
Hence, using the fact that I2n - 1 > I2n and I2n > I2n + 1, show that
22 .4 2 ...(2n) 2
  2n  
  .¤
2  2n  1 1.32 .52 ...(2n  1) 2 (2n  1) 2
« Proof »
4U94-1a)!
Find
i.
ii.
4x  12
dx
 6x  13
1
 x2  6x  13 dx .¤
x
2


« (i) 2log e x2  6x  13  c (ii)
4U94-1b)!
Evaluate

3
3
2
¤©BOARD OF STUDIES NSW 1984 - 1997
9  u2 du .¤
1
2
tan 1
x3
c »
2
4 UNIT MATHEMATICS – INTEGRATION – HSC
«
4U94-1c)!
Evaluate

2
1
9
3
 
 »
23
4
11  2t
dt .¤
(2t  1)(3  t)
« ln 18 »
4U94-1d)!
Evaluate

0 e2xsin x dx .¤
«
1
5
e
2
 1 »
4U94-6a)!
i.
2x

Given that sinx 
for 0  x  , explain why
2

  sin x
dx
e
2

ii.
Show that
iii.
Hence show that

=

2 e sin xdx
0


 -2x
2 e  dx .
0


2 e sin xdx .
0

0 esin xdx <

(e  1) .¤
e
« Proof »
4U93-1a)!
Evaluate
x
dx by using the substitution x + 1 = u².¤
3 (x  1)  x  1

8
« 7 »
4U93-1b)!
Evaluate

2
0

d

by using the substitution t  tan .¤
2  cos 
2
«

»
3 3
4U93-1c)!
Evaluate
1
2 sin1x dx .¤
0

«
4U93-1d)!
i.
Find real numbers a, b and c such that
ii.
Hence find
 (x
2

3

1 »
12
2
4x  3
ax  b
c
 2

.
(x  1)(x  2) x  1 x  2
4x  3
dx .¤
 1)(x  2)
2


1 ln x2  1  2tan1x  ln x  2  c »
« (i) a = 1, b = 2, c = -1 (ii) 2
4U93-4b)!
Suppose k is a constant greater than 1. Let f(x) 

assume f 
2  0 .]
¤©BOARD OF STUDIES NSW 1984 - 1997
1
 . [You may
where 0  x 
k
2
1  (tan x)


i.

Show that f(x)  f 
2  x  1 for 0  x  .
ii.

Sketch y = f(x) for 0  x  .
2
4 UNIT MATHEMATICS – INTEGRATION – HSC
2
[There is no need to find f x but assume y = f(x) has a horizontal tangent at x = 0. Your
graph should exhibit the property of (b)(i).]
iii.
Hence, or otherwise, evaluate

dx
2
0 1  (tan x) k

.¤
y
1

2
« (i) Proof (ii)
4U92-1a)!
Find:
x
(iii)

»
4
 tan  sec  d .
2x  6
 x  6x  1dx .¤
2
i.
ii.
2
« (i) 21 tan 2   c (ii) log x 2  6x  1  c »
4U92-1b)!
5
2
Evaluate

3
2
dx
by using the substitution u = x - 2.¤
(x  1)(3  x)
«
4U92-1c)!
Evaluate
 »
3
5(1  t)
 (t  1)(3  2t) dt .¤
1
0
« log e
4U92-1d)!
 xe
i.
Find
ii.
Evaluate
x2
dx .
 2x e
1
0
4
»
3
3 x2
dx .¤
« (i)
1
2
e x  c (ii) 1 »
2
4U92-4a)!
Each of the following statements is either true or false. Write TRUE or FALSE for each statement and
¤©BOARD OF STUDIES NSW 1984 - 1997
4 UNIT MATHEMATICS – INTEGRATION – HSC

2
i.
sin7  d  0

ii.
iii.
2

0 sin7  d  0
1
1ex dx  0
2

2 (sin8  0
iv.

v.
For n = 1, 2, 3, ....,
cos8) d  0
dt  1 dt .¤
0 1  t n 0 1  t n1
1
« (i) True (ii) False (iii) False (iv) True (v) True »
4U91-1a)!
Find:
i.
ii.
t 1
dt ;
t3
ex
 e2x  9dx , using the substitution u = ex.¤

x
1 1
1 tan 1 e  c »
(ii)

c
3
t 2t 2
3
« (i)  
4U91-1b)!
1
i.
Evaluate
5
 (2t 1)(2  t) dt .
0
ii.

2
By using the substitution t = tan
d

and (i), evaluate 
.¤
3sin   4 cos 
2
0
« (i) ln 6 (ii)
4U91-1c)!
1
ln 6 »
5

2
Let I n   sinn x dx , where n is a non-negative integer.
0

2

i.
Show that I n  (n  1) sin n2 xcos2 x dx when n  2.
ii.
n1
I when n  2.
Deduce that I n 
n n 2
iii.
Evaluate I4.¤
0
« (i) Proof (ii) Proof (iii)
4U90-2a)!
Find the exact values of:
3
i.

2
¤©BOARD OF STUDIES NSW 1984 - 1997
x 1
dx
x  2x  5
2
3 »
16
4 UNIT MATHEMATICS – INTEGRATION – HSC
2

ii.
4  x 2 dx ¤
0
« (i) 2 5  13 (ii)

1»
2
4U90-2b)!
Find:
dx
 (x 1)(x  2) ;
 cos x dx , by writing cos x  (1 sin
i.
2
3
3
ii.
« (i)
1
3
2
x) cosx , or otherwise.¤
ln x 1  16 ln(x 2  2) 
1
x
tan 1
 c (ii) sin x  13 sin 3 x  c »
3 2
2
4U90-2c)!
x
 (1  t
Let I n 
) dt, n  1,2,3, ...
2 n
0
1
2n
(1  x 2 ) n x 
I  1.
2n  1
2n  1 n
Observe that (1  t2 ) n1  t2 (1  t2 )n 1  (1  t2 ) n .¤
Use integration by parts to show that I n 
Hint:
« Proof »
4U89-2a)!
Evaluate:
3
i.
x
2
0
3
ii.
 x lnx dx ;
2
1
2
iii.
2
dx ;
9

4  x 2 dx .¤
3
« (i)
4U89-2b)!
i.
ii.
A
Bx  C
4x2 5x  7
 2
in the form
.
2
(x 1)(x  x  2)
x 1 x  x 2
0
4x 2  5x  7
Hence evaluate 
dx .¤
(x 1)(x 2  x  2)
1
Write
« (i) 
4U88-1a)!
Find the exact value of:
1
i.
2x
 1  2x dx ;
0
e
ii.
 (ii) 9 ln 3  26 (iii)   3 »
3 2
9
9
(log e x)2
1 x dx ;
¤©BOARD OF STUDIES NSW 1984 - 1997
2
6x  3
 2
(ii) 2 ln 2 »
x 1 x  x  2
4 UNIT MATHEMATICS – INTEGRATION – HSC
1
2
iii.
 cos
1
x dx .¤
0
« (i) 1  21 log e 3 (ii)
4U88-1b)!
Use the substitution t  tan
1

3
(iii)  1 
»
3

2

2
d

to find the exact value of 
.¤
sin   2
2
0
«

»
3 3
4U87-1i)
Prove that:
t 1
1
dt   loge 2 ;
2
t
2
1
6
4 dx
 9
4 (x  1)(x  3)  2loge  5 .¤
2
a.
b.

« Proof »
4U87-1ii)
a.
b.
Use the substitution x 
2
sin to prove that
3
2
3

0

4  9x2 dx  .
3
Hence, or otherwise, find the area enclosed by the ellipse 9x2 + y2 = 4.¤
4 units2 »
3
« (a) Proof (b)
4U87-1iii)
a.

2
n  1
Given that I n   cosn x dx , prove that I n  
 I n  2 where n is and integer and n  2.
 n 
0

2
b.
Hence evaluate  cos5x dx .¤
0
« (a) Proof (b)
8
»
15
4U86-1i)
1
Evaluate

0
x
dx .¤
(2  x)
«


2
4 2 -5 »
3
4U86-1ii)
1
Use integration by parts to show that
 tan
0
1
x dx    1 loge 2 .¤
4 2
« Proof »
4U86-1iii)
¤©BOARD OF STUDIES NSW 1984 - 1997
4 UNIT MATHEMATICS – INTEGRATION – HSC
2
Find numbers A, B, C such that
1
Hence evaluate
x
B
C
.
A

2
2x  3 2x  3
4x  9
x2
0 4x2  9 dx .¤
« A 
4U86-1iv)
1
2
Using the substitution t = tan(  ), or otherwise, show that

3
1
3
3 1 3
log e 5 »
,B ,C=- ; 
4
8
8 4 16
1
 1  sin d 
3  1 .¤
0
« Proof »
4U85-1i)
Find:
3
 x cos x dx ;
a.
0
1
b.
x
2
0
1
dx .¤
 4x  5
« (a) -2 (b) 0.1419 (to 4 decimal places) »
4U85-1ii)
Find real numbers A, B, C such that
1
2
Hence show that
x
A
B
C



.
2
2
(x 1) (x  2) x 1 (x 1) x  2
 3
x
 (x  1) (x  2) dx  2log  2   1 .¤
e
2
0
« A = -2, B = -1, C = 2, Proof »
4U85-1iii)!
a

a

Use the substitution x = a - t, where a is a constant, to prove that f(x)dx  f(a  t)dt . Hence, or
0
1
otherwise, show that
 x(1  x)
99
dx 
0
0
1
.¤
10 100
« Proof »
4U84-1i)
Show that:
e4
4  xloge x dx  7e8  e2 ;
a.
e
e4
b.
dx
 x log
e
e
x
 2 log e 2 .¤
« (a) Proof (b) Proof »
4U84-1ii)
Evaluate:
3
a.
x 1
dx
2
1
x
0
¤©BOARD OF STUDIES NSW 1984 - 1997
4 UNIT MATHEMATICS – INTEGRATION – HSC
4
b.

4
x6
dx
x 5

6
c.
 sin 4x cos 2x dx .¤

6
« (a) log e 2 
¤©BOARD OF STUDIES NSW 1984 - 1997

(b) 21 13 (c) 0 »
3
4 UNIT MATHEMATICS – VOLUMES – HSC
Volumes
4U97-3a)!
y
C(0, 2)
(3, 2)
O
x
In the diagram, the shaded region is bounded by the x axis, the line x = 3, and the circle with centre
C(0, 2) and radius 3. Find the volume of the solid formed when the region is rotated about the y
axis.¤
«
4U97-5a)!
i.
8
 units3 »
3
Sketch the graph of y = -x2 for 2  x  2.
ii.
Hence, without using calculus, sketch the graph of y = e-x for 2  x  2.
iii.
The region between the curve y = e-x , the x axis, the y axis, and the line x = 2 is rotated
around the y axis to form a solid. Using the method of cylindrical shells, find the volume of
the solid.¤
2
2
Y
-2
2X
y
1
-2
« i)
4U96-3a)!
¤©BOARD OF STUDIES NSW 1984 - 1997
-4
ii)
O
2 x


iii) 1 
1
units3 »
4

e
4 UNIT MATHEMATICS – VOLUMES – HSC
y
1
S
0
1
4
x
x + y2 = 0
-1
The shaded region is bounded by the lines x = 1, y = 1, and y = -1 and by the curve x  y2  0 . The
region is rotated through 360 about the line x = 4 to form a solid. When the region is rotated, the line
segment S at height y sweeps out an annulus.
i.
Show that the area of the annulus at height y is equal to (y4  8y2  7) .
ii.
Hence find the volume of the solid.¤
« i) Proof ii)
296
 units3 »
15
4U95-3b)!
The circle x² + y² is rotated about the line x = 9 to form a ring.
When the circle is rotated, the line segment S at height y sweeps out an annulus.
y
4
s
-4
4
0
-4
x
x=9
The x coordinates of the end-points are x1 and - x1, where x1 = 16  y2 Error! Switch argument
not specified..
i.
ii.
Show that the area of the annulus is equal to 36 16  y2 Error! Switch argument not
specified.
Hence find the volume of the ring. ¤
« (i) Proof (ii) 2882 units3 »
4U94-3b)!
¤©BOARD OF STUDIES NSW 1984 - 1997
4 UNIT MATHEMATICS – VOLUMES – HSC
y
E
t=0
0
-2
.
2
4
6
x
The graph shows part of the curve whose parametric equations are
x  t2  t  2 , y  t3  t , t  0 .
i.
Find the values of t corresponding to the points O and E on the curve.
ii.
The volume V of the solid formed by rotating the shaded area about the y axis is to be
calculated using cylindrical shells.
Express V in the form V  2 f ( t ) dt .
b
a
Specify the limits of integration a and b and the function f(t). You may leave f(t) in
unexpanded form.
Do NOT evaluate this integral. ¤
« (i) t = 1, 2 (ii) V  2 (t 2  t  2)(t 3  t)(2t  1) dt »
2
1
¤©BOARD OF STUDIES NSW 1984 - 1997
4 UNIT MATHEMATICS – VOLUMES – HSC
4U93-6a)!
The solid S is generated by moving a straight edge so that it is always parallel to a fixed plane P. It is
constrained to pass through a circle C and a line segment L. The circle C, which forms a base for S,
has radius a and the line segment L is distance d from C. Both C and L are perpendicular to P and sit
on P in such a way that the perpendicular C at its centre O bisects L.
l
S
G
C
E
x
F
O
a
d
P
i.
ii.
Calculate the area of the triangular cross-section EFG which is parallel to P and distance x
from the centre O of C.
Calculate the volume of S. ¤
« (i)
a 2  x2 d units2 (ii) 21 a 2d units3 »
4U92-5a)!
2a
a
x
a
a
The solid S is a rectangular prism of dimensions a  a  2a from which right square pyramids of base
a  a and height a have been removed from each end. The solid T is a wedge that has been obtained
by slicing a right circular cylinder of radius a at 45 through a diameter AB of its base.
Consider a cross-section of S which is parallel to its square base at distance x from its centre, and a
corresponding cross-section of T which is perpendicular to AB and at distance x from its centre.
i.
The triangular cross-section of T is shown on the diagram.
Show that its area is
¤©BOARD OF STUDIES NSW 1984 - 1997
1 (a 2  x2 ) .
2
4 UNIT MATHEMATICS – VOLUMES – HSC
ii.
iii.
iv.
Draw the cross-section of S and calculate its area.
Express the volumes of S and T as definite integrals.
What is the relationship between the volumes of S and T? (There is no need to evaluate
either integral). ¤
C
A
« (i) Proof (ii)
x
a
B
A = a2 - x2 (iii) VS  2 (a 2  x 2 ) dx ,
a
0
VT 
¤©BOARD OF STUDIES NSW 1984 - 1997

a
0
(a  x ) dx . (iv) VT  12 VS »
2
2
4 UNIT MATHEMATICS – VOLUMES – HSC
4U91-5b)!
Figure 1
Water surface
E
D
A
F
C
G
B
H
O
Figure 2
Figure 3
C
C'
F
D
a
a
E
F
H
G
h
O
H
Water
A drinking glass having the form of a right circular cylinder of radius a and height h, is filled with
water. The glass is slowly tilted over, spilling water out of it, until it reaches the position where the
water’s surface bisects the base of the glass. Figure 1 shows this position.
In Figure 1, AB is a diameter of the circular base with centre C, O is the lowest point on the base, and
D is the point where the water’s surface touches the rim of the glass.
Figure 2 shows a cross-section of the tilted glass parallel to its base. The centre of this circular section
is C and EFG shows the water level. The section cuts the lines CD and OD of Figure 1 in F and H
respectively.
Figure 3 shows the section COD of the tilted glass.
i.
ii.
a
(h  x) , where OH = x.
h
x 
ax
-1
Use Figure 2 to show that C'F =
and HC'G = cos h .
h
Use Figure 3 to show that FH =
iii.
Use (ii) to show that the area of the shaded segment EGH is
iv.

x
x
x 2
a 2 cos 1      1     .
 h  h
 h 


1
1
Given that  cos  d  cos   1  2 , find the volume of water in the tilted glass of
Figure 1. ¤
« (i) Proof (ii) Proof (iii) Proof (iv)
¤©BOARD OF STUDIES NSW 1984 - 1997
2a 2 h
units3 »
3
4 UNIT MATHEMATICS – VOLUMES – HSC
4U90-3a)!
y
y = e - x²
S
t
-a
x
a
The region under the curve y  ex and above the x axis for a  x  a is rotated about the y axis to
form a solid.
i.
Divide the resulting solid into cylindrical shells S of radius t as in the diagram. Show that
2
each shell S has approximate volume V  2tet t where t is the thickness of the shell.
ii.
Hence calculate the volume of the solid.
iii.
What is the limiting value of the volume of the solid as a approaches infinity? ¤
2
« (i) Proof (ii) (1  e a ) units3 (iii)  units3 »
4U89-5b)!
2
P
O
Z
T
a
Y
X
A
B
Let ABO be and isosceles triangle, AO = BO = r, AB = b.
Let PABO be a triangular pyramid with height OP = h and OP perpendicular to the plane of ABO as in
the diagram. Consider a slice S of the pyramid of width a as in the diagram. The slice s is
perpendicular to the plane of ABO at XY with XY || AB and XB = a. Note that XT || OP.
i.
r  a  b ah a , when a is small. (You may assume that the
Show that the volume of S is 
  
r
r
slice is approximately a rectangular prism of base XYZT and height a.)
ii.
Hence show that the pyramid PABO has volume
¤©BOARD OF STUDIES NSW 1984 - 1997
1
hbr .
6
4 UNIT MATHEMATICS – VOLUMES – HSC
iii.
2
Suppose now that AOB =
and that n identical pyramids PABO are arranged about O as
n
centre with common vertical axis OP to form a solid C. Show that the volume Vn of C is

n
1
3
2
given by Vn  r hn sin .
iv.
Note that when n is large, the solid C approximates a right circular cone. Using the fact that
sinx
approaches 1 as x approaches 0, find lim Vn . Hence verify that a right circular cone of
n 
x
1 2
radius r and height h has volume r h . ¤
3
« (i) Proof (ii) Proof (iii) Proof (iv) 13 r 2 h »
4U88-7a)!
H
I
20cm
L
M
xcm
4cm
hcm
i.
16cm
J
K
FIGURE NOT TO SCALE
A trapezium HIJK has parallel sides KJ = 16 and HI = 20cm. The distance between these
sides is 4cm. L lies on HK and M lies on IJ such that LM is parallel to KJ. The shortest
distance from K to LM is h cm and LM has length x cm. Prove that x = 16 + h.
ii.
20cm
12cm
10cm
16cm
FIGURE NOT TO SCALE
The diagram above is of a cake tin with a rectangular base with sides of 16cm and 10cm. Its
top is also rectangular with dimensions 20cm and 12cm. The tin has depth 4cm and each of
its four side faces is a trapezium. Find its volume. ¤
« (i) Proof (ii) 794 23 cm3 »
4U87-5i)
a.
b.
c.
On a number plane shade the region representing the inequality (x - 2R)2 + y2  R2.
Show that the volume of a right cylindrical shell of height h with inner and outer radii x and
x + x respectively is 2xhx when x is sufficiently small for terms involving (x)2 to be
neglected.
The region (x - 2R)2 + y2  R2 is rotated about the y axis forming a solid of revolution called
a torus. By summing volumes of cylindrical shells show that the volume of the torus is given
by: V = 42R3. ¤
¤©BOARD OF STUDIES NSW 1984 - 1997
4 UNIT MATHEMATICS – VOLUMES – HSC
y
2R
R
(x - 2R)2 + y2  R2
« (a)
4U86-5i)
a.
b.
3R
x
(b) Proof (c) Proof »
The coordinates of the vertices of a triangle ABC are (0, 2), (1, 1), (-1, 1) respectively and H
is the point (0, h). The line through H, parallel to the x-axis meets AB and AC at X and Y.
Find the length of XY.
The region enclosed by the triangle ABC, defined in (a) above, is rotated about the x axis.
Find the volume of the solid of revolution so formed. ¤
« (a) (4 - 2h) units (b)
8
units3 »
3
4U85-5ii)
a
a.
1
2 2
Using the substitution x = asin, or otherwise, verify that  (a  x ) dx 
2
0
2
b.
Deduce that the area enclosed by the ellipse
1 2
a .
4
2
x
y
 2  1 is ab.
2
a
b
c.
H
h
The diagram shows a mound of height H. At height h above the horizontal base, the
horizontal cross-section of the mound is elliptical in shape, with equation
x2 y2
 2  2 ,
2
a
b
h2
, and x, y are appropriate coordinates in the plane of the cross-section.
H2
8abH
Show that the volume of the mound is
.¤
15
where   1
« Proof »

4U84-5)

¤©BOARD OF STUDIES NSW 1984 - 1997
4 UNIT MATHEMATICS – VOLUMES – HSC
y
S
t
-a
i.
ii.
t
0
a
x
The diagram shows the area A between the smooth curve y = f(x), -a  x  a, and the x axis.
(Note that f(x)  0 for -a  x  a and f(-a) = f(a) = 0). The area A is rotated about the line
x = -s (where s > a) to generate the volume V. This volume is to be found by slicing A into
thin vertical strips, rotating these to obtain cylindrical shells, and adding the shells. Two
typical strips of width t whose centre lines are distance t from the y axis are shown.
a.
Show that the indicated strips generate shells of approximate volume 2 f(-t)(s t)t, 2 f(t)(s + t)t, respectively.
b.
Assuming that the graph of f is symmetrical about the y axis, show that V = 2sA.
Assuming the results of part (i), solve the following problems.
a.
A doughnut shape is formed by rotating a circular disc of radius r about and axis in
its own plane at a distance of s(s > r) from the centre of the disc. Find the volume
of the doughnut.
b.
The shape of a certain party jelly can be represented by rotating the area between
the curve y = sinx, 0  x  , and the x axis about the line x = 
volume generated. ¤
¤©BOARD OF STUDIES NSW 1984 - 1997