MATH 101, Section 212 (CSP)
Week 3: Marked Homework Solutions
2011 Jan 27
1. [12] The intersection points are given by 2 y = 8 − y 2 iff y 2 +2 y − 8 = 0 iff ( y +4)( y − 2) = 0 iff y = − 4 or y = 2, so the intersection points are ( − 8 , − 4) and (4 , 2).
(a) Using x as the variable of integration, the region must be divided into two, at x = 4, and the vertex of the parabola at x = 8 must be determined. Vertical rectangles of infinitesimal width dx indicate that the integrands should be functions of x . The area of a typical infinitesimally thin vertical rectangle is dA = ( y
T
− y
B
) dx, where y
T is the y -coordinate of the top curve and y
B is the y -coordinate of the bottom curve, both functions of
= +
A = R x x
. In the region to the left of
=
=4 x =
−
8
− dA
1
+ R x =8 x =4 dA
2
= R
4
−
8
( 1
2 x x
+
= 2, the top curve is
√
√
8 − x . The area of the region is
8 − x ) dx + R
4
8 y
T
= 1
2 x and the bottom curve is y
T y
B
8 − x . In the region to the right of x = 4, the top curve is now
8 − x and the bottom curve is still y
B
= −
2
√
8 − x dx
4 8
= 1
4 x 2 − 2
3
(8 − x ) 3 / 2
−
8
+ − 4
3
(8 − x ) 3 / 2
4
= 16 16 64 128 32
4
−
3
−
4
−
3
+ 0 +
3
= 36 .
(b) Using y as the variable of integration, the area can be calculated with a single integral. Horizontal rectangles of infinitesimal height dy indicate that the integrand should be functions of y . The area of a typical infinitesimally thin horizontal rectangle is dA = ( x
R
− x
L
) dx, where x
R is the x -coordinate of the right curve and x
L both functions of y . The right curve is x
R
= 8 − y 2 is the x -coordinate of the left curve, and the left curve is x
L
= 2 y . The area of the region is
2
R
2
−
4
[ (8 − y
2 ) − 2 y ] dy = 8 y − 1
3 y
3 − y
2
−
4
= 36 .
2. [8] The intersection point is given by x + 2 = x 2 iff x 2 − x − 2 = ( x + 1)( x − 2) = 0 iff x = − 1 or 2, but x = − 1 is outside the region, so the intersection point at x = 2 divides the region into two parts, both of which contribute positive area.
1

The total area is
R
3
0
| ( x + 2) − x 2 | dx =
=
=
R
2
0
| ( x + 2) − x 2 | dx + R
3
2
| ( x + 2) − x 2 | dx
R
2
0
[( x + 2) − x
2
] dx +
2
R
3
2
[ x
2
1
2 x
2 + 2 x − 1
3 x
3
0
+ 1
3 x
3
− ( x + 2)] dx
− 1
2 x
2 − 2 x
3
2
= 31
.
6
3. [6] A typical infinitesimally thin slice of the solid is generated by an infinitesimally thin vertical strip under the graph of
10 radius dV = A y
(
= x )
√
25
− x 2 dx = 100 π
25
− x 2
V = R x =4 x =3 dV = R
4
3 dx
100 π
25
− x 2 dx y =
= 100 π
√
25
− x 2
R
3
10
4 dx
, of width
25
− x 2
.
dx
. The total volume of the solid is
. The slice generated is a disk of
, therefore has a cross sectional area of A ( x ) = πy 2 = 100 π
25
− x 2 and volume
4. [8] Find the intersections of the two curves: y 2 + 1 = 2 y + 1 iff y 2 = 2 y iff y 2 − 2 y = 0 iff y ( y − 2) = 0 iff y = 0 or y = 2. Using cross-sections, a typical infinitesimally thin slice of the solid is a washer (or annular ring) generated by an infinitesimally thin horizontal strip between the two graphs, of height dy . The outer radius is the distance from the farthest curve x = 2 y + 1 to the axis of rotation x = 1: r o
= (2 y + 1) − (1) = 2 y, and the inner radius is the distance from the nearest curve x = y 2 x = 1: r i
= ( y
2 + 1) − (1) = y
2
.
+ 1 to the axis of rotation
The washer therefore has infinitesimal volume dV = π ( r 2 o
− r i
2 ) dy = π (4 y 2 − y 4 ) dy, and the solid has total volume
V = R y =2 y =0 dV = R
2
0
π (4 y 2 − y 4 ) dy
= π
=
4
3 y
3 − 1
5 y
5
64
15
π ( ≈ 13 .
404) .
2
0
2

5. [8] Using cross-sections, a typical infinitesimally thin slice of the solid is a washer generated by an infinitesimally thin vertical strip between the two graphs y = 1 and y = e
− x , of width dx . The outer radius is the distance from the axis of rotation y = 2 to the lower curve y = e
− x : r o
= (2) − ( e − x ) , and the inner radius is the distance from y = 2 to the upper curve y = 1: r i
= (2) − (1) .
The washer has infinitesimal volume dV = π ( r 2 o
− r i
2 ) dx = π [(2 − e − x ) 2 − 1 2 ] dx = π (3 − 4 e − x + e −
2 x ) dx, and the solid has total volume
V = R x =2 x =0 dV =
= π (3 x + 4 e − x
R
2
0
−
π (3 − 4 e − x
1
2 e −
2 x )] 2
0
= π [(6 + 4 e −
2 − 1
2 e −
4
+ e −
) − (0 + 4 −
2 x
1
2
) dx
)
= π ( 5
2
+ 4 e −
2 − 1
2 e −
4 ) .
6. [8] The base of a typical square has length
` = (8 − x
2 ) − ( x
2 ) = 8 − 2 x
2
, and therefore its area is
A ( x ) = `
2 = (8 − 2 x
2 ) 2
(see the figure below). The volume of a typical infinitesimally thin slice is then dV = (8 − 2 x
2 ) 2 dx,
3

and the total volume of the solid is
V = R x =2 x =
−
2
= 2 R
2
0
(8 dV
− 2
= x
2
R
) 2
2
−
2
(8 dx
− 2 x 2 ) 2 dx
(even integrand, symmetric limits of integration)
= 2 R
2
0
(64 − 32 x
2 + 4 x
4 ) dx
2
= 2 64 x −
2048
=
15
32
3 x
3 +
( ≈ 136 .
5) .
4
5 x
5
0
7. [10] A cross-section perpendicular to the base and perpendicular to the y -axis can be identified by its y -coordinate. A typical infinitesimally thin slice has thickness dy and is an equilateral triangle. The base of the equilateral triangle is horizontal in the xy -plane with its right end at x = 1 − y and its left end on the y -axis, i.e.
x = 0. Therefore the base has length b = (1 − y ) − (0) = 1 − y.
Then the area of an equilateral triangle with base of length b is
A ( y ) = 1
2
( b )
√
3
2 b =
√
3 b 2
4
=
√
3
4
(1 − y ) 2 , and the slice has infinitesimal volume dV = A ( y ) dy =
√
3
4
(1 − y ) 2 dy =
√
3
4
(1 − 2 y + y
2 ) dy.
The total volume is
V =
=
=
=
R
4 y =1 y =0
√
3 dV = y − y
2
√
3
4
√
3
4
R
+
0
1
1
3
√
3
4 y
3
(1 − 2 y + y
2 ) dy
1
0
1
3
1 − 1 2
√
3 =
12
+ 1
3
1 3 −
( ≈ 0 .
144) .
0 − 0 2 + 1
3
0 3
4