Math 241

Solution of Homework #7

Section 10.4 #5. θ is from 0 to π. The area is #6. θ is from =

π 2

π 1 2 r dθ 0 2

=

π 1 2 θ dθ 0 2

1 = 6 θ 3 |π = 1 π 3 0 6

π 2

to π. The area is

1−cos 2θ )dθ 2

π 2

π 1 2 r dθ 2

π 2

=

π 2

π 1 (1+sin θ)2 dθ 2 cos 2θ )dθ 4

π 4

=

π 2

π 1 (1+2 sin θ+sin2 2 sin 2θ π |π 8 2

θ)dθ

π 1 (1 2

+ 2 sin θ +

=

π 3 (4

π 4

+ sin θ −

1 2 r dθ 2

3 = 4 θ − cos θ −

π 4

= 3π + 1 8

#8. θ is from 0 to π . The area is 4

1 4 = 4 (θ − 1 sin 8θ)|0 = 8

π

0

=

0

1 (sin 4θ)2 dθ 2

=

0

1 1−cos 8θ dθ 2 2

π 16

#29. r = sin θ is a circle with center (0, 1/2) and radius 1/2. r = cos θ is a circle with center (1/2, 0) and radius 1/2. Notice that for r = sin θ, 0 ≤ θ ≤ π and π ≤ θ ≤ 2π generate the same circle. It is similar for r = cos θ. Look at the region bounded by these two curves. θ is from 0 to r = sin θ. For

0

π 4

π . 2

For 0 ≤ θ ≤

π 2 π 4

π , 4

the boundary of the region is

π 4

≤ θ ≤ π , the boundary of the region is r = cos θ. Therefore, the area is 2

π 2 π 4

1 (sin θ)2 dθ 2

+

1 (cos θ)2 dθ 2

=

0

π 4

1−cos 2θ dθ 4

+

1+cos 2θ dθ 4

=

π 8

−

1 4

#37. From #29, we know these two curves intersect at two points. One is the origin. To find the other point, set r = sin θ = cos θ = 0. Since cos θ = 0, we can divide the equation by cos θ to get

sin θ cos θ

= tan θ = 1. Therefore, θ is

π 4

or

5π . 4

These two arguments give the

1 same point ( 1 , 1 ), or ( √2 , π ) in terms of polar coordinates. 2 2 4

π 3 π 3 π 3

#45.

0

dr r2 + ( dθ )2 dθ =

0

(3 sin θ)2 + (3 cos θ)2 dθ =

0

3dθ = π

√ √ 2π 2π √ dr θ #48. 0 r2 + ( dθ )2 dθ = 0 1 + θ2 dθ = [ 2 1 + θ2 + 1 ln(θ + 1 + θ2 )]|2π 0 2 √ √ = π 1 + 4π 2 + 1 ln(2π + 1 + 4π 2 ) 2

Section 12.1 #3. The distance of a point to the xz-plane is the absolute value of its y value. Therefore, Q is closest to the xz-plane. A point lies in the yz...