Submitted by: Submitted by gsandyj
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Category: Science and Technology
Date Submitted: 09/01/2011 08:07 PM
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...