Thomas Calculus 13th [Solutions]

Section 15.4 Double Integrals in Polar Form 1105
30. If f ( x, y ) is positive at some point P in R or on the boundary of R then by the continuity of f there is a
disk of positive radius around P (or if P is on the boundary, the intersection of such a disk with R)
on which
f ( x, y ) is positive. This sub-region will make a positive contribution to the area f ( x, y)
dA , and since
f ( x, y ) is never negative, f ( x, y)
dA will be greater than 0. This contradicts our assumption that
R
R
f ( x, y) dA 0 , so f ( x, y ) is positive nowhere on R and is thus equal to 0 at every point of R.
15.4 DOUBLE INTEGRALS IN POLAR FORM
R
1.
2.
2 2 2
x y 9 r 9 2 , 0 r 9
2 2 2 2 2 2
2
x y 1 r 1, x y 4 r 4 , 1 r 4
2 2
3. y x y x 3 y r 3 r
4 4 4 4
, , 1 csc , 0 csc
4.
5.
x 1 r sec , y 3x 0 , 0 r sec
3 3
2 2 2
x y 1 r 1, x 2 3 r 2 3 sec , y 2 r 2 csc ;
2 3 sec 2 csc 0 , 1 r 2 3 sec ;
6 6
6 2 , 1 r 2 3 csc
6.
2 2 2
x y 2 r 2, x 1 r sec ;
2 sec or
3
3 3 3 , sec r 2
7.
2 2
x y x r r
2 2 cos , 0 2 cos
2 2
8.
2 2
x y y r r
2 2 sin 0 , 0 2 sin
9.
2
1 1 x 1
dy dx r dr d 1
1 0 0 0 2 d
0 2
10.
2
1 1 y 2 2 /2 1 3 1
/2
x y dx dy r dr d d
0 0 0 0 4 0 8
11.
2
2 4 y 2 2 /2 2 3
/2
x y dx dy r dr d d
0 0 0 0 0
4 2
12.
2 2
a a x a
2 2 dy dx r dr d a d a
a a x 0 0 2 0
2 2 2 2
13.
6 y x dx dy
/2 6 csc r
2 dr d
/2 2 d
2
/2
0 0 /4 0 /4 /4
cos 72 cot csc 36 cot 36
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