Multivariate Calculus - Bruce E. Shapiro

LECTURE 18. DOUBLE INTEGRALS IN POLAR COORDINATES 151
With this change of variables, when r = 5, u = −25; when r = 7, u = −49. Then
the inner integral is then
∫ 7
5
e −r2 rdr = − 1 2
Finally, we can calculate the double integral,
V =
∫ −49
−25
e u du
= − 1 2 eu | −49
−25
= − 1 (
e −49 − e −25)
2
= 1 (
e −25 − e −49)
2
≈ 6.94 × 10 −12
∫ π/2 ∫ 7
0
5
e −r2 rdrdθ
∫ π/2
≈ 6.94 × 10 −12 dθ
0
( π
)
≈ (6.94 × 10 −12 )
2
≈ 1.1 × 10 −11
Example 18.6 Find the integral
√
4 − x 2 − y 2 dA
s
where S is the first-quadrant sector of the circle of radius 2 centered at the orgin
between y = 0 and y = x.
Solution. The domain is the same 45 degree sector of a circle as in example 18.3.
Since
the integral is
f(r cos θ, r sin θ) = √ 4 − (r cos θ) 2 − (r sin θ) 2
= √ 4 − r 2
√
4 − x 2 − y 2 dA =
∫ 2 ∫ π/2
√
4 − r 2 rdθdr
S
=
= π 2
0
∫ 2
0
r √ ( ∫ )
π/2
4 − r 2 dθ dr
0
∫ 2
0
0
r √ 4 − r 2 dr
Math 250, Fall 2006 Revised December 6, 2006.