Multivariate Calculus - Bruce E. Shapiro

158 LECTURE 19. SURFACE AREA WITH DOUBLE INTEGRALS
Solution. The domain is the set
{(x, y) : 0 ≤ y ≤ 3, 0 ≤ x ≤ √ 9 − y 2 }
Therefore the area is
A =
∫ 3
0
∫ √ 9−y 2
0
√
1 + f 2 x + f 2 y dxdy
Differentiating f(x, y) = (9 − y 2 ) 1/2 gives f x = 0 and
f y = (1/2)(9 − y 2 ) −1/2 (−2y) = −y(9 − y 2 ) −1/2
and therefore
A =
=
=
=
=
= 3
∫ 3
0
∫ 3
0
∫ 3
0
∫ 3
0
∫ 3
∫ √ 9−y 2
0
∫ √ 9−y 2
0
√
1 + f 2 x + f 2 y dxdy
√
1 + [−y(9 − y 2 ) −1/2 ] 2 dxdy
∫ √ 9−y 2 √
1 + y 2 (9 − y 2 ) −1 dxdy
0
∫ √ √
9−y 2
0
0 0
∫ 3
0
∫ √ 9−y 2
∫ √ 9−y 2
0
1 + y2
9 − y 2 dxdy
√ 9
9 − y 2 dxdy
1
√
9 − y 2 dxdy
Since the integrand is only a function of y, we can move it from the inner integral
to the outer integral:
∫ 3 ∫ √ 9−y
1
2
A = 3 √ dxdy
9 − y 2
0
= 3
= 3
0
∫ 3
0
∫ 3
0
1
√
9 − y 2
√
9 − y 2 dy
dy = 9
Example 19.3 Find a formula for the surface area of a sphere of radius a.
Solution. The equation of a sphere of radius a centered at the origin is
x 2 + y 2 + z 2 = a 2
Revised December 6, 2006. Math 250, Fall 2006