Multivariate Calculus - Bruce E. Shapiro

LECTURE 19. SURFACE AREA WITH DOUBLE INTEGRALS 159
Solving for z
z = ± √ a 2 − x 2 − y 2
The total area is the sum of the area of the top half and the area of the bottom half
of the plane; the equation of the sphere in the top half of the plane is
f(x, y) = √ a 2 − x 2 − y 2
Differentiating
f x =
−x
√
a 2 − x 2 − y 2 and f y =
−y
√
a 2 − x 2 − y 2
The area of the top half of the sphere is then
A =
=
=
=
√
1 + fx 2 + fy 2 dxdy
C
√
1 +
c
c
c
= a
x 2
a 2 − x 2 − y 2 + y 2
a 2 − x 2 − y 2 dxdy
√
a 2 − x 2 − y 2
a 2 − x 2 − y 2 + x 2
√
c
a 2
a 2 − x 2 − y 2 dxdy
1
√
a 2 − x 2 − y 2 dxdy
a 2 − x 2 − y 2 + y 2
a 2 − x 2 − y 2 dxdy
where C is the circle of radius a in the xy-plane. In polar coordinates, then
∫ 2π ∫ a
1
A = a √
0 0 a 2 − r rdrdθ 2
Letting u = a 2 − r 2 we have du = −2rdr; when r = 0, u = a 2 and when r = a,
u = 0.Thus the area of the top half of the sphere is
A = a
= − a 2
= − a 2
= −a
∫ 2π ∫ 0
1
0 a
∫ 2 2π ∫ 0
0
∫ 2π
0
∫ 2π
0
√u (−du/2)dθ
u −1/2 dudθ
a 2 u 1/2
0
dθ
1/2 ∣
a 2
[0 − (a 2 ) 1/2 ]dθ
∫ 2π
= a 2 dθ = 2πa 2
Therefore the area of the sphere is 2A = 4πa 2
0
Math 250, Fall 2006 Revised December 6, 2006.