Find the area of the surface given by the part of the sphere

Find the area of the surface given by the part of the sphere x^2 + y^2 + z^2 = 9 that lies above the plane z = 1.

Solution

Solution :

The surface area equals
(1 + (z_x)² + (z_y)²) dA
= (1 + (-x/(9 - x² - y²))² + (-y/(9 - x² - y²))²) dA, since z = (9 - x² - y²)
= (1 + (x² + y²)/(9 - x² - y²)) dA
= (9/(9 - x² - y²)) dA
= 3 dA/(9 - x² - y²).

Since the region of integration is the interior of x² + y² = 8
(from the intersection of the sphere with z = 1), converting to polar yields
( = 0 to 2) (r = 0 to 8) 3 * (r dr d)/(9 - r²)
= 2 (r = 0 to 8) 3r(9 - r²)^(-1/2) dr
= 2 * -3(9 - r²)^(1/2) {for r = 0 to 8}
= 2 * 3(3 - 1)
= 12.