Evaluate the surface integral double integrals F ds for the

Evaluate the surface integral double integral_s F ds for the given vector field F and the oriented surface S. In other words, find the flux of F across S. For closed surfaces, use the positive (outward) orientation. F(x, y, z) = xz i + x j + y k S is the hemisphere x^2 + y^2 + z^2 = 25, 0, oriented in the direction of the positive y-axis

Solution

Using Cartesian Coordinates: Solving for y yields y = (25 - x^2 - z^2), since y 0.
==> r(x, z) = <x, (25 - x^2 - z^2), z> with x^2 + z^2 25.

r_x x r_z = <-x/(25 - x^2 - z^2), -1, -z/(25 - x^2 - z^2)>

==> n = -(r_x x r_z) = <x/(25 - x^2 - z^2), 1, z/(25 - x^2 - z^2)>
so n points out of S.

Hence, the flux s F · dS equals
<xz, x, (25 - x^2 - z^2)> · <x/(25 - x^2 - z^2), 1, z/(25 - x^2 - z^2)> dA
= [x^2 z/(25 - x^2 - z^2) + x + z] dA

Convert to polar coordinates (x = r cos , y = r sin ):
(r = 0 to 5) ( = 0 to 2) [(r cos )^2 * (r sin )/(25 - r^2) + r cos + r sin ] * r d dr
= (r = 0 to 5) ( = 0 to 2) [r^4 cos^2() sin /(25 - r^2) + r^2 cos + r^2 sin ] d dr
= (r = 0 to 5) [(-1/3) r^4 cos^3()/(25 - r^2) + r^2 sin - r^2 cos ] {for = 0 to 2} dr
= 0.