Question

surface integral in hemisphere region

Answer 1

ok u might wanna sit back and get comfortable now

Answer 2

First, you know the relevant equation for surface integrals:
double integral over region (s) of f(x,y,z) dS = double integral over region (D) of f( r(u,v) ) * |r_u x r_v| dA

I would suggest rewriting x, y, and z in terms of spherical coordinates, knowing that rho must be 2 since it is the radius of the hemisphere: x = 2sin(phi)cos(theta), y = 2sin(phi)sin(theta), z=2cos(phi). You can then rewrite the bounds of the hemisphere S as 0 < phi < pi/2 and 0 < theta < 2pi.

You can then write a position vector r(phi, theta) = 2sin(phi)cos(theta) i + 2sin(phi)sin(theta) j + 2cos(phi) k. You can then take the partials with respect to phi and theta, cross them, and find the magnitude of the resulting vector. This would be written as |r_phi x r_theta|, which equals 4sin(phi) in this case. (The general solution is rho^2 * sin(phi).)

Plug this back in to the relevant equation I first mentioned, double integral over region (s) of f(x,y,z) dS = double integral over region (D) of f( r(u,v) ) * |r_u x r_v| dA
Make use of your limits of integration for the double integral, substitute the integrand's x, y, and z in terms of phi and theta (this is your f( r(u,v) ) ), and don't forget to multiply in the value you got for |r_u x r_v|, or 4sin(phi). This should give you the final answer of 16pi