1. ## Surface Integral

Evaluate the following surface integral of the given function on the given surface.

g(x,y,z)=z on the first-octant part of the sphere $\displaystyle x^2+y^2+z^2=1$

My work so far:

$\displaystyle z=\sqrt{1-(x^2+y^2)}$

The domain: $\displaystyle (x^2+y^2)\leq 1$

$\displaystyle \frac{\partial{z}}{\partial{x}}=\frac{-x}{\sqrt{1-(x^2+y^2)}}$

$\displaystyle \frac{\partial{z}}{\partial{y}}=\frac{-y}{\sqrt{1-(x^2+y^2)}}$

$\displaystyle \iint (z)*\frac{1}{\sqrt{1-(x^2+y^2)}}dA$

Here is the point where I get stuck. In all my previous problems my g(x,y,z) only contained x's and y's. So, if that were the case here, I would just convert everything to polar coordinates, plug in the bounds and solve the integral. In this case though, there is a z in the integrand so I'm not sure how to continue. Could someone please help me out of a jam? Thanks

2. ## Re: Surface Integral

Originally Posted by dbakeg00
In this case though, there is a z in the integrand so I'm not sure how to continue. Could someone please help me out of a jam? Thanks
Substitute $\displaystyle z=\sqrt{1-x^2-y^2}$ . If $\displaystyle S$ is the given surface and $\displaystyle D$ its projection onto the $\displaystyle xy$ plane, then $\displaystyle \iint_Sz\;dS=\iint_D\;dxdy=\textrm{Area}\;(D)=\pi/4$ .

3. ## Re: Surface Integral

That worked. Thanks man, I appreciate the help.