bounds on a surface integral

[mathstresser]

Evalute the surface integral.

\(\displaystyle \L\\\int_{S}\int_{S} x^2+y^2+z^2 dS\)

S: cylinder together with its top and bottom disks
\(\displaystyle \L\\\ x^2+ y^2 = 9, z=0 and z=2\)

\(\displaystyle x=x
y=sqrt(9-x^2)
z=z\)

I get \(\displaystyle \L\\\int_{S}\int_{S} (x^2 + z+2 + (9-x^2))(sqrt(1+ 0^2 +(\frac{-1}{2sqrt(9-x})^2)\)

\(\displaystyle \L\\\int_{0}^{2}\int_{?}^{?} (z^2 + 9)(sqrt(1+ \frac{1}{4(9-x)})) dxdz\)

I don't know the bounds for dx. I thought maybe (0,3) because of the radius. Or (0,x) which would be
\(\displaystyle \L\\ (0,sqrt(9-y^2))\)

But then how would I evaluate the integral?

 

[peacefreak77]

I would try to switch to polar or cylindrical coordinates if possible.

I would be of more help, but I'm not entirely sure what you mean by "surface" integral.