# Surface Integral Help

• June 13th 2008, 10:50 PM
Chris L T521
Surface Integral Help
It's been 2 years since I've looked at this stuff, and I'm trying to make sense of all of this. How would I evaluate:

$\iint_\sigma \bigg(x^2+y^2\bigg)z \,dS$

Where $\sigma$ is the portion of the cone $z=\sqrt{x^2+y^2}$ between the planes $z=1$ and $z=2$.

My biggest problem is setting it up properly.

Any help would be appreciated.

--Chris
• June 14th 2008, 03:13 AM
CaptainBlack
Quote:

Originally Posted by Chris L T521
It's been 2 years since I've looked at this stuff, and I'm trying to make sense of all of this. How would I evaluate:

$\iint_\sigma \bigg(x^2+y^2\bigg)z \,dS$

Where $\sigma$ is the portion of the cone $z=\sqrt{x^2+y^2}$ between the planes $z=1$ and $z=2$.

My biggest problem is setting it up properly.

Any help would be appreciated.

--Chris

Convert to cylindrical polars, then the area element of the cone in question is $\sqrt{2}r d\theta dr$ and the integral becomes:

$I=\sqrt{2} \int_{r=1}^2 \int_{\theta=0}^{2 \pi} r^4 ~d\theta ~dr$

RonL