Could someone take a look at this please? Thanks(adsbygoogle = window.adsbygoogle || []).push({});

=====

Q. Evaluate IntegralA.ndS for the following case:

A=(6z, 2x+y, -x) and S is the entire surface of the region bounded by the cylinder x^2 + z^2 = 9, x=0, y=0, z=0 and y=8.

=====

Using Gauss' (or Divergence) Theorem:

IntegralA.ndS = IntegralDiv AdV

Div A= 1, therefore

Integral 1 dV

When I last did this question, I said that dV = pi*r^2*h where r^2=9 and h=8. However, I've also got a factor of 1/4 to give me the final answer of 18pi. Where the hell did the quarter come from?! I'm guessing it has something to do with the limits and the fact that it's bounded by x=0, y=0 and z=0, but I don't understand why! Anyone offer some advice?!

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Surface Integral

**Physics Forums | Science Articles, Homework Help, Discussion**