Originally Posted by

**galactus** I agree with PH, rectangular is best if possible. Here's the cylindrical if you must:

For the z limits with r and theta held fixed:

$\displaystyle z+y=1$

$\displaystyle z=1-rsin({\theta})$

For the r limits with z and theta held fixed:

$\displaystyle 2x^{2}+y^{2}=1$

If you sub in $\displaystyle x=rcos({\theta})\;\ and\;\ y=rsin({\theta})$ and

solve for r you arrive at:

$\displaystyle r=\frac{1}{\sqrt{cos^{2}({\theta})+1}}$

Of course, the first octant is 0 to $\displaystyle \frac{\pi}{2}$

$\displaystyle \int_{0}^{2\pi}$$\displaystyle \int_{0}^{\frac{1}{\sqrt{cos^{2}({\theta})+1}}}$$\displaystyle \int_{0}^{1-rsin({\theta})}rdzdrd{\theta}$

This will be hard to do by hand. I would use technology if available to evaluate.