Volume involving a triple integral

**Shapeshift** · January 19th 2011, 06:11 AM

Find the volume determined by z less than or equal to 6-x^2-y^2 and z greater than or equal to sqrt(x^2+y^2).

I'm not sure how to find the limits of integration for this question. I set the two inequalities to each other and found that x^2+y^2=9 and x^2+y^2=4, but am a little stumped on where to continue from there. I'm thinking at some point I will need to convert to cylindrical coordinates, but am still not sure how to go about it. Any help would be appreciated.

**DrSteve** · January 19th 2011, 07:22 AM

In cylindrical the equations are $z=6-r^2$ and $z=r$ . These intersect in the circle $r=2$ .

So in cylindrical, we have $0\leq \theta \leq 2\pi , 0\leq r\leq 2$ and $r\leq z\leq 6-r^2$ .

**Shapeshift** · January 19th 2011, 04:59 PM

Thanks so much! I ended up getting 32pi/3, which I hope is right!

**DrSteve** · January 19th 2011, 07:41 PM

Yep - that's what I got.

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