# Thread: Triple Integral Problem ><

1. ## Triple Integral Problem ><

Hello there!

I have a question I've been stuck on for a couple days...would really appreciate some help! =) =) =)

Find the volume of the region bounded by...
the cone x^2 + y^2 = z^2 (z >= 0)
and the paraboloid z = 2x^2 + 2y^2.

2. $\displaystyle V=\iint_D \left( \int_{2x^2+2y^2}^{\sqrt{x^2+y^2}}dz\right)dxdy$ where $\displaystyle D$ is the projection on the xy-plane: $\displaystyle \sqrt{x^2+y^2}=2x^2+2y^2$

Switch to polar coordinates and we get $\displaystyle r=2r^2 \Longleftrightarrow r=0,\ r=\frac{1}{2}$

So we have $\displaystyle D'=\left\{0\leq r\leq \frac{1}{2},\ 0\leq \varphi \leq 2\pi \right\}$ and the Jacobian is $\displaystyle r$.

3. Thanks for the speedy response! =D

Another question: You've given me the limits for the third integral, I've found the limits for the second integral, how do I find them for the first? I try to put the two equations in terms of y and let them equal, but then I can't get rid of an x or z completely.

Much appreciated. =)

4. I've already given you all the limits you need. The area $\displaystyle D$ is transformed to $\displaystyle D'$ when switching to polar coordinates, so the integral we're left with is:

$\displaystyle \iint_D \left( \int_{2x^2+2y^2}^{\sqrt{x^2+y^2}}dz\right)dxdy=\ii nt_D(\sqrt{x^2+y^2}-(2x^2+2y^2))dxdy=\iint_{D'}(r-2r^2)r\ drd\varphi$