
Volume of Solid
Find the volume of the solid above the cone $\displaystyle z=\sqrt{x^2+y^2}$ and below the sphere $\displaystyle x^2+y^2+z^2=1$
Obviously use polar coordinates here.
So the two equations are $\displaystyle z=r$ and $\displaystyle z=\sqrt{1r^2}$.
I'm trying to draw the region of integration in the xy plane and getting stuck. I think the cone comes down to a single point at (0,0), so maybe that can be disregarded. The sphere has a projection of a unit circle. So is the region of integration the unit circle?
Also what are the bounds for the integral?

ok so we have $\displaystyle V=\iiint_{E}dV$
$\displaystyle E$ is the region that we want.
so we convert to spherical coordinates.
it is obvious that $\displaystyle 0\le p\le 1$ and $\displaystyle 0\le \theta\le 2\pi$
to find the values for $\displaystyle \phi$ we have
$\displaystyle z=p\cos(\phi)=\sqrt{(p\sin(\phi)\cos(\theta))^2+(p \sin(\phi)\sin(\theta))^2}$
which is
$\displaystyle \cos(\phi)=\sin(\phi)\Longrightarrow \phi=\frac{\pi}{4}$
so we have
$\displaystyle \int_0^{2\pi}\int_0^{\frac{\pi}{4}}\int_0^1 p^2\sin(\phi)dp d\phi d\theta$ and i am sure that you can solve it from here.

That spherical method is just what I was gonna post. That's what I arrived at also.
Oh well, here's another way, but equivalent in rectangular.
$\displaystyle \int_{\frac{1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}$$\displaystyle \int_{\sqrt{\frac{1}{2}x^2}}^{\sqrt{\frac{1}{2}x^2}}$$\displaystyle \int_{\sqrt{x^2+y^2}}^{\sqrt{1x^2y^2}}dzdydx$
Now, do it in cylindrical coordinates.
$\displaystyle \int_{0}^{2{\pi}}$$\displaystyle \int_{0}^{\frac{1}{\sqrt{2}}}$$\displaystyle \int_{r}^{\sqrt{1r^{2}}}rdzdrd{\theta}$
Thanks for the fix PH. Everytime I tried to edit last night I kept getting a "this page can't be displayed" window, so I just gave up.