Volume of Solid

euler · November 29th 2006, 04:32 PM

Find the volume of the solid above the cone $z=\sqrt{x^2+y^2}$ and below the sphere $x^2+y^2+z^2=1$

Obviously use polar coordinates here.

So the two equations are $z=r$ and $z=\sqrt{1-r^2}$ .

I'm trying to draw the region of integration in the x-y 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?

**putnam120** · November 29th 2006, 05:39 PM

ok so we have $V=\iiint_{E}dV$

$E$ is the region that we want.
so we convert to spherical coordinates.
it is obvious that $0\le p\le 1$ and $0\le \theta\le 2\pi$

to find the values for $\phi$ we have

$z=p\cos(\phi)=\sqrt{(p\sin(\phi)\cos(\theta))^2+(p \sin(\phi)\sin(\theta))^2}$
which is

$\cos(\phi)=\sin(\phi)\Longrightarrow \phi=\frac{\pi}{4}$

so we have

$\int_0^{2\pi}\int_0^{\frac{\pi}{4}}\int_0^1 p^2\sin(\phi)dp d\phi d\theta$ and i am sure that you can solve it from here.

**galactus** · November 29th 2006, 06:49 PM

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.

$\int_{\frac{-1}{\sqrt{2}}}^{\frac{1}{\sqrt{2}}}$ $\int_{-\sqrt{\frac{1}{2}-x^2}}^{\sqrt{\frac{1}{2}-x^2}}$ $\int_{\sqrt{x^2+y^2}}^{\sqrt{1-x^2-y^2}}dzdydx$

Now, do it in cylindrical coordinates.

$\int_{0}^{2{\pi}}$ $\int_{0}^{\frac{1}{\sqrt{2}}}$ $\int_{r}^{\sqrt{1-r^{2}}}rdzdrd{\theta}$

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.

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