I am working on a problem that requires me to find the volume of the solid bounded by the sphere $x^2 + y^2 + z^2 = 2$ and the paraboloid $x^2 + y^2 = z$. I know that to do this, I must use triple integration.
Thus far, I know that we need to reevaluate the given equations and put them into spherical coordinates in order to achieve our bounds. When we've gotten our bounds, it's just simple triple integration to computer the volume.
So, for our sphere, I said that $x^2 + y^2 + z^2 = 2$ is equivalent to $\rho = sqrt(2)$.
Then, for our sphere, I get a little messed up. I'm trying to find $\phi$ boundaries here, but I get the equation $\rho^2sin^2\phi = \rho cos\phi$ Then, the $\rho$ can cancel out, making it $\rho sin^2\phi = cos\phi$, but my issue is that the $\rho$ doesn't cancel out completely so I can't get the bounds for $\phi$.
I'm very new to doing this type of thing, so any help would be greatly appreciated. Anything I'm missing or didn't know about? Thank you very much!
Best Answer
You want the volume above the paraboloid below the sphere, so first observe that $z$ goes from the paraboloid to the sphere. Let's use cylindrical coordinates:
$x=r\cos \theta $
$y=r\sin \theta $
$z=z$
Then $dV=rdzdrd\theta $ and we have so far
$\int \int \int^{\sqrt {{2-r^{2}}}} _{r^{2}}rdzdrd\theta $.
To find the limits on $r$ and $\theta $ project the curve of intersection of the given surfaces, onto the $x-y$ plane. It is immediate that the region is a circle of radius $1$ so we see that $r$ goes from $0$ to $1$ and $\theta $ goes from $0$ to $2\pi $. Thus, we have
$\int_{0}^{2\pi } \int_{0}^{1} \int^{\sqrt {{2-r^{2}}}} _{r^{2}}rdzdrd\theta =\frac{\pi}{6}\left(8\sqrt{2}-7\right)$