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!

  • $\begingroup$ Are you sure that $x^2+y^2=2$ is really a sphere in $R^3$ ? $\endgroup$ – Emilio Novati Jul 30 '15 at 19:51
  • $\begingroup$ $x^2+y^2=2$ is not the equation of a sphere, but of a cylinder. $\endgroup$ – Aretino Jul 30 '15 at 19:51
  • $\begingroup$ @Aretino I'm sorry, I made a mistake in writing the problem out. I've edited it to fix the error. $\endgroup$ – user3472798 Jul 30 '15 at 20:03
  • $\begingroup$ If you draw a sketch of the surfaces you'll see that $\phi$ can take any value in $[0,\pi]$, whereas $\rho$ runs from $0$ to a certain upper bound $\rho_\max$ which depends on $\phi$. You just need to find $\rho_\max$ as a function of $\theta$. $\endgroup$ – Aretino Jul 30 '15 at 20:16

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 $


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

$\int_{0}^{2\pi } \int_{0}^{1} \int^{\sqrt {{2-r^{2}}}} _{r^{2}}rdzdrd\theta =\frac{\pi}{6}\left(8\sqrt{2}-7\right)$

  • $\begingroup$ The intersection of the sphere and the paraboloid is at $z=1$ , so it's not $0<r<1$? $\endgroup$ – Emilio Novati Jul 30 '15 at 20:32
  • $\begingroup$ @EmilioNovati: yes of course. thanks. I wll fix it. $\endgroup$ – Matematleta Jul 30 '15 at 21:53

Just use Cavalieri's principle.

You want to compute the volume of the region that is generated by the rotation of $$ A=\left\{(x,y)\in\mathbb{R}^2: x\in[0,1],\,x^2\leq y\leq\sqrt{2-x^2}\right\} $$ around the $y$-axis, hence: $$ V=\pi\int_{0}^{1}y\,dy + \pi\int_{1}^{\sqrt{2}}(2-y^2)\,dy = \color{red}{\frac{\pi}{6}\left(8\sqrt{2}-7\right)}.$$


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