# Volume of intersection of a sphere and a paraboloid

Could I calculate the volume of the intersection of $$x^2+y^2+z^2=8$$ and $$4z=x^2+y^2+4$$ using spherical coordinates and treating the paraboloid as a function, as in the following example?

$$f(x,y)=\frac{1}{4}(x^2+y^2)+1=\frac{1}{4}(r^2\sin^2(\theta)\cos^2(\varphi))+r^2\sin^2(\theta)\sin^2(\varphi))+1=\frac{1}{4}(r^4\sin^2(\theta)) + 1$$

Volume = $$\int_0^{2\pi} \int_0^\pi \int_0^\sqrt{8} (\frac{1}{4}(r^4\sin^2(\theta))+1)\cdot r^2\sin(\theta) drd\theta d\varphi$$

Should that work? I tried it and didn't get the correct result but then I'm sick and might have seen my error.

Edit: I think I am mixing something up here. Basically I just add up the values of $$f$$ but still integrate over a spherical volume. So the idea to restrict the integrated volume with a function doesn't work - at least not like that. Right?

• Are you asked specifically to use spherical coordinates? I think cylindrical are better here. – Alekos Robotis Jan 7 '19 at 14:58
• How are you getting $r^4\sin^2{\theta}$? I think it should be $r^2\sin^2{\theta}$. – Calvin Godfrey Jan 7 '19 at 15:11
• you are right @CalvinGodfrey but the whole idea is wrong anyway and yeah, with cylindrical coordinates where the parabolioid basically is the boundaries of $\int dr$ is simpler. It was more about my idea, but that's just rubbish. – xotix Jan 8 '19 at 21:45

## 1 Answer

The volume of a set is the sum of the volume elements into that set! Simply $$\int_\Omega\mathbb dV$$. The problem is then determine the integration limits for the set $$\Omega$$. As suggested in the comments to your question, it seems easier set the integral in cylindrical coordinates. The limiting surfaces are $$r^2+z^2=8$$ for the sphere and $$4z=r^2+4$$ fo the paraboloid. The volume element is $$r\,\mathbb dz\,\mathbb dr\,\mathbb d\theta$$

You are almost in the right path when you consider the surfaces as functions. We have to do so, but for them to determine the integration limits.

We need first the intersection of these surfaces, obtained solving the system of equations formed with the equations for the surfaces

$$\begin{matrix} r^2+z^2=8\\ 4z=r^2+4 \end{matrix}$$

$$z^2+4z-12=0$$ with solutions $$z=-6,2$$. For $$z=-6$$ we have $$r^2=-28$$ with no solutions. For $$z=2$$ we have $$r^2=4$$ giving $$r=\pm2$$. And dropping $$-2$$ the intersection is a circle with $$r=2$$ into the plane $$z=2$$

Then, $$r$$ runs from $$0$$ to $$2$$, $$z$$ from $$r^2/4+1$$ to $$\sqrt{8-r^2}$$ (positive root because we are over the plane $$z=2$$) and $$\theta$$ runs from $$0$$ to $$2\pi$$.

$$V=\int_0^{2\pi}\int_0^2\int_{r^2/4+1}^{\sqrt{8-r^2}}r\,\mathbb dz\,\mathbb dr\,\mathbb d\theta$$