Let $E$ be the smaller of the two solid regions bounded by the surfaces $z=x^2+y^2$ and $x^2+y^2+z^2=6$. Evaluate $\iiint(x^2+y^2)dV$.

Question

I tried solving and came up with something like the integral in the image given $$\int_{\theta=0}^{2\pi} \int_{r=0}^{\sqrt{2}} \int_{z=r^2}^{\sqrt{6-r^2}} r^3 \, dz \, dr \, d\theta$$ But I cant find a provided answer anywhere, and wanted to ask about the approach and the answer of the question above

My approach

Answer 1

Your setup using cylindrical coordinates is fine. The integrals w.r.t. $z$ and $\theta$ are trivial. The last integral w.r.t. $r$ can be done by parts,

$$\int_0^{\sqrt2} r^3 \left(\sqrt{6-r^2} - r^2\right) \, dr = uv \bigg|_{r=0}^{\sqrt2} - \int_{r=0}^{\sqrt2} v \, du$$

where

$$u = r^2 \implies du = 2r\,dr \\ dv = r \sqrt{6-r^2} \, dr \implies v = -\frac13 \left(6-r^2\right)^{3/2}$$

and the remaining integral $\int v \, du$ can be done in the same way we found $v$ above.

---

In spherical coordinates $(x,y,z)=(\rho\cos\theta\sin\varphi,\rho\sin\theta\sin\varphi,\rho\cos\varphi)$, the givens surfaces' equations would be

$$z = x^2 + y^2 \implies \rho \cos\varphi = \rho^2 \sin^2\varphi \implies \rho = \cot\varphi\csc\varphi\\ x^2 + y^2 + z^2 = 6 \implies \rho = \sqrt6$$

from which we find the polar angle $\varphi$ where the sphere and paraboloid intersect:

$$\cot\varphi \csc\varphi = \sqrt6 \implies \cos\varphi = \sqrt6 \, (1-\cos^2\varphi) \implies \cos\varphi = \sqrt{\frac23}$$

Now, we have to split up the integration region along the cone $\varphi = \cos^{-1}\sqrt{\frac23}$ because $\rho$ is not bounded above by the same surface to either "side" of this cone.

Edit: Elaborating on the above point, it's easier to see why the split must be done if we consider a cross-section of $E$ taken along the $z$-axis. For instance, fixing $y=0$, the slice looks like the figure below. (The dashed line is $z=\sqrt2\,x$.)

As $\varphi$ sweeps from the sphere's north pole (straight up along the $z$-axis in the figure) to the dashed line, the distance from the origin to the circle is exactly $\sqrt6$. Past this line, we must use the distance from the origin to the parabola, $\cot\varphi\csc\varphi$.

You should find that

$$\left\{\int_{\theta=0}^{2\pi} \int_{\varphi=0}^{\color{red}{\cos^{-1}\sqrt{\tfrac23}}} \int_{\rho=0}^{\color{blue}{\sqrt6}} + \int_{\theta=0}^{2\pi} \int_{\varphi=\color{red}{\cos^{-1}\sqrt{\tfrac23}}}^{\tfrac\pi2} \int_{\rho=0}^{\color{blue}{\cot\varphi\csc\varphi}}\right\} \rho^4 \sin^3\varphi \, d\rho \, d\varphi \, d\theta$$

gives the same result as the much, much simpler cylindrical setup.