Compute the triple integral over the region $x^2+y^2+z^2 \leq 2x$, $z\leq 0$

Question

Compute the triple integral over the region $D = \{x^{2} + y^{2} + z^{2} \leq 2x,\, z\leq 0\}$: $$ \iiint_{D}\left(y^{2}z + x\right){\rm d}V $$

I'm struggling a great deal with setting up the integration bounds. From playing around, I think the best way to go about it is using spherical coordinates as opposed to cylindrical coordinates. It's easy to make a quick sketch of the $xy$ plane to deduce that we have to be below the line $y=2x$, however then we must find the angle of intersection with the line $y=2x$ with the circle $x^2+y^2=1$, and using polar coordinates, we can find that it corresponds to $\arctan2$ and then of course $\arctan2 + \pi$, and by $z \leq 0$, we ought to obtain that $\frac{\pi}{2}\leq\phi \leq \pi$. I'm unsure about the $R$ bound (and everything previously stated to be honest), but using the provided inequality, we would obtain that $R \leq2\cos\theta \sin\phi$, but by my sketch it would make more sense to have $0 \leq R \leq1$.

Thus, my suggested integral would look something like: $$\int_{\arctan{2}+\pi}^{\arctan{2}}d\theta\int_{\frac{\pi}{2}}^{\pi}d\phi\int_{0}^1(R^2 \sin{\phi})(R^3\sin^2{\theta}\sin^2{\phi}\cos{\phi+R\cos{\theta}\sin{\phi}})dR$$ which is obviously a nightmare. Any help/ hints on how to establish the bounds would be very appreciated!

Answer 1

First, let $x\to x+1$ to simplify the integral

$$I=\iiint_{D}y^2z+x \ dV= \iiint_{D’}y^2z+x+1 \ dV= \frac{2\pi}3+\iiint_{D’}y^2z\ dV $$ where $D’= \{$ $x^2+y^2+z^2 \leq 1$, $z\leq 0\}$, $\iiint_{D’}x \ dV= 0$ due to symmetry and $\iiint_{D’} dV=\frac{2\pi}3 $ is the half-sphere volume. Then, evaluate the remaining integral in spherical coordinates \begin{align} \iiint_{D’}y^2z\ dV=& \ \frac12\iiint_{D’}(x^2+y^2)z \ dV \\ =&\ \frac12\int_{\pi/2}^\pi\int_0^{2\pi}\int_0^1 \cos\phi\sin^3\phi \ r^5 \ dr d\theta d\phi = -\frac\pi{24} \end{align}

Thus, $I= \frac{2\pi}3-\frac\pi{24}=\frac{5\pi}{8}$.

Answer 2

You should recognize — complete the square — that $x^2+y^2+z^2=2x$ is a sphere of radius $1$ centered at $(1,0,0)$. To make life more convenient so that I can easily use spherical coordinates, I am going to switch the $xyz$-axes to the $zxy$-axes. Thus, the problem becomes this:

Integrate $x^2y+z$ over the region $D' = \{x^2+y^2+z^2\le 2z, y\le 0\}$. In spherical coordinates, the sphere is given by $\rho = 2\cos\phi$, $0\le\phi\le\pi/2$. (You can see this algebraically, by writing $\rho^2 = 2\rho\cos\phi$, or by basic right triangle geometry.) So we have $$\int_{-\pi}^0\int_0^{\pi/2}\int_0^{2\cos\phi} (\rho^3\sin^3\phi\cos^2\theta\sin\theta + \rho\cos\phi)\rho^2\sin\phi\,d\rho\,d\phi\,d\theta.$$ Note that $\int_{-\pi}^0 \cos^2\theta\sin\theta\,d\theta = -2/3$ and so the first term integrates to
$$-\frac23\int_0^{\pi/2}\int_0^{2\cos\phi} \rho^5\sin^4\phi\,d\rho\,d\phi = -\frac{\pi}{24}.$$ The second term gives $$\pi\int_0^{\pi/2}\int_0^{2\cos\phi} \rho^3\cos\phi\sin\phi\,d\rho\,d\phi = \pi\int_0^{\pi/2}4\cos^5\phi\sin\phi\,d\phi = \frac{2\pi}3.$$ Thus, the final answer is $\boxed{\pi\left(\frac23-\frac1{24}\right) = \frac{5\pi}8}$.