Volume enclosed by $(x^2+y^2+z^2)^2=x$

Question

I need to calculate the volume of solid enclosed by the surface $(x^2+y^2+z^2)^2=x$, using only spherical coordinates.

My attempt: by changing coordinates to spherical: $x=r\sin\phi\cos\theta~,~y=r\sin\phi\sin\theta~,~z=r\cos\phi$ we obtain the Jacobian $J=r^2\sin\phi$. When $\phi$ and $\theta$ are fixed, $r$ varies from $0$ to $\sqrt[3]{\sin\phi\cos\theta}$ (because $r^4=r\sin\phi\cos\theta$). Keeping $\theta$ fixed, we let $\phi$ vary from $0$ to $\pi$. Thus the volume equals:

$$V=\int\limits_{0}^{\pi}\int\limits_{0}^{\pi}\int\limits_{0}^{\sqrt[3]{\sin\phi\cos\theta}}r^2\sin\phi ~dr ~d\phi ~d\theta=0$$ Which is obviously wrong. What am I doing wrong?

Answer 1

Succinctly, the bounds of integration on $\theta$ should be $-\pi/2\leq\theta\leq\pi/2$, which yields the very believable value of $\pi/3$. The comments seem to indicate some confusion about this, so let's explore it a bit further.

A standard exercise related to spherical coordinates is to show that the graph of $\rho^2=x$ is a sphere. Now we've got $\rho^4=x$ so it makes sense to suppose that we have a slightly distorted sphere - perhaps, an oblate spheroid or close. In both cases, we have $$x=\text{a non-negative expression}.$$ Thus, the graph must lie fully in the half-space on one side of the $yz$-plane. Using some snazzy contour plotter, we see that it looks like so:

The contour line wrapped around the equator is the intersection of the object with the $xy$-plane. If we look at this from the above (along the $z$-axis) we can see how a polar arrow might trace out that contour:

From this image, it's pretty clear that we need $\theta$ to sweep from the negative $y$-axis to the positive $y$-axis. A natural way to make that happen is let $\theta$ range over $-\pi/2\leq\theta\leq\pi/2$.

To polish the problem off, we rewrite the equation $(x^2+y^2+z^2)^2=x$ in spherical coordinates as $$\rho^4 = \rho\sin(\varphi)\cos(\theta),$$ or $$\rho = \sqrt[3]{\sin(\varphi)\cos(\theta)}.$$ This gives us the upper bound of $\rho$ in the spherical integral:

$$ \int\limits_{-\pi/2}^{\pi/2}\int\limits_{0}^{\pi}\int\limits_{0}^{\sqrt[3]{\sin\varphi\cos\theta}}r^2\sin\varphi ~dr ~d\varphi ~d\theta=\frac{\pi}{3}. $$

Answer 2

We may as well consider the surface $(x^2+y^2+z^2)=z$ which is a rotational surface $S$ around the $z$-axis. Put $x^2+y^2=:\rho^2$. Then $S$ is given by the equation $$(\rho^2+z^2)^2=z\geq0\ .$$ It follows that $$\rho^2(z)=\sqrt{z}-z^2\qquad(0\leq z\leq1)\ ,$$ so that $$V=\pi\int_0^1\rho^2(z)\>dz=\pi\>\left({2\over3}z^{3/2}-{1\over3}z^3\right)\biggr|_0^1={\pi\over3}\ .$$