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

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 $\sqrt{\sin\phi\cos\theta}$ (because $r^4=r\sin\phi\cos\theta$). Keeping $\theta$ fixed, we let $\phi$ vary from $0$ to $\pi$. Thus the volume equals:

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

• Just use the usual rectangular coordinates. For fixed $x$ between 0 and 1, the slice through this solid is a simple two-dimensional disk. – user940 Jan 27 '16 at 16:58
• I believe the bounds on $\theta$ should be $-\pi/2<\theta<\pi/2$, yielding the very believable value of $\pi/3$. You can see this because $x$ must be non-negative. – Mark McClure May 28 '16 at 18:42

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 $$\rho = \sqrt{\sin(\varphi)\cos(\theta)}.$$ This gives us the upper bound of $\rho$ in the spherical integral:

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

• hi, can I know how you did the graph? Thank you – User 123 May 31 '16 at 7:49

It's a solid of revolution. Let $h=\sqrt{y^{2}+z^{2}}$, then $(x^{2}+h^{2})^{2}=x$.

$\therefore \; h^{2}=\sqrt{x}-x^{2} \:$ where $\, 0\leq x \leq 1$.

\begin{align*} V &= \int_{0}^{1} \pi h^{2} dx \\ &= \pi \int_{0}^{1} \left( \sqrt{x}-x^{2} \right) dx \\ &= \pi \left[ \frac{2}{3} x^{\frac{3}{2}}- \frac{1}{3} x^{3} \right]_{0}^{1} \\ &= \frac{\pi}{3} \end{align*}

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 $$V=\pi\int_0^1\rho^2(z)\>dz=\pi\>\left({2\over3}z^{3/2}-{1\over3}z^3\right)\biggr|_0^1={\pi\over3}\ .$$

Byron Schmuland's suggestion is the easiest way to find this volume,

but using spherical coordinates and symmetry to find the volume bounded by $(x^2+y^2+z^2)^2=z\;\;$

gives $\;\;\displaystyle V=\int_0^{2\pi}\int_0^{\frac{\pi}{2}}\int_0^{\sqrt{\cos\phi}}\rho^2\sin\phi\;d\rho d\phi d\theta=\frac{\pi}{3}$.