# Triple integral - volume of sphere

I'm trying to work out the volume of the sphere given by $x^2+y^2+z^2=x$ which is of course obvious but in trying to do it by calculating the integral $$\int \int \int _{\Omega } dV .$$ Now the problem is in struggling to the limits of the integrals - where of course $\Omega$ is the domain of the triple integral.

I'm trying to do this using spherical coordinates but I don't think I'm getting the limits right. Can anyone tell me how to correctly get the limits for the integrals.

## 4 Answers

Let's do this in spherical coordinates. For spherical coordinates, I'll use triples $(r,\theta,\phi)$ where $r$is length, $\theta$ is the angle a vector's projection to the $xy$-plane makes with the $x$-axis, and $\phi$ is a vector's angle from the $z$-axis.

Now, I don't like balls that aren't centered at $(0,0,0)$, so let's translate the ball so it's centered at the origin. The trick here is completing the square: $$x^2 - x + y^2 + z^2 = 0$$ $$x^2 - x + \frac{1}{4} + y^2 + z^2 = \frac{1}{4}$$ $$\bigg(x - \frac{1}{2}\bigg)^2 + y^2 + z^2 = \frac{1}{4}$$ so the ball has radius $\frac{1}{2}$ and is centered at $(\frac{1}{2}, 0, 0)$. Translation doesn't change volume, so the volume is the same as the volume of a ball of radius $\frac{1}{2}$ centered at $(0,0,0)$.

Now, to sweep out a whole sphere of fixed radius $r$ exactly once, we'll have $\theta$ ranging from $0$ to $2\pi$, $\phi$ ranging from $0$ to $\pi$; we'll integrate these spheres from $0$ to $\frac{1}{2}$:

$$\int_0^\frac{1}{2} \int_0^{2\pi} \int_0^\pi r^2\sin\phi\ d\phi d\theta dr$$

Bonus: $$\cdots = \int_0^\frac{1}{2} r^2\ dr \int_0^{2\pi}\ d\theta \int_0^\pi \sin\phi\ d\phi$$

• Your limits don't match the equation of the sphere in the question. – Doug M Feb 28 '17 at 19:58
• @DougM Thanks, good catch. – Neal Feb 28 '17 at 20:00

$$\iiint_{\left(x-\frac{1}{2}\right)^2+y^2+z^2=\frac{1}{4}}1\,d\mu =\frac{1}{8}\iiint_{u^2+v^2+w^2=1}1\,d\mu$$ by the change of variables $x=\frac{u+1}{2},y=\frac{v}{2},z=\frac{w}{2}$. Now you may switch to spherical coordinates or just perform integration along shells to find $V=\frac{1}{8}\cdot\frac{4\pi}{3}=\color{red}{\frac{\pi}{6}}$.

By definition, the r-limit runs from 0 to the radius (which can be easily determined), the $\theta$-limit runs from 0 to $2\pi$, and the $\phi$-limit runs from 0 to $\pi.$

So since the sphere is $(x-\frac{1}{2})^2+y^2+z^2=\frac{1}{4}$, the radius is $\frac{1}{2}.$

So we have

$\int_{0}^{\frac{1}{2}}\int_0^{2\pi}\int_0^{\pi}r^2\sin\phi d\phi d\theta dr=\boxed{\frac{\pi}{6}}.$

$x^2 +y^2 + z^2 = x\\ (x - \frac 12) +y^2 + z^2 = \frac 14$

Rather that traditional spherical coordinates, might I suggest.

$x = \rho \cos \theta \cos \sin\phi + \frac 12\\ y = \rho \sin \theta \cos \sin\phi\\ z = \rho \cos\phi$

And then your integral is

$\int_0^{2\pi}\int_0^{\pi}\int_0^{\frac 12} \rho^2\sin\phi \ d\rho \ d\phi \ d\theta$

If you insist on traditional spherical coordinates.

Limits:

$x^2 + y^2 + z^2 = x\\ \rho^2 = \rho \cos\theta \sin\phi\\ \rho(\rho - \cos\theta \sin\phi) = 0$

This establishes our limits for $\rho$

$\phi$ is able to range from $0$ to $\pi$

The entire sphere is to the right of the $yz$ plane

$\int_{-\frac{\pi}{2}}^{\frac {\pi}{2}}\int_{0}^{\pi}\int_0^{\cos\theta\sin\phi} \rho^2\sin\phi \ d\rho \ d\phi \ d\theta$