# 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.

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.

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

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

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

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