Triple integration for the volume of a given sphere

Question

I have a problem which I've had a look on "Maths Stack Exchange" and other resources to help, but still am stuck, so any help would be most appreciated.

My Problem:

• Set up a triple integral for the volume of the sphere $S_{R}$, where $$S_{R} = {(,,)∈ℝ^3|^{2}+^{2}+^{2}=^{2}}$$, with $R>0$ is the radius of the sphere.

Help:

• I don't understand what type of coordinates I'm meant to be using for this, as I chose to use spherical coordinates, but others seem to use cylindrical coordinates.

• I know that the triple integral will look eventually like $$\int\int\int_{S_{R}}dV$$is this correct, where my limits will be (using spherical coordinates): $${\int^{r}}{\int^{\theta}}{\int^{\phi}}dr\,d\theta\,d\phi$$

• Also I don't understand how to compute my limits, as I'm only given the radius $R>0$, and haven't been given any range for the $x,y,z$ values. As I'm just meant to set up the triple integral, will this mean I'm not meant to be computing the limits?

Any help with this would be really great, and thank you in advance for any help and further understanding of my problem.

Answer 1

This is by far easiest in spherical coordinates, as long as you use the correct volume element of $dr\,\, r \sin \theta \, dr\,\, r\,d\phi = r^2dr\,d\theta\,d\phi$ rather than simply writing $dr\,d\theta\,d\phi$.

$$ \int_{r=0}^R\int_{\theta = 0}^{\pi} \int_{\phi=0}^\pi r^2\sin\theta\,dr\,d\theta\,d\phi = \\ 2\pi\int_{r=0}^R\int_{\theta = 0}^{\pi} r^2\sin\theta\,dr\,d\theta = \\ 2\pi\int_{r=0}^R \left[ -r^2\cos\theta \right]_{\theta = 0}^\pi dr = -2\pi\int_{r=0}^R \left[ \cos\pi - \cos 0\right]\,dr \\ = 4\pi \int_{r=0}^R r^2dr = \frac43\pi R^3 $$

It is a bit tougher in cylindrical coordinates, with volume element $r\,dr\,d\phi\,dz$, because the limits on $z$ are less simple: $$ \int_{r=0}^R\int_{\phi = 0}^{2\pi} \int_{z=-\sqrt{R^2-r^2}}^{\sqrt{R^2-r^2}} r\,dr\,d\phi\,dz = \\ 2\int_{r=0}^R\int_{\phi = 0}^{2\pi} r\sqrt{R^2-r^2} \,dr\,d\phi =\\ 4\pi\int_{r=0}^Rr\sqrt{R^2-r^2} \,dr\,d\phi =\\ 4\pi \left[-\frac13(R^2-r^2)^{3/2} \right]_0^R \\ =\frac43 \pi R^3 $$ It is toughest of all in Cartesian coordinates, but still can be done without truly scary integrals.

Answer 2

Unsurprisingly, spherical coordinates are the natural choice here because the region we are integrating over is a sphere.

• $\rho$, the distance from a point on the sphere to the origin, ranges from $0$ to $R$, the radius of the sphere
• $\phi$, the angle downward from the positive $z$-axis, ranges from $0$ to $\pi$ (representing the negative $z$ axis)
• $\theta$, the angle in the $xy$-plane, ranges from $0$ to $2\pi$

This gives $$V=\int_{0}^{2\pi}\int_{0}^{\pi}\int_{0}^{R}\rho^{2}\sin{\phi}\ d\rho \ d\phi \ d\theta$$

Note that by symmetry, the volume of the sphere is $8$ times the volume in any octant, for example the first octant, so this is also $$V=8\int_{0}^{\pi/2}\int_{0}^{\pi/2}\int_{0}^{R}\rho^{2}\sin{\phi}\ d\rho \ d\phi \ d\theta$$

If you want to use cylindrical coordinates, observe that the equation $x^{2}+y^{2}+z^{2}=R^{2}$ is equivalent to $z^{2}=R^{2}-r^{2}$, where $r=\sqrt{x^{2}+y^{2}}$ is the cylindrical coordinate. This gives $$V=\int_{0}^{2\pi}\int_{0}^{R}\int_{-\sqrt{R^{2}-r^{2}}}^{\sqrt{R^{2}-r^{2}}}r\ dz\ dr\ d\theta$$