How can I get the surface of a sphere with triple integrals?

Question

So, I want to get the surface area of a sphere, $S$, using a triple integral (yeah, I know we usually associate triple integrals to volumes but I will continue). Since we are dealing with the surface alone, we should sum (integrate over) all points in 3D space that are located at a constant distance $r= R$ from the centre.

$$S=\iiint_{r=R}dxdydz$$

If I turn this to spherical coordinates, the transformation implies a determinant for the Jacobian like $|J| = r^2\sin\theta$. So my triple integral turns into

$$S = \int_0^{2\pi}\int_{0}^{\pi}\int r^2\sin\theta \, dr d\theta d\varphi = \int_0^{2\pi}d\varphi\int_{0}^{\pi}\sin\theta d\theta \int r^2dr = 4\pi\int r^2dr$$

My question is what are the bounds of this final integral?

On one hand, since $r=R$ should stay constant, that means that $dr = 0$, so $S = 0$, which is clearly wrong. I could also say that the limits are $R$ and $R$, but then the integral will also be zero. I can also immagine the interval as a very thin sliver like so:

$$\int_{R}^{R+dR} r^2 dr = \frac{1}{3}[(R+dr)^3-R^3] = \frac{1}{3} (R^3+3R^2dr+3Rdr^2+dr^3-R^3)=R^2dr$$

Which is very weird again.

I know this is because some basic misunderstanding. But I would like to know exactly why I'm wrong and how should I coherently do it. Perhaps some Dirac delta is involved somewhere? Perhaps Green's Theorem has something to do with converting a triple integral into a double integral in an enclosed surface? How?

Answer 1

We might pick the bounds $R$ and $R+\Delta R$. That is:

$$4\pi \int_R^{R+\Delta R} r^2\,dr = \frac{4\pi}{3}((R+\Delta R)^3 - R^3)\approx 4\pi R^2\Delta R$$ That is the volume of a spherical shell with thickness $\Delta R$.

If we divide by the thickness and take the limit for $\Delta R \to 0$, we get the usual result $4\pi R^2$ for the surface area of the sphere.

Answer 2

We can apply the Gauss divergence theorem, which is a special case of Stokes' theorem.

The surface integral for $S$ is: $$S=\iint \hat r\cdot dS =\iiint\nabla \cdot \hat r\,dV$$ where $r$ is a vector to a point in the sphere and $\hat r$ is a unit vector in the same direction.

Then we have $$\nabla\cdot \hat r = \frac\partial{\partial x}\Big(\frac x{\sqrt{x^2+y^2+z^2}}\Big)+\ldots =\Big(\frac 1r -\frac{x^2}{r^3}\Big)+\ldots =\frac 3r-\frac{x^2+y^2+z^2}{r^3}=\frac 2 r$$

It follows that: $$S=\iint \hat r\cdot dS =\iiint \nabla\cdot \hat r\,dV=\iiint\frac 2r\cdot r^2\sin\theta\,dr\,d\theta\,d\phi = 4\pi R^2$$