Surface area of a sphere limits

Question

If I am finding the surface area of a sphere in spherical coordinates my intergral would be like this:

$$\int^{\pi}_0 \int^{2\pi}_0 R^2 \sin (\theta) d\phi d\theta =4\pi R^2$$ But if I do the following:

$$\int^{\pi}_0 \int^{2\pi}_0 R^2 \sin (\theta) d\theta d\phi =0$$

What makes the answers different? In both cases I am integrating over the same total area! What does the last integral represent?

Answer 1

I think the confusion is in your integration limits, each integral symbol with its limits corresponds to one and only one variable! Geometrically what you want is: $$ \int^{2\pi}_0\left( \int^{\pi}_0 R^2 \sin (\theta) d\theta\right) d\phi =2R^2\int^{2\pi}_0d\phi = 4\pi R^2 $$ The problem arises when you integrate $\sin\theta$ from $0$ to $2\pi$, but in spherical coordinates $\theta\in[0,\pi]$.

Answer 2

$$\int^{\pi}_0 \int^{2\pi}_0 R^2 \sin (\theta) d\phi d\theta =4\pi R^2$$ is meaningless if you don't specify on which variable you integrate.

I would write : $$\int^{\pi}_{\theta = 0} \int^{2\pi}_{\phi=0} R^2 \sin (\theta) d\phi d\theta =4\pi R^2$$

If you switch the order of $d\theta$ and $d\phi$, it doesn't change the fact that it is $\theta$ which is integrated from $0$ to $\pi$ and $\phi$ from $0$ to $2\pi$.

Answer 3

Think about where the $\sin(\theta)$ came from: instead of integrating over the surface area of the sphere, you are integrating over a rectangle in the $\theta-\phi$ plane. The $\sin(\theta)$ term, the determinant of the Jacobian of the map fro the rectangle to the sphere, rescales the infinitesimal area element $d\theta d\phi$ so that you get infinitesimal surface area on the sphere (near the north and south pole, the infinitesimal square $d\theta d\phi$ gets "squished" onto the sphere into a region of much smaller surface area than near the equator.)

When $\theta > \pi$, your second integral asserts that $d\theta d\phi$ gets mapped into a patch on the sphere with negative surface area -- this makes no sense. The correct Jacobian determinant is $|\sin\theta|$ (geometrically, the size of the patch is the same on the front as on the back of the sphere) and you can verify that $$\int_0^\pi \int_0^{2\pi} R^2 |\sin{\theta}| d\theta d\phi = 4\pi R^2.$$