Surface area of a sphere limits 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?

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 
$$\int_0^\pi \int_0^{2\pi} R^2 |\sin{\theta}| d\theta d\phi = 4\pi R^2.$$
