Volume of sphere with triple integral Using the same notations as in this picture :

The element of volume is: $r^2 \sin(\theta) \, dr \, d\theta \, d\phi$
If I try to create the volume visually, I begin with integrating $r$ between $0$ and $R$ to get the radius.
Now I can either:


*

*integrate $\theta$ between $0$ and $\pi$ to have the area of half a circle, and then integrate $\phi$ between $0$ and $2\pi$ to have the full sphere volume

*integrate $\theta$ between $0$ and $2\pi$ to have the area of a circle, and then integrate $\phi$ between $0$ and $\pi$ to have the full sphere volume
Of course the second method won't work because $$\int_0^{2\pi} \sin(\theta)d\theta = 0.$$
But I don't understand why it works visually but it fails when I write it down. I'm not sure there is something to understand but it still disturbs me... 
 A: To your question about choosing the bounds of integration, it is customary to use $\theta \in [0,\pi]$. It is done because of the issue with integrating the $\sin (\theta)$ as you mentioned. What you are dealing with is the 3D equivalent of integrating a sine-function. If you evaluate:
$$\int_0^{2\pi} \sin(\theta)d\theta$$
You obtain:
$$\left[ \cos(\theta) \right ]_0^{2\pi}=[1-1]=0$$
This is because the integral evaluates the negative regions of the sinusoid as negative area. See the pink regions of this image?

However, in the case of the sphere, the entire volume enclosed is positive volume. So you want to count the negative values of the sinusoid as positive, giving the total enclosed area. Thus, you are actually integrating:
$$\int_0^{2\pi} |\sin(\theta)|d\theta$$
This function looks like:

So to your point about the sinusoidal integral, you would assume symmetry and make the bounds different ala this:
$$\int_0^{2\pi} |\sin(\theta)|d\theta=2\int_0^{\pi} \sin(\theta)d\theta$$
If you apply this, you obtain the correct volume integration.
