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I understand triple integrals in cartesian and 2D/3D cylindrical polar coordinates (at least I think I do) because I can visualize their coordinate grids. Take the usual $x-y$ coordinate grid, stack them, and you have the 3D cartesian grid. In 3D cylindrical polar, you do the same thing since $z=z$. So just stack up a bunch of circles of constant $r$ that intersect with lines of constant $\theta$.

This makes understanding triple integrals in these coordinates easier since you project the shadow of the object on one of the planes, figure out the bounds in that 2D domain like you do in double integrals, and then figure out the bounds in the vertical dimension. How do you figure out the bounds in spherical coordinates. Onto what domain do you project your 3D object? So if $dV = dzdydx$, you project your object onto the $x-y$ plane, figure out the bounds in that 2D domain, then figure out the bounds in the $z$. If $dV = \rho ^2 \sin\phi d\rho d\theta d\phi$, your domain would be the "$\theta - \phi$" plane. But what plane exactly is this?

So for a sphere centered at the origin of radius $R$, the bounds corresponding to $d\rho d\theta d\phi$ from outside in are $0-\pi$, $0-2\pi$, and $0 - R$. What are the bounds if you shift that sphere up so that the sphere is centered at $z = R/2$?

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There is no plane to project onto. The closest thing to that (in the sense of thinking of a portion of $3$-dimensional space as being made of $2$-dimensional sheets stacked up) is to think of $(\theta, \phi)$ coordinate patches for a given radius $\rho$.

These patches provide a "grid" on the surface of a sphere of radius $\rho$, and you stack them up in the radial direction. I put scare quotes around "grid" because the lines are not actually straight; they're longitude and co-latitude, respectively.

The advantages of using spherical coordinates are all lost if you have some region that is spherically symmetric about a different center. (This is an unsatisfying answer to your last question.)

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    $\begingroup$ I will come back and post a graphic illustrating this coordinate patch when I have some time. $\endgroup$ – Sammy Black Nov 19 '14 at 23:28
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I prefer to think of it as slicing the region with the half-plane where $\theta$ is fixed to be constant, and then working in a $zr$-plane.


The new sphere is given by: \begin{align*} x^2 + y^2 + (z - \tfrac{R}{2})^2 &= (\tfrac{R}{2})^2 \\ r^2 + (z^2 - Rz + (\tfrac{R}{2})^2) &= (\tfrac{R}{2})^2 \\ r^2 + z^2 &= Rz \\ \rho^2 &= R\rho\cos\phi \\ \rho &= R\cos\phi \end{align*} It's easy to see that $\theta$ ranges from $0$ to $2\pi$. Slicing the region at some fixed $\theta$ and viewing our cross-section in the $rz$-plane, our slice is bounded by $r = 0$ and the right-half of the circle $r^2 + (z - \tfrac{R}{2})^2 = (\tfrac{R}{2})^2 \iff \rho = R\cos\phi$. It's easy to see that $\phi$ ranges from $0$ to $\pi/2$. Hence, the volume of the desired region is: $$ V = \int_0^{2\pi} \int_0^{\pi/2} \int_0^{R\cos\phi} \rho^2 \sin\phi \, d\rho \, d\phi \, d\theta $$


EDIT: I misread the radius to be $R/2$ instead of $R$. Oh well. Hopefully this is still helpful.

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The best way to approach it would be to move your coordinate system to maintain the spherical symmetry. If your sphere isn't at the origin, it doesn't make a whole lot of sense to use spherical coordinates centered on the origin.

So, make the change of coordinates $\rho' = \rho + R/2$, and integrate over $\rho', \theta, \phi.$

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