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## Homework Statement

Find [itex] VR_{z}^{2} = \int \!\!\! \int \!\!\! \int_{E} (x^{2} + y^{2})dV[/itex] given a constant density lying above upper half of [itex]x^{2}+y^{2} = 3z^{2}[/itex] and below [itex]x^{2}+y^{2}+z^{2} = 4z[/itex].

## Homework Equations

## The Attempt at a Solution

Why does it say upper half of [itex]x^{2}+y^{2} = 3z^{2}[/itex]? It's not like there are two possible halves.

Here's what I've done so far but I'm a little unsure about some theory.

For [itex]x^{2}+y^{2} = 3z^{2}[/itex]

[itex] \tan^{2}\phi = 3[/itex] which means [itex]\phi=\pm\frac{\pi}{3}[/itex]

I've seen most questions don't actually have a - value for [itex]\phi[/itex] because it only works in a range of [itex] 0 [/itex] and [itex]\pi[/itex] right? The movement itself is taken care of by [itex]\theta[/itex]? So I can just get rid off that negative [itex] \frac{\pi}{3}[/itex]?

For [itex]x^{2}+y^{2}+z^{2} = 4z[/itex]:

[itex]\rho=4cos\phi[/itex] and then usually questions will say [itex]0\leq\rho\leq4\cos\phi[/itex] but I don't really know how we get it ranges from 0... ?

Assuming I did everything right I'll then get:

[itex]E=\{(r,\theta,\phi):0\leq\theta\leq2\pi,\, 0\leq\phi\leq\frac{\pi}{3},\, 0\leq \rho \leq4 \cos\phi\}[/itex]

Which I then plug into an integral like this:

[itex] \int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4\cos\phi} (x^{2} + y^{2}) \,\rho^{2}\sin\phi d\rho d\phi d\theta[/itex]