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## Homework Statement

Find the centroid of the solid bounded below by the cone [tex]z = \sqrt{3(x^2+y^2)}[/tex] and bounded above the sphere [tex]x^2+y^2+z^2=36[/tex].

## Homework Equations

Let G be the given solid and denote its volume by [tex]V_{G}=\int \int \int_{G} 1 dV.[/tex]

[tex]\frac{\bar{x}= \int \int \int_{G} x dV}{V_{G}}[/tex],[tex]\frac{\bar{y}= \int \int \int_{G} y dV}{V_{G}}[/tex],[tex]\frac{\bar{z}= \int \int \int_{G} z dV}{V_{G}}[/tex]

## The Attempt at a Solution

I know I need to use spherical coordinates making the solid G given by:

[tex]x = r sin \phi cos \theta[/tex]

[tex]y = r sin \phi sin \theta[/tex]

[tex]z = r cos \phi[/tex]

Now, for the limits of integration (the area where I struggle most with these integrals!), I think [tex]0 \leq r \leq 6[/tex] and [tex]0 \leq \theta \leq 2\pi[/tex] [tex]0 \leq \phi \leq \frac{3\pi}{4}[/tex]

This is stupid, but where do I go from here if this is all correct?