Find the centroid of the solid bounded below by the cone

[wilcofan3]

Homework Statement

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

Homework Equations

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

$$\frac{\bar{x}= \int \int \int_{G} x dV}{V_{G}}$$,$$\frac{\bar{y}= \int \int \int_{G} y dV}{V_{G}}$$,$$\frac{\bar{z}= \int \int \int_{G} z dV}{V_{G}}$$

The Attempt at a Solution

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

$$x = r sin \phi cos \theta$$
$$y = r sin \phi sin \theta$$
$$z = r cos \phi$$

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

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

 

[tiny-tim]

Hi wilcofan3!

(try using the X² tag just above the Reply box )

Find the centroid of the solid bounded below by the cone $$z = \sqrt{3(x^2+y^2)}$$ and bounded above the sphere $$x^2+y^2+z^2=36$$.
…
I know I need to use spherical coordinates making the solid G given by:

$$x = r sin \phi cos \theta$$
$$y = r sin \phi sin \theta$$
$$z = r cos \phi$$

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

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

(erm … 3π/4 is 135º)

Now multiply by the volume element (ie the r θ φ equivalent of dxdydz), and integrate.