# Thread: Multiple Integrals (Volume + Centroid)

1. ## Multiple Integrals (Volume + Centroid)

A) Find the volume bounded by sphere rho=rt(6) and the paraboloid
z=x^2 + y^2
B) Locate the centroid of this region

Any help, tips, work, or similar problems would be greatly appreciated. =)

2. I start you off. The upper surface $x^2+y^2+z^2=6^2$ and the lower surface $z=x^2+y^2$. When they intersect the region over we are integrating is the circle $x^2+y^2=6^2$. Now use polar change of variable.

3. Originally Posted by ThePerfectHacker
I start you off. The upper surface $x^2+y^2+z^2=6^2$ and the lower surface $z=x^2+y^2$. When they intersect the region over we are integrating is the circle $x^2+y^2=6^2$. Now use polar change of variable.

When you say polar change of variable you mean things like
x = rcos theta
y = rsin theta
z = z
right?
and then i would take the triple integral of that?
I always have trouble setting up the problem. =(
Like setting the integral ranges.

4. Originally Posted by numberonenacho
A) Find the volume bounded by sphere rho=rt(6) and the paraboloid
z=x^2 + y^2
B) Locate the centroid of this region
The upper surface (sphere) is given by $z=\sqrt{36 - x^2-y^2}$.
Thus, the volume is,
$\int_A \sqrt{36-x^2-y^2} - (x^2+y^2) \ dy~dx$ where $A$ is the disk of radius $6$.
Using polar change of variable,
$\int_0^{2\pi} \int_0^6 (\sqrt{36-r^2} - r^2)rdr~d\theta$

5. When I tried to evaluate the integral, I can do the R part of it, but theres no place for me to put the theta value. Shouldn't there be two different variables in a double integration problem like this?

6. Is that ${\rho}=\sqrt{6}$?.

If so, try:

$\int_{0}^{2\pi}\int_{0}^{\sqrt{2}}\int_{r^{2}}^{\s qrt{6-r^{2}}}rdzdrd{\theta}$

7. hmm okay. So when I evaluate it

how do I do the second part of the integral. That rdr complicates everything.