# Multiple Integrals (Volume + Centroid)

• Nov 19th 2007, 04:11 PM
numberonenacho
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. =)
• Nov 19th 2007, 04:29 PM
ThePerfectHacker
I start you off. The upper surface $\displaystyle x^2+y^2+z^2=6^2$ and the lower surface $\displaystyle z=x^2+y^2$. When they intersect the region over we are integrating is the circle $\displaystyle x^2+y^2=6^2$. Now use polar change of variable.
• Nov 20th 2007, 04:31 PM
numberonenacho
Quote:

Originally Posted by ThePerfectHacker
I start you off. The upper surface $\displaystyle x^2+y^2+z^2=6^2$ and the lower surface $\displaystyle z=x^2+y^2$. When they intersect the region over we are integrating is the circle $\displaystyle x^2+y^2=6^2$. 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.
• Nov 20th 2007, 04:44 PM
ThePerfectHacker
Quote:

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 $\displaystyle z=\sqrt{36 - x^2-y^2}$.
Thus, the volume is,
$\displaystyle \int_A \sqrt{36-x^2-y^2} - (x^2+y^2) \ dy~dx$ where $\displaystyle A$ is the disk of radius $\displaystyle 6$.
Using polar change of variable,
$\displaystyle \int_0^{2\pi} \int_0^6 (\sqrt{36-r^2} - r^2)rdr~d\theta$(Wink)
• Nov 24th 2007, 09:17 AM
numberonenacho
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?
• Nov 24th 2007, 09:49 AM
galactus
Is that $\displaystyle {\rho}=\sqrt{6}$?.

If so, try:

$\displaystyle \int_{0}^{2\pi}\int_{0}^{\sqrt{2}}\int_{r^{2}}^{\s qrt{6-r^{2}}}rdzdrd{\theta}$
• Nov 24th 2007, 10:00 AM
numberonenacho
hmm okay. So when I evaluate it

http://img75.imageshack.us/img75/9914/37828357zt8.gif

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