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. =)

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- Nov 19th 2007, 04:11 PMnumberonenachoMultiple 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 PMThePerfectHacker
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 PMnumberonenacho
- Nov 20th 2007, 04:44 PMThePerfectHacker
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 AMnumberonenacho
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 AMgalactus
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 AMnumberonenacho
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.