(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

So my question is as follows: Find the volume of the solid bounded by z = 3x^2 + 3y^2 and z = 6sqrt(x^2 + y^2)

3. The attempt at a solution

I drew the graphs of these out, with the z = 3x^2 + 3y^2 being a circular paraboloid w/ vertex at (0,0,0) and extending in positive z direction.

The z = 6sqrt(x^2+y^2) is a cone centered on the z axis w/ vertex at (0,0,0).

I started by solving z = 6sqrt(x^2+y^2) for x^2 + y^2 and then plugging that into my other z = function and got that they intersect at z = 12. After plugging 12 in for z in the equations, i got both of them x^2 + y^2 = 4.

Circle, radius 2.

So, when setting up my triple integral, I have the triple integral of dzdydx. I believe my bounds on x are -2 -> 2.

Now, for my y bounds, I just solved the x^2 + y^2 = 4 for y, found that y = sqrt(4-x^2). I figured since it was a circular figure at this point, the y bounds are -sqrt(4-x^2) -> sqrt(4-x^2).

For my z bounds, I figured that it went from the lower function to the higher function, so from 3x^2 + 3y^2 -> 6sqrt(x^2+y^2).

My question comes to this....is any of this right at all? How do I actually integrate this?

I know I am going to need to probably change it into cylindrical coordinates, but then I run into the problem of how do I find my r integration bounds...

I know this is probably a jumble, but it's what I've got and hopefully some light can be shed on this.

thanks

-twiztidmxcn

**Physics Forums | Science Articles, Homework Help, Discussion**

Dismiss Notice

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Homework Help: Triple Integral - Volume Question

**Physics Forums | Science Articles, Homework Help, Discussion**