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)
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.