# Triple Integral - Volume Question

• twiztidmxcn
You just need to finish it off by setting up the integral in cylindrical coordinates. Remember, in cylindrical coordinates, the projection of a solid figure in the xy-plane is a circle from the origin to the point of interest. In this case, the projection is a circle of radius 2. So the integral is ∫∫∫ dz r dr dθ where the limits of integration are 0 to 2 for r, 0 to 2π for θ, and for z, the same limits you already found, 3x^2 + 3y^2 to 6sqrt(x^2+y^2).

## Homework Statement

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.

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

twiztidmxcn said:

## Homework Statement

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).
Yes, that's correct.

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.

Good. In other words, if you project the entire region into the xy-plane, it covers a circle of radius 2 and your integral must cover all of that.

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).
Good. That's correct.

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).
Yes, that's also correct.

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...
It would probably have been easier to set up the integral in cylindrical coordinates from the start. Your figure, projected to the xy-plane forms a circle with center at (0,0) and radius 2. To cover that, take $\theta$ ranging from 0 to $2\pi$, r ranging from 0 to 2. Now express the bounds on z in terms of r and $\theta$.

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
Actually, it's very well done and clear.