Triple Integral - Volume Question

  • #1

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.

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
 

Answers and Replies

  • #2
HallsofIvy
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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.

Circle, radius 2.
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 [itex]\theta[/itex] ranging from 0 to [itex]2\pi[/itex], r ranging from 0 to 2. Now express the bounds on z in terms of r and [itex]\theta[/itex].

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.
 

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