I am still unsure of my ability to evaluate the volume of a solid using triple integrals. Here is my question:

Now I know that the intersection of the two paraboloids is

9 = x^2 + y^2.

But I am unsure how to set up the triple integral. I was thinking of splitting the volume into two halves.

But as I am writing this I am not sure that I have to do that. I would say I could set up the triple integral with the following limits of integration as so:

Let E be the solid in question. Then

E = {(x,y,z) | -3 <= x <= 3, -(9-x^2)^1/2 <= y <= (9-x^2)^1/2, x^2 + y^2 <= z <= 36 - 3x^2 - 3y^2 }

How does this look? First I would integrate with respect to z, then y then x.

I was thinking that I could split the solid into two triple integrals as such:

E = {(x,y,z) | -3 <= x <= 3, -(9-x^2)^1/2 <= y <= (9-x^2)^1/2, x^2 + y^2 <= z <= 9 }

and

E = {(x,y,z) | -3 <= x <= 3, -(9-x^2)^1/2 <= y <= (9-x^2)^1/2, 9 <= z <= 36 - 3x^2 - 3y^2 }

But I don't see a problem with the first way I set up the triple integral. However I haven't tried to integrate it as of yet.

Any thoughts on the problem would be appreciated. Thankyou.

I really wouldn't do this volume in cartesian coordinates. This one's just begging for cylindrical coordinates (it's got rotational symmetry about the z-axis), so, without further ado:

z = x^2 + y^2 = r^2
z = 36 - 3x^2 - 3y^2 = 36 - 3r^2

r^2 = 36 - 3r^2
r = 3
z = 9

So the integral intersects when r = 3 and z = 9. It's not easy to figure out that it'd be real difficult do this in only one integral, so let's split the solid up into two smaller solids and add them.

Volume 1: The solid enclosed by z = r^2 and the plane z = 9.
Volume 2: The solid enclosed by the plane z = 9 and z = 36 - r^2

I'll do Volume 1, and you can have your hand at Volume 2.

Let's start with the obvious and go from there:

[tex]\int\!\!\!\int\!\!\!\int_{V_1}1\,dV[/tex]

So we have some kind of triple-integral over some volue. Since we're doing this is cylindrical coordinates,

[tex]dV = r\,dz\,dr\,d\theta[/tex]
with [itex]dz\,dr\,d\theta[/itex] arranged in whichever order we want.

Just because the math works out the easiest, I'll choose the order I wrote. So we have,

Yes. I thought so too. However I usually try to set up the integral using cartesian coordinates then I show the conversion to polar coordinates.

I ended up doing the question in polar coordinates but I don't remember my answer.

I also ended up doing the question in one volume instead of two. I used the upper limit of

36 - 3r^2

and a lower limit of

r^2.

However, intuitively, I thought that splitting up the integral into two separate integrals was the way to go. And I think that is what I should have done in the first place.

I don't see any reason to "break the solid up into two smaller solids and add them". The parabola z= 36- 3x^{2}- 3y^{2}= 36- 3r^{2} is always above the parabola z= x^{2}+ y^{2}= r^{2} so that will be the upper and lower limits on the "z" integration. There is complete circular symmetry so the θ integration will be from 0 to 2π. Finally, all of the volume lies inside the circle or intersection (which has radius 3) so the r integration will be from 0 to 3.
In cylindrical coordinates the volume is given by
[tex]\int_{\theta=0}^{2\pi}\int_{r=0}^3\int_{z=r^2}^{36-3r^2}rdzdrd\theta [/tex].

That reduces very quickly to [tex]2\pi\int_0^3r(36-3r^2-r^2)dr [/tex] or
[tex]2\pi\int_0^3(36r- 4r^3)dr [/tex]