# Thread: [SOLVED] Double Integrals - Volume Between Paraboloids

1. ## [SOLVED] Double Integrals - Volume Between Paraboloids

I'm studying for my calculus exam and having difficulty with this question.

Find the volume lying between the paraboloids $z = x^2+y^2$ and $3z=4-x^2-y^2$.

I know to switch to polar coordinates, so $z=r^2$ and $3z=4-r^2$. The solution to the question says that they intersect where $3(x^2+y^2)=4-(x^2+y^2)$, which I understand is just because you're setting z=z and solving, and that r can also be subbed in to be $3r^2=4-r^2$. But then the solution says they intersect at that equation where z=z, "i.e. on the cylinder $x^2+y^2=1$. Is this just solving for $r$ from the $3r^2=4-r^2$?

I suppose my question is, what is the process I follow to solve this question? The solution integrates over the disc of radius 1, sets the intersection equation to zero with everything on one side, and integrates. But I don't know why.

The final answer is $\frac{2pi}{3}$ cubic units.
Thanks so much for your help.

2. Originally Posted by wagonwould
I'm studying for my calculus exam and having difficulty with this question.

Find the volume lying between the paraboloids $z = x^2+y^2$ and $3z=4-x^2-y^2$.

I know to switch to polar coordinates, so $z=r^2$ and $3z=4-r^2$. The solution to the question says that they intersect where $3(x^2+y^2)=4-(x^2+y^2)$, which I understand is just because you're setting z=z and solving, and that r can also be subbed in to be $3r^2=4-r^2$. But then the solution says they intersect at that equation where z=z, "i.e. on the cylinder $x^2+y^2=1$. Is this just solving for $r$ from the $3r^2=4-r^2$?

I suppose my question is, what is the process I follow to solve this question? The solution integrates over the disc of radius 1, sets the intersection equation to zero with everything on one side, and integrates. But I don't know why.

The final answer is $\frac{2pi}{3}$ cubic units.
Thanks so much for your help.
Edit- I just kind of realized you were asking why as opposed to how. Un momento, I will explain

By definition of polar co-ordinates our integral transforms from

$\int \int \int dxdydz$

to

$\int \int \int rdrd\theta dz$

So now we need bounds for r, theta and z.

In the case of this problem our Z is bounded by 2 functions. Let us label them Z1 and Z2. So to get the area between these 2 functions we need to subtract the smaller Z from the bigger Z. What this is in effect, is the integral of

$\int_{Z1}^{Z2} dz$

But now we need our bounds for R. Well, if we put out equations together, this yields

$x^2+y^2 =1$

Which is a circle of radius 1. What this means is, we are integrating over this domain in the XY plane. Notice that

$r^2=x^2 +y^2=1$

Therefore, r runs from the origin (0) to 1.

Now we need bounds for our theta...But remember, this is a circle! So how many degrees are there in a circle? That would be 360 or

$0<=\theta<=2\pi$

So you see, what we are essentially doing is integrating from 0-->360 over a circle with radius R projected upward into the Z plane.

Does this make sense?

3. Thanks so much AllanCuz, that makes a lot more sense. I really appreciate you taking the time to answer the question. I don't have enough message posts on here yet to respond to you, but yes, it's from Adams Calculus A Complete Course. AM 2411 at UWO? I recognized your username haha.
Thanks!

4. Originally Posted by wagonwould
Thanks so much AllanCuz, that makes a lot more sense. I really appreciate you taking the time to answer the question. I don't have enough message posts on here yet to respond to you, but yes, it's from Adams Calculus A Complete Course. AM 2411 at UWO? I recognized your username haha.
Thanks!
....LOL

I figured you were using the text because when I needed to confer I was doing it right (i'm quite tired lol) i noticed the question was verbatim out of the text.

Are you coming to my help session tomorrow?

5. Thanks for answering even though you're tired! I tried switching over to visitor messages on your profile...hope that works.