# [SOLVED] Double Integrals - Volume Between Paraboloids

• Apr 11th 2010, 07:43 PM
wagonwould
[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 $\displaystyle z = x^2+y^2$ and $\displaystyle 3z=4-x^2-y^2$.

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

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 $\displaystyle \frac{2pi}{3}$ cubic units.
Thanks so much for your help.
• Apr 11th 2010, 07:53 PM
AllanCuz
Quote:

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

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

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

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 $\displaystyle \frac{2pi}{3}$ 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

$\displaystyle \int \int \int dxdydz$

to

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

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

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

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

$\displaystyle x^2+y^2 =1$

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

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

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

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

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?
• Apr 11th 2010, 08:33 PM
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!
• Apr 11th 2010, 08:35 PM
AllanCuz
Quote:

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?
• Apr 11th 2010, 08:39 PM
wagonwould
Thanks for answering even though you're tired! I tried switching over to visitor messages on your profile...hope that works.