# Double iintegral (volume) this is redicuious

1. Sep 7, 2008

### evilpostingmong

1. The problem statement, all variables and given/known data

Find the volume of the solid bounded by y=2 y=x and z=4-y^2

2. Relevant equations

3. The attempt at a solution
I DO NOT UNDERSTAND WHY I'M WRONG! limits of integration for x=2 to 0 limtis of integration for y=2 to x and 4-y^2 is the integrand I ended up with 4/3 but the answer (according to the answer sheet) is 4! WHY

2. Sep 7, 2008

### Ithryndil

I think your limits of integration are wrong

The area in the xy plane is a triangle. I can't really draw it but x doesn't go from 2 to 0 if I read your limits of integration right. Also, y isn't going to go from 2 to x either.

Last edited: Sep 7, 2008
3. Sep 7, 2008

### evilpostingmong

Wait I found it. limits for x are y to 0 and for y 2 to 0. If I went from 2 to x for y then that would be like integrating for x again, since the integral lines would be parallel to y.

4. Sep 7, 2008

### Ithryndil

I think that would work...however I believe y should go from 0 to 2 otherwise you might get a negative answer.

Here's what I have:

$$\int^{2}_{0}$$$$\int^{2}_{x}4 - y^{2}dydx$$

With that I integrate with respect to y first and then x.

Last edited: Sep 7, 2008
5. Sep 7, 2008

### tiny-tim

Hi evilpostingmong!
I don't think well in 3D …

but that isn't bounded, is it?

6. Sep 7, 2008

### Ithryndil

Technically you're correct...I assume that they mean it's bounded by the axis planes...

7. Sep 7, 2008

### tiny-tim

hmm … I always assume someone copied the question wrong

until I'm reassured!

8. Sep 7, 2008

### Ithryndil

If it's not bounded then...the volume is infinite if I am not mistaken.

9. Sep 7, 2008

### evilpostingmong

It's alright, I did it out. I got 4 so these limits (x=y to x=0 and y=2 to y=0) are correct.
I appreciate the help, but the limits you put were the same as mine (the ones I put at the first post) which would net you 4/3.
Yes it is bounded by the xy plane.

10. Sep 7, 2008

### Ithryndil

When I did the integral with my limits I got 4.

11. Sep 7, 2008

### evilpostingmong

Then I made a stupid mistake when I did the integral itself, which I'm known for, lol I just
did it my original way (the one that we thought was wrong) and yes it is 4. Ur rite.

12. Sep 7, 2008

### Ithryndil

One mistake I made was getting to a point where I had 16x/3 and not plugging in the 2 for the x and getting a 16/3 instead of a 32/3 which changes things.