Double iintegral (volume) this is redicuious

[evilpostingmong]

Homework Statement

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

Homework Equations

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

 

[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.

 

[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.

 

[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.

 

[tiny-tim]

Hi evilpostingmong!

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

I don't think well in 3D …

but that isn't bounded, is it?

 

[Ithryndil]

Hi evilpostingmong!


I don't think well in 3D …

but that isn't bounded, is it?

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

 

[tiny-tim]

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

hmm … I always assume someone copied the question wrong …

until I'm reassured!

 

[Ithryndil]

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

 

[evilpostingmong]

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.

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.

 

[Ithryndil]

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

 

[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.

 

[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.