# How to setup the limits for the integral?

• June 27th 2009, 09:03 AM
AniMuS
How to setup the limits for the integral?
http://img261.imageshack.us/img261/6067/3it.jpg

This is the sketch i made. Maybe its wrong, because i think it gives a very difficult integral like this. I'll look into it on the meanwhile.

http://img520.imageshack.us/img520/5048/object.jpg
• June 27th 2009, 10:08 AM
Amer
Quote:

Originally Posted by AniMuS
http://img261.imageshack.us/img261/6067/3it.jpg

This is the sketch i made. Maybe its wrong, because i think it gives a very difficult integral like this. I'll look into it on the meanwhile.

http://img520.imageshack.us/img520/5048/object.jpg

the limits

$\int_{-2}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}}\int_{0}^{y+2} dzdxdy$

after you determine the boundaries for z drop the graph to the x-y plane you will have a circle x^2+y^2=4 then determine the boundary of x as I write then for y
• June 27th 2009, 10:24 AM
AniMuS
If i'd put that answer on the test, i'd have 0. :)

The order of the integration must be the one they want.

Anyway i found i drew the plane wrong. Both intersections of the plane with the cilinder should be parallel to the x axis. So that makes it much easier.

Now that i drew it well, i think it's this. Correct me if im wrong please

$\int_{0}^{2}\int_{2-z}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dxdydz$
• June 27th 2009, 10:30 AM
Amer
Quote:

Originally Posted by AniMuS
If i'd put that answer on the test, i'd have 0. :)

The order of the integration must be the one they want.

Anyway i found i drew the plane wrong. Both intersections of the plane with the cilinder should be parallel to the x axis. So that makes it much easier.

Now that i drew it well, i think it's this. Correct me if im wrong please

$\int_{0}^{2}\int_{2-z}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dxdydz$

I think z change from 0 to 4 not 2 since you have the two curve

2+y=z and y=2 they intersect when z=4
• June 27th 2009, 10:40 AM
AniMuS
Yes, you're right. Hadn't seen that. Thanks :)