How to setup the limits for the integral?

AniMuS · #1 $Old$ June 27th, 2009, 11:03 AM

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.

Amer · #2 $Old$ June 27th, 2009, 12:08 PM

Quote:

Originally Posted by AniMuS $View Post$

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.

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

AniMuS · #3 $Old$ June 27th, 2009, 12:24 PM

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

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

Amer · #4 $Old$ June 27th, 2009, 12:30 PM

Quote:

Originally Posted by AniMuS $View Post$

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

AniMuS · #5 $Old$ June 27th, 2009, 12:40 PM

Yes, you're right. Hadn't seen that. Thanks