# Thread: How to setup the limits for the integral?

1. ## How to setup the limits for the integral?

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.

2. Originally Posted by AniMuS

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

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

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

3. 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

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

4. 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

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

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