How to setup the limits for the integral?

Jun 2009
22
0


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.

 
Last edited:
May 2009
1,096
396
Jordan


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
 
Jun 2009
22
0
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

\(\displaystyle \int_{0}^{2}\int_{2-z}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dxdydz\)
 
May 2009
1,096
396
Jordan
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
 
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