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

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- Jun 27th 2009, 09:03 AMAniMuSHow 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 - Jun 27th 2009, 10:08 AMAmer
- Jun 27th 2009, 10:24 AMAniMuS
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

$\displaystyle \int_{0}^{2}\int_{2-z}^{2}\int_{-\sqrt{4-y^2}}^{\sqrt{4-y^2}} dxdydz$ - Jun 27th 2009, 10:30 AMAmer
- Jun 27th 2009, 10:40 AMAniMuS
Yes, you're right. Hadn't seen that. Thanks :)