# Double integral problem

## Homework Statement

Hi there. I have this problem with double integral, which says: calculate using a double integral the volume limited by the given surfaces: $$x^2+y^2=4$$, $$z=4-y$$ $$z=0$$

Its a cylinder cut by a plane.

At first I've did this integral: $$\displaystyle\int_{-2}^{2}\displaystyle\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}(4-y)dydx$$

The thing is that when I solve this the terms null them selves and it gives zero. There is evidently something that I'm doing wrong, I've realized that I'm taking off a half of the volume to the other, so I think I'm defining one part of the area over which I'm making the integral as negative. Is this true? And the thing is: How do I solve this properly? I've introduced this double integral into mathematica and it solved it and gave $$16\pi$$. So, its clear I'm doing something wrong

I also thought on trying other way: $$2\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{\sqrt[ ]{4-y^2}}(4-y)dxdy=2\displaystyle\int_{0}^{2}(4x-yx)_0^{\sqrt[ ]{4-y^2}}dy=2\displaystyle\int_{0}^{2}(4\sqrt[ ]{4-y^2}-y (\sqrt[ ]{4-y^2}))dy}$$
But it gets more complicated that way.

Bye there, and thanks for helping.

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LCKurtz
Homework Helper
Gold Member

## Homework Statement

Hi there. I have this problem with double integral, which says: calculate using a double integral the volume limited by the given surfaces: $$x^2+y^2=4$$, $$z=4-y$$ $$z=0$$

Its a cylinder cut by a plane.

At first I've did this integral: $$\displaystyle\int_{-2}^{2}\displaystyle\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}(4-y)dydx$$
You have set it up correctly. Without seeing your work I can only guess that you have a sign error somewhere to cause the cancellation.

Having said that, still I would suggest you set it up in polar coordinates in the first place.

Thanks.