How to properly solve a double integral for calculating volume?

[Telemachus]

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.

 

[LCKurtz]

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.

 

[Telemachus]

Thanks.