# Thread: Problems with double integrals

1. ## Problems with double integrals

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 themselves 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.

2. Originally Posted by Ulysses
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 themselves 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.
Why not use polar coordinates

$\displaystyle\int_{-2}^{2}\displaystyle\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}(4-y)dydx=\int_{0}^{2}\int_{0}^{2\pi} [4-r\sin(\theta)]r\,dr\, d\theta$

3. I'm not sure about that, I haven't used changes of variables yet. I don't understand why still giving a wrong result. The result is evidently $16\pi$, I can see the graph that the volume its equal to the cylinder of radius 2 and height equal 4. I did it this way now, but stills giving me wrong.

$2\displaystyle\int_{0}^{2}\displaystyle\int_{0}^{\ sqrt[ ]{4-y^2}}(4-y)dxdy=2\displaystyle\int_{0}^{2}(4x-yx)_0^{\sqrt[ ]{4-y^2}}dy=2\displaystyle\int_{0}^{2}(4\sqrt[ ]{4-y^2}-y (\sqrt[ ]{4-y^2}))dy}$

Look what this integral gives: 2 integral_0&#94;2 y&#40;4-y&#94;2&#41;&#94;&#40;1&#47;2&#41; dy - Wolfram|Alpha
2 integral_0&#94;2 4&#40;4-y&#94;2&#41;&#94;&#40;1&#47;2&#41; dy - Wolfram|Alpha

It gives [math8\pi-16/3[/tex]

I don't know whats wrong now. Mm I see, I think I've changed the symmetry, I should solve the first I've planted.