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Double integral problem

  • Thread starter Telemachus
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  • #1
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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: [tex]x^2+y^2=4[/tex], [tex]z=4-y[/tex] [tex]z=0[/tex]

Its a cylinder cut by a plane.

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

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 [tex]16\pi[/tex]. So, its clear I'm doing something wrong

I also thought on trying other way: [tex]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}[/tex]
But it gets more complicated that way.

Bye there, and thanks for helping.
 

Answers and Replies

  • #2
LCKurtz
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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: [tex]x^2+y^2=4[/tex], [tex]z=4-y[/tex] [tex]z=0[/tex]

Its a cylinder cut by a plane.

At first I've did this integral: [tex]\displaystyle\int_{-2}^{2}\displaystyle\int_{-\sqrt[ ]{4-x^2}}^{\sqrt[ ]{4-x^2}}(4-y)dydx[/tex]
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.
 
  • #3
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Thanks.
 

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