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$D=\{(y−2)^2+x^2\le1,0≤z≤2\} $

I need to calculate the volume.

so I'ts a cylinder with radius 1, and shifted by 2 units in the y axe. The problem here is that the circle doesn't touch the origin! so in cylindrical coordinates $y$ should be $y=2+rsen(θ)$, is this correct ?

$\int_{-sin^{-1}{(1/\sqrt{5})}}^{sin^{-1}{(1/\sqrt{5})}}\int_{1}^{2sin\theta+1/2(\sqrt{16sin^2\theta-12})}\int_{0}^{2}rdzdrd\theta$

Is this theoretically right?

enter image description here

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    $\begingroup$ Do you really want to calculate it using a triple integral? Why not with base-area$\times$ height? $\endgroup$ – b00n heT Jan 16 '19 at 11:11
  • $\begingroup$ Yes with triple integral. I'm just curious :P $\endgroup$ – NPLS Jan 16 '19 at 11:12
  • $\begingroup$ What does it means $.... =1\le 0$ ? $\endgroup$ – Emilio Novati Jan 16 '19 at 11:12
  • $\begingroup$ I fixed it , yeah $\endgroup$ – NPLS Jan 16 '19 at 11:13
  • $\begingroup$ For the purpose of this exercise (which I must confess I don't see the point of), you can forget the $z$-coordinate, can't you? It becomes the problem of finding the area of a circle using polar coordinates. $\endgroup$ – TonyK Jan 16 '19 at 12:27
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If you want use a triple integral than note that, since we don't have a cylindrical symmetry, the use of cylindrical coordinates is not the best choice.

It is simpler to use Cartesian coordinates, noting that: $$ -\sqrt{1-(y-2)^2}\le x \le \sqrt{1-(y-2)^2} $$

for: $1\le y\le3$

so the integral is: $$ \int_1^3\int_{-\sqrt{1-(y-2)^2}}^{\sqrt{1-(y-2)^2}}\int_0^2 dzdxdy $$

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  • $\begingroup$ yeah, sure! but I just want to see how to do it in cylindrical coordinates, I'm just curious how to parametrize a shifted cylinder integrating it $\endgroup$ – NPLS Jan 16 '19 at 12:06

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