Triple integral over a shifted cylinder

calculusdefinite integralsintegrationmultivariable-calculus

$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?

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Best Answer

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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