# triple integral over a shifted cylinder

$$D=\{(y−2)^2+x^2\le1,0≤z≤2\}$$

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?

• Do you really want to calculate it using a triple integral? Why not with base-area$\times$ height? – b00n heT Jan 16 '19 at 11:11
• Yes with triple integral. I'm just curious :P – NPLS Jan 16 '19 at 11:12
• What does it means $.... =1\le 0$ ? – Emilio Novati Jan 16 '19 at 11:12
• I fixed it , yeah – NPLS Jan 16 '19 at 11:13
• 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. – TonyK Jan 16 '19 at 12:27

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