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I am struggling with a tricky triple integral. It is as follows:

Let $K \subset \mathbb{R^3}$ be the body given by the inequalities: \begin{equation} x \geq 0, y^2 + z^2 \leq 1, x+y\leq 2. \end{equation} Calculate $\iiint_K \sqrt{y^2+z^2}dV$.

I think that I have to transfer coordinates. Something like polar coordinates, but I cannot seem to get a good transformation.

Help would be greatly appreciated!

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    $\begingroup$ Hint: cylinders. $\endgroup$ – Sean Roberson Jun 29 '17 at 19:44
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Hint. Note that $$I:=\iiint_K \sqrt{y^2+z^2}dV=\iint_{y^2+z^2\leq 1} \sqrt{y^2+z^2}\left(\int_{x=0}^{2-y}dx\right)dydz\\=\iint_{y^2+z^2\leq 1} (2-y)\sqrt{y^2+z^2} dydz$$ Then use the polar coordinates $y=r\cos(t)$, $z=r\sin(t)$, $$I=\int_{r=0}^1\int_{t=0}^{2\pi} r^2(2-r\cos(t)) dr dt.$$ Can you take it from here?

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  • $\begingroup$ Ah okay. When I drew it I already noticed that cyllindrical coordinates might help. It looks like the Area is some sort of cilinder along the x-axis. $\endgroup$ – user444389 Jun 29 '17 at 19:49
  • $\begingroup$ Ah okay thanks. What now remains is working out the integral. I get $\frac{4*\pi}{3}$. $\endgroup$ – user444389 Jun 29 '17 at 19:57
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    $\begingroup$ I got the same result ;-) $\endgroup$ – Robert Z Jun 29 '17 at 19:59

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