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This problem is similar to this however, in my case my cylinder is not at (0,0).

Use double integrals to determine the volume of the solid bounded by: $$x^2+y^2\leq2y \qquad z\leq2-x^2-y^2 \qquad z=0$$

When $z=0$ we have the following plot

$$(\sqrt2)^2=x^2+y^2 \qquad x^2+(y-1)^2=1$$

Thus, the integration region is defined by the intersection. Which method should I use to integrate?

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  • $\begingroup$ polar integration $\endgroup$ – Saketh Malyala Jun 30 '17 at 4:16
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In cylindrical coordinates the bounds are given by \begin{align} 0\le\ &z\ \le 2-r^2 \\ 2\cos(\theta)\le\ &r\ \le \sqrt{2} \\ -\pi/4 \le \ &\theta\ \le \pi/4 \end{align} The volume of the region is just the triple integral $$\int_{-\pi/4}^{\pi/4}\int_{2\cos(\theta)}^{\sqrt{2}}\int_{0}^{2-r^2}r\ dzdrd\theta$$ I will leave the computation to you.

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