I am stuck on the problem of finding volume of the figure given by $0 \leq z \leq 2, x^{2} + y^{2} \leq 2, x^{2} + y^{2} + z \leq 2x$. I have tried three different coordinates but the problem is $2x$ in the last relation. Thanks ahead for help!
Add. Sorry that I didn't mention this before. But I want a full solution without "brute-force" computation. I did try many different tricks, so if any of those works, it means that I do not know how to work out the details.
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Use difference of squares to get rid of the $2x$, and you should have the intersection of a cylinder with a paraboloid, although the paraboloid will not be centered at the origin.
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Hint: Try completing the square $$x^2 + y^2 + z \leq x \iff x^2 - 2x + 1 + y^2 + z \leq 1 \iff (x - 1)^2 + y^2 \leq 1 - z$$
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You may use the Cylindrical Coordinates for doing the Volume. As @azhi illustrated your flat region is as follows on $z=0$:
You can find that both surfaces intersects each other when $\theta=\pi/4$ and this point is being shown above. I consider, you are looking at the bottom of the area. Now we can consider two areas $D_1$ and $D_2$ and use the symmetric which rules the whole volume and write:
$$V=2(V|_{D_1}+V'|_{D_2})=2\left(\int_{\theta=0}^{\pi/4}\int_{r=0}^1\int_{z=0}^{2r\cos\theta-r^2}rdzdrd\theta+\int_{\theta=\pi/4}^{\pi/2}\int_{r=0}^{2\cos \theta}\int_{z=0}^{2r\cos\theta-r^2}rdzdrd\theta\right)$$
Denote by $D$ the bottom boundary of the solid (its volume be $V$), which can be expressed as set $\{(x,y)\in \mathcal{R}^2 \mid x^2+y^2\leq 2\} \cap \{(x,y)\in\mathcal{R}^2 \mid x^2+y^2\leq 2 x\}$. Observing that the solid is symmetric with respect to the plane $y=0$, so we have: The Total Volume of Solid= 2 $ \times$ that of the part with $y\geq 0.$ Now denote $D_+:=D\cap \{y>0\}$, then we have
$$\begin{aligned}V&=2\iint_{D_+} (2x-x^2-y^2)dxdy\\ &=2\int_0^1\bigg(\int_{1-\sqrt{1-y^2}}^{\sqrt{2-y^2}} (2x-x^2-y^2)dx\bigg)dy ~~( D_+ \text{is $x-$ simple})\\ &=2-\pi/4. \end{aligned}$$
May this can help you.