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I hope that this question has not been answered yet.I want to find the volume of the following set. I think it is a torus but I am not entirely sure.The set of points is: $\{(x,y,z)\in\mathbb{R}^3:x^2+(2-\sqrt{y^2+z^2})\leq1\}$ My first idea was that it is a circle that is somehow shifted by to units along one axis.I am a bit puzzeled about the following term: $(2-\sqrt{y^2+z^2})$ what does it do? Thank you very much in advance for your help!

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  • $\begingroup$ I think this is no torus. It is unusally, that $x$ enters quadratically and $y$ and $z$ do not. $\endgroup$ – Matthias Dec 2 '14 at 16:00
  • $\begingroup$ Maybe there is a typo? $\endgroup$ – Matthias Dec 2 '14 at 16:04
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Your set is $x^2+1 \le \sqrt {y^2+z^2}$ or $(x^2 + 1)^2 \le y^2+z^2$ This is unbounded in $y$ and $z$, so you cannot compute its volume (or just state that it is infinite).

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It is not a torus.

I'll rearrange the inequality: $$\sqrt{y^2+z^2}\ge x^2+1$$ which is the set of points in $\Bbb R^3$ which distance to the $X$ axis is greater than or equal to $1$ plus the square of their $x$ coordinate.

Making $z=0$ we get $$|y|\ge x^2+1$$ that gives two symmetrical parabolas and their "inside". The set of points that you have asked about is obtained as the revolution of these two parabolas along the $X$ axis, some sort of an infinite doughnut (the radius of the hole is $1$).

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Assume there's a typo (please ignore this answer if there is no typo) and that the equation should read $$x^2+(2-\sqrt{y^2+z^2})^2\leq 1$$ which is the equation of a torus with radius of locus of cross-section centre (is there better term for this?) of $c=2$, and radius of cross-section of $s=1$.

By Pappus' centroid theorem, the volume of the torus is given by the cross section multiplied by the distance travelled by its centroid, i.e.

$$\begin{align} V&=\pi s^2\cdot 2\pi c \\ &=\pi(1)^2\cdot 2\pi(2)\\ &=4\pi^2\qquad\blacksquare \end{align}$$

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