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How can I calculate the volume of this region in cylindrical coordinates?

$D=\{2x^2+y^2=z^2+4,|z| \le 2\}$

I think I got this wrong :

$$\operatorname{Volume} = 2\int_{0}^{2\pi}\int_{0}^{2} \int_{0}^{\sqrt{2r^2\cos^2\theta+r^2\sin^2\theta-4}}rdzdrd\theta$$

The problem is that the region is an ellipse and not a circle. I can integrate for $x^2+y^2=z^2+4$. Could you explain to me how to do it?

I think, what's wrong is also the Jacobian :

$x=2r\cos\theta$

$y=\frac{1}{\sqrt{2}}r\cos\theta$

$|J|=2\sqrt{2}$ ?

so maybe z range like this : $z=\{0,\sqrt{8r^2\cos^2\theta+\frac{1}{2}r^2\sin^2\theta-4}\}$

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  • $\begingroup$ The z-range is from -2 to +2. I'm not sure which coordinate transformation you picked for the Jacobian. It doesn't look like cylindrical coordinates in which we just have $x=r\cos\theta,\ y=r\sin\theta$. Either way, an ellipse is not easily described in regular polar coordinates. Instead I would suggest to use e.g. $x=a u\cos v,\ y=b u\sin v,\ z=z$, where $a$ and $b$ are the elliptical semi axes that depend on $z$, $v$ runs from $0$ to $2\pi$ but is not the polar angle, and $u$ runs from $0$ to $1$. $\endgroup$ – Klaas van Aarsen Jan 16 '19 at 20:22
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It's a single sheet hyperboloid and its central axis of symmetry is the $z$-axis.

Easiest way to find its volume is to consider that at every $z$ we have an ellipse, and integrate over the areas of those ellipses.

For a constant $z$ we have the ellipse: $$\frac{x^2}{\frac{z^2+4}{2}}+\frac{y^2}{z^2+4}=1$$ It has semi axes $a=\sqrt{\frac{z^2+4}{2}}$ and $b=\sqrt{z^2+4}$.

Since the area of an ellipse is $\pi a b$, the volume is: $$\text{Volume}=\int_{-2}^2 \pi a b\,dz = \int_{-2}^2 \pi\cdot\frac{z^2+4}{\sqrt 2}\,dz$$

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