Question on Change of Variables I have to calculate $\lambda^3(Z)$ with $Z=\{(x,y,z) \in \Bbb R^3\mid  \frac{x^2}{a^2}+\frac{y^2}{b^2} \le1,\:\:\: 0\le z\le h\}$ via the transformation $$\phi(\mu, \nu, z) = (k\cdot cosh(\mu)\cdot cos(\nu),\:\:\: k\cdot sinh(\mu)\cdot sin(\nu),\:\:\: z)^T $$
(I also have to determine $k\gt 0$.)
I've already calculated the Jacobi-determinant of $\phi$: $D\phi = k^2sinh^2(\mu) cos^2(\nu) +k^2cosh^2(\mu)sin^2(\nu)$ but now I'm lost on how to continue. (I know i have to apply the Theoreme on Change of Variables but I don't know how to do it). Any tipps would be greatly appreciated. Thanks in advance!
 A: The new coordinates are elliptical cylindrical. The curves for $\mu$ constant are ellipses. Note that the region to calculate its volume it's a cylinder of base an ellipse centered in the origin and with semiaxes $a$ and $b$. It's possible to choose $k$ and some value of $\mu$ to fit that ellipse into a constant value of $\mu$. It can be done as follows. Can easily be checked that:
$$\frac{x^2}{k^2\cosh^2\mu}+\frac{y^2}{k^2\sinh^2\mu}=1$$
For every value of $x,y$ and $\mu$. So, for constant $\mu=\mu_0$ the points $(x,y)$ satisfying the equation form an ellipse. This can be well the edge of the cylinder base:
$$\frac{x^2}{k^2\cosh^2\mu_0}+\frac{y^2}{k^2\sinh^2\mu_0}=\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$$
With
$$\begin{cases}
  k^2\cosh^2\mu_0=a^2 \\
  k^2\sinh^2\mu_0=b^2
\end{cases}$$
$$\begin{cases}
  k^2=a^2-b^2\\
  \mu_0=\text{arctanh}(b/a)
\end{cases}$$
Now $Z=\{\mu\in[0+\infty),\nu\in[0,2\pi),z\in\mathbb R\mid  \mu \le\mu_0,\:\:\: 0\le z\le h\}$ (Notice that the jacobian can be expressed as $k^2(\sinh^2\mu+\sin^2\nu)$)
$$Volume(Z)=\int_0^h\int_0^{2\pi}\int_0^{\mu_0}k^2(\sinh^2\mu+\sin^2\nu)\mathbb d\mu\mathbb d\nu\mathbb dz=$$
$$=h\int_0^{2\pi}\int_0^{\mu_0}k^2(\sinh^2\mu+\sin^2\nu)\mathbb d\mu\mathbb d\nu=$$
$$=hk^2\int_0^{2\pi}\int_0^{\mu_0}\sinh^2\mu\mathbb d\mu\mathbb d\nu+hk^2\int_0^{2\pi}\int_0^{\mu_0}\sin^2\nu\mathbb d\mu\mathbb d\nu=$$
$$=hk^22\pi(1/4)(\sinh(2\mu_0)-2\mu_0)+hk^2\pi\mu_0=hk^2\pi(1/2)\sinh(2\mu_0)$$
Substituting $k$ and $\mu_0$ by their values leads to:
$$Volume(Z)=h(a^2-b^2)\pi(1/2)\sinh(2\text{arctanh}(a/b))=h(a^2-b^2)\pi(1/2)\frac{2ab}{a^2-b^2}$$
$$Volume(Z)=\pi abh$$
For a cylinder can be calculated $Volume=Area\_of\_Base·Height$ With $Area\_of\_Base=\pi ab$ as corresponds to an ellipse. Then $Volume=\pi abh$. It works!!
