In the following example I need to find Jordan measure of the following set:$$A=\{(x,y,z)\mid z>0,1-x^2=z, y^2+z^2=z\}.$$ Here we have two cylinders, but I am not sure to find Jordan measure, is there formula or something?

  • $\begingroup$ Just calcule like Calculus course, the integral of constant function $1$ over the set $A$, it's the jordan measure of $A$. $\endgroup$ – Mephisto Jan 23 at 19:51
  • $\begingroup$ @Trevor The intersection of the two surfaces is a curve $\endgroup$ – Raffaele Jan 23 at 20:31

To find the limits of integral,

Take $y = r\cos\theta, z = r\sin\theta$ given the orientation of cylinder.

Equation of cylinder is $r = \sin\theta, 0 \leq \theta \leq \pi$

$-\sqrt{1-r\sin\theta} \leq x \leq \sqrt{1-r\sin\theta}$

$0 \leq r \leq \sin\theta$

$0 \leq \theta \leq \pi$

Order of integral $dx, $ then $dr$ and then $d\theta$.

In cartesian,

${-\sqrt{1-z}} \leq x \leq {\sqrt{1-z}}$

$-\sqrt{z-z^2} \leq y \leq \sqrt{z-z^2}$

$0 \leq z \leq 1$

  • $\begingroup$ Right, so I am only calculating integral with given bounds in order $dr,d\theta,dx$? $\endgroup$ – Trevor Jan 23 at 20:39
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    $\begingroup$ first $dx$ then $dr$ then $d\theta$. That is why $x$ is in terms of $r$ and $\theta$. $\endgroup$ – Math Lover Jan 23 at 20:41
  • $\begingroup$ Do you want set up in cartesian, if you prefer that? $\endgroup$ – Math Lover Jan 23 at 20:41
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    $\begingroup$ Yes, yes, because $sin$ is positive only in first two quadrants. $\endgroup$ – Trevor Jan 23 at 20:44
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    $\begingroup$ I think it is much simpler in cartesian. That is why I asked. Note $z-z^2 = z(1-z)$. So you get $\sqrt{1-z}$ twice that multiplies for a simple integral wrt dz. You can choose dx or dy first in any order. dz is last. $\endgroup$ – Math Lover Jan 23 at 20:51

The curve $L$ is the intersection of the two surfaces (see the picture below) $$\begin{cases} y^2+z^2=z\\ z=1-x^2\\ \end{cases} $$ Its parametric equations are

$$L=\left(\cos t,\frac{1}{2} \sin 2 t,\sin ^2t\right);\;t\in[0,2\pi]$$

$$\frac{dL}{dt}=\left(-\sin t,\cos 2t,\sin 2t\right)$$

Length of $L$ is given by $$L=\int_{0}^{2\pi}\sqrt{\left\lVert \frac{dL}{dt}\right\rVert^2}\,dt$$ $$L=\int_{0}^{2\pi} \sqrt{1+\sin ^2t}\,dt\approx 7.64$$

$\int \sqrt{1+\sin ^2t}\,dt$ is an elliptic integral and the value that can be found is only an approximated value.

Hope this helps


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