# Jordan measure of two intersecting cylinders [closed]

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?

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

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

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

$$0 \leq z \leq 1$$

• Right, so I am only calculating integral with given bounds in order $dr,d\theta,dx$? – Trevor Jan 23 at 20:39
• first $dx$ then $dr$ then $d\theta$. That is why $x$ is in terms of $r$ and $\theta$. – Math Lover Jan 23 at 20:41
• Do you want set up in cartesian, if you prefer that? – Math Lover Jan 23 at 20:41
• Yes, yes, because $sin$ is positive only in first two quadrants. – Trevor Jan 23 at 20:44
• 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. – 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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