Arc length of curve of intersection between cylinder and sphere

Question

Given the sphere $x^2+y^2+z^2 = \frac{1}{8}$ and the cylinder $8x^2+10z^2=1$, find the arc length of the curve of intersection between the two.

I tried parametrizing the cylinder (the task specifies this as a hint). My attempt:

$$x(t) = \frac{1}{\sqrt{8}} \sin(t)$$ $$z(t) = \frac{1}{\sqrt{10}} \cos(t)$$

Plugging this into $x^2+y^2+z^2 = \frac{1}{8}$, I solve for $y$ to get

$$y = \sqrt{\frac{\cos(2t)+1}{4\sqrt{5}}}$$

I then tried integrating $|x(t), y(t), z(t)|$ from $0$ to $2\pi$ with no luck. I suspect my parametrization is wrong as my expression for $y$ looks rather ugly. Any ideas?

Answer 1

The intersection of the cylinder with the sphere produces two curves, each of which is a great circle.

Since the radius of the sphere is $\dfrac{1}{\sqrt{8}}$, the length of each great circle is $\dfrac{2 \pi}{\sqrt{8}} = \dfrac{\pi}{\sqrt{2}}$.

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Using SymPy to verify:

>>> from sympy import *
>>> t = Symbol('t')
>>> x = sin(t) / sqrt(8)
>>> y = cos(t) / sqrt(40)
>>> z = cos(t) / sqrt(10)
>>> x_dot = diff(x,t)
>>> y_dot = diff(y,t)
>>> z_dot = diff(z,t)
>>> v = Matrix([x_dot, y_dot, z_dot])
>>> v.T * v
Matrix([[sin(t)**2/8 + cos(t)**2/8]])
>>> simplify(sin(t)**2/8 + cos(t)**2/8)
1/8

Note that the Euclidean norm of the velocity vector is $\dfrac{1}{\sqrt{8}}$ and, thus, independent of $t$.

Answer 2

From $8x^2 + 10z^2 = 1$,you get $z^2 = \frac{1}{10}.(1-8x^2)$. Substitute this in the other equation $ x^2+y^2+z^2 = \frac{1}{8}$ you get

$$x^2 + 5y^2 = \frac{1}{8}$$

This is the curve of intersection, now parameterize this ellipse with

$x = \frac{1}{\sqrt{8}} sint$

and

$ y = \frac{1}{\sqrt{40}} cost$

$ z = \frac{1}{\sqrt{10}} cost$

Now arc length $ L= \int_0^{2\pi} \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 +(\frac{dz}{dt})^2} dt$

$$ = \int_0^{2\pi} \sqrt{\frac{1}{8} cos^2t + \frac{1}{40} sin^2t+\frac{1}{10} sin^2t}dt$$

$$ = \int_0^{2\pi} \sqrt{\frac{1}{40}}.\sqrt{ 5cos^2t + sin^2t +4sin^2t} dt$$

$$=\int_0^{2\pi} \sqrt{\frac{5}{40}} dt$$

$$L= \frac{1}{2\sqrt{2}}\int_0^{2\pi} dt\tag 1$$

$$ L = \frac{2\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$$

Answer 3

Apparently I was right.

If we define $r(t) = (x, y, z)$ where

$$x = \frac{1}{\sqrt{8}}sin(t)$$ $$z = \frac{1}{\sqrt{10}}cos(t)$$ $$ y = \pm \sqrt{\frac{cos(2t)+1}{4\sqrt{5}}}$$

we find

$$|r(t)| = \sqrt{(\frac{1}{\sqrt{8}}sin(t))^2 + (\frac{1}{\sqrt{10}}cos(t))^2+(\frac{\sqrt{cos(2t)+1}}{4*\sqrt{5}})^2} = \frac{1}{2\sqrt{2}}$$

$$\int_{0}^{2\pi} |r(t)| dt = \int_{0}^{2\pi} \sqrt{(\frac{1}{\sqrt{8}}sin(t))^2 + (\frac{1}{\sqrt{10}}cos(t))^2+(\frac{\sqrt{cos(2t)+1}}{4*\sqrt{5}})^2} dt $$ $$ = \int_{0}^{2\pi} \frac{1}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$$

The correct answer is $\pi\sqrt{2}$, I suspect the reason lies in the fact that $$ y = \pm \sqrt{\frac{cos(2t)+1}{4\sqrt{5}}}$$, meaning we must also take into account the negative value here (seems we must multiply our integral by 2). I used wolfram to my advantage to figure out the integral, my approach by hand was a failure as it got quite messy.

This is from my calculus 2 class, I'm not familiar with elliptic integrals yet, but I thank you for the assistance Satish.