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?
Best Answer
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$
$$ = \int_0^{2\pi} \sqrt{\frac{1}{8} cos^2t + \frac{1}{40} sin^2t+\frac{1}{10} sin^2t}dt$$
$$ = \int_0^{2\pi} \sqrt{\frac{1}{40}}.\sqrt{ 5cos^2t + sin^2t +4sin^2t} dt$$
$$=\int_0^{2\pi} \sqrt{\frac{5}{40}} dt$$
$$L= \frac{1}{2\sqrt{2}}\int_0^{2\pi} dt\tag 1$$
$$ L = \frac{2\pi}{2\sqrt{2}} = \frac{\pi}{\sqrt{2}}$$