2
$\begingroup$

Let we have two planes $p_1$ and $p_2$ that intersect by an angle $\beta$. Let $p_1$ be on distance of $a$ from the center $C$ of sphere with radius $1$ and $a<1$. Let intersection of the two planes be on the distance of $b$ from C, $b<1$.

I want to find area of the surface of the sphere which is bounded by the planes and is in the opposite quadrant of where the center $C$ is.

enter image description here

I tried doing integral by running along intersection of planes and cut slices on the sphere, as I know if a slice is whole circle (belt) it's area is $2\pi dx$, so since I don't want whole circle it should be proportional to central angle that part makes (grey lines). $$\int_{-\sqrt{1-b^2}}^{\sqrt{1-b^2}}\alpha(x)dx$$ Now I want to calculate the angle, I get this formula: $$\alpha(x)=\arccos\frac{a}{r}-\arccos\frac{a}{b}+\pi-\beta-arcsin\frac{a}{b}-\arcsin\frac{\sin(\beta+arcsin\frac{a}{b})b}{r}$$ Where $r=\sqrt{1-x^2}$ is the radius of a particular belt.

I don't know to solve this integral and I wander if there is an easier way?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.