# Surface Integral of Sphere between 2 Parallel Planes

A circular cylinder radius $$r$$ is circumscribed about a sphere of radius $$r$$ so that the cylinder is tangent to the sphere along the equator. Two planes each perpendicular to the axis of the cylinder, intersect the sphere and the cylinder in circles.

WTS: area of the part of the sphere between two parallel planes depends only on the sphere radius and distance between the planes, and not on the particular position of the planes.

Prove that the area of the part of the sphere between the two planes is equal to the part of the cylinder between the two planes.

$$\underline{Attempt}$$

I constructed this diagram.

However in order to prove the question statement, it means that the arc length of the sphere must be equal to the length of the straight line of the cylinder wall, which doesn't make sense to me. Nonetheless, I first tried to find the area of the sphere between the planes with using spherical coordinates. $$A_{sphere} = \int_0^{2\pi} \int_{\frac{\pi}{2}-\alpha}^{\frac{\pi}{2}+\alpha} r^2 \sin \phi \, d\phi d\theta = 4\pi r^2 \sin \alpha$$

And then I am stuck because I don't know how to obtain the height between the planes in terms of the angles.

• You're done. What is $\sin\alpha$? By the way, you shouldn't really assume the planes are symmetric about the plane $z=0$. You should have $z=z_0$ and $z=z_0+h$. – Ted Shifrin Nov 11 '18 at 2:32
• @TedShifrin : thanks, I see it now. I kept focusing on how to obtain the height using angle Beta because angles Alpha and Beta were not equal. – NetUser5y62 Nov 11 '18 at 3:10

You should be able to find the distance between the planes in terms of the angle $$\phi$$ and the radius $$r$$