# Surface area of a spherical cap

1. Oct 7, 2012

### soothsayer

1. The problem statement, all variables and given/known data
Calculate the area of a circle of radius r (distance from center to circumference) in the two-dimensional geometry that is the surface of a sphere of radius a. Show that this reduces to πr2 when r << a

2. Relevant equations
Surface area of a spherical cap = 2πah = π(r2 + h2)

3. The attempt at a solution
I've thrown all the calculus I've known at this problem and couldn't crack it. I immediately realized this problem was trying to get me to calculate the area of a spherical cap in the limit where the radius of the base of the cap was much smaller than the radius of the sphere, but I tried a straightforward double integral in spherical coordinates and couldn't get it to come out right. I tried to same thing integrating over infinitesimally thin rings from the top of the sphere to the circumference of the circle. No dice. I even found the surface area for the spherical cap, which I posted above, and attempted to just show that it reduced to πr2 when r << a and couldn't even prove that much. I couldn't get a formula with r and a together. None of the handfuls of derivations I found for the surface area of a spherical cap helped me.

2. Oct 8, 2012

### slider142

The surface area integral you performed should have a cosine term in it if you used spherical coordinates, specifically cos(r/a). The entire formula should only contain a and r as variables. Replace the cosine term by the 2nd degree Taylor polynomial for cosine about r/a = 0 to get an approximation of cosine for small values of r/a that have negligible 4th degree and higher contributions. The formula should then immediately simplify to πr2.

Last edited: Oct 8, 2012
3. Oct 8, 2012

### voko

How about finding the surface of revolution of a circle's arc?

4. Oct 8, 2012

### soothsayer

That's it. I simply forgot to expand into the second term of the Taylor series. My friend in the class came along while I was in the library and alerted me of this earlier today. Thanks for the help!