# How to calculate the surface area of a spherical cap?

I need to find the surface area of a spherical cap.I searched the google, but most of them were based on calculus.(I studying grade 11 so, I haven't learned calculus yet!).

This question was given in my textbook. In the book, they have only covered the basics of sphere's surface area.

My guess is to calculate the area of the circumscribing cylinder with the height given(2 cm).

Area of the part of circumscribing cylinder=$(2*pi*r*2)cm^2$
Is this true? In case if this is true, it seems to be logically impossible to say that a piece of a spherical cap with height 2 cm in the top is equal to a piece in the center of the sphere with the same height!!!

Thank you.

• Is the last 2 supposed to be an exponent or a factor? Do you mean $2\pi r^2$ or $4 \pi r$? Jan 28, 2018 at 4:34
• The 2 refers to the height which is multiplied, 4πr Jan 28, 2018 at 4:38
• That's the right answer. I don't know how to prove it without calculus, though. Jan 28, 2018 at 4:47

The following is due to Archimedes of Syracuse (287 BC - 212 BC).

First, remeber that the lateral surface area of a right conical frustum is $$\pi(r_1+r_2)S=2\pi mS\tag 1$$ where $r_1$ is radius of the base, $r_2$ is the radius of the top, $m=\frac{r_1+r_2}2$, the "mid-radius", and $S$ is the lateral (slant) height.

Now, consider the following figure.

Here a horizontal slice of the black sphere has been replaced by a red frustum. Also there is a blue cylinder containing the sphere. Consider the blue slice of the cylinder matching the frustum in its height and matching the sphere in its radius. We want to show that the lateral surface area of the blue cylindrical slice equals the lateral surface area of the red frustum.

The lateral surface area of the blue cylindrical slice is$$2\pi rd.$$

How come that $2\pi rd=2\pi mS$? The red filled triangle is similar to the triangle $MNC$. So $$\frac{m}{r'}=\frac d S,$$ from where $$d=\frac{mS}{r'}.$$ Substituting this into the formula for the lateral cylindrical surface, we get $$2\pi \frac{r}{r'}mS.$$ Compare this to $(1).$

Then think: If the spherical slice is very thin then $r'$ is very close to $r$ and the lateral surface area of the frustum is very close to the lateral surface area of the spherical slice... This is all independent from the actual position of the slice we chose.

Then slice the whole spherical cap. The lateral surface area of the slices will equal the corresponding lateral surface area of the cylinder.

Calculus is needed only if the argumentation above seems to be shaky. But the genius did not need calculus.