**DERIVING THE SURFACE AREA OF A CONE**

Here's a problem. Given a right circular cone with base radius **r** and
height **h**, what is its surface area?

There's a page on the Web that shows a way to derive the formula using calculus. It's a good demonstration of the method of successive approximations, which is the foundation of integration.

While that's fine if you want to understand higher math, there's another method that uses nothing more than simple plane geometry. All it takes is one crucial insight.

The cone's surface is made of two parts: the base and the side. Our formula will be the sum of the two areas:

A_{cone} = A_{base} + A_{side}

Let's start with the simple part: the base. It's just a disk of radius **r**.
Its area is pi · **r**^{2}. Our area formula is now

A_{cone} = pi · **r**^{2} + A_{side}

Now let's look at the side. Don't let the line identified by **p** scare
you. It's the length of the side of the cone, and can be determined by the Pythagorean
theorem [**r**^{2} + **h**^{2} = **p**^{2}].
Taking the square root of both sides gives **p** = (**r**^{2}
+ **h**^{2})^{1/2}.

We can transform the slanted cone side into a circular disk minus a sector. To do this,
cut the side along **p** and then flatten it. You'll end up with something
like this:

Okay, we have a flattened disk minus a sector. So how do we figure out the size of this
shape? We know that the area of a disk with radius **p** is pi · **p**^{2}.
But that doesn't take into account the missing sector. What is its area?

Here's where the insight comes in. Note the dotted line in the figure above. Consider
it as a measure of how much of the disk is actually there (as opposed to missing). If it's
1, then we have a complete disk; if it's 1/2, then we have half a disk. (Precisely, it's
the ratio of the area of the disk-minus-sector to the full disk.) To find the area of the
disk-minus-sector, take the area of the full disk [pi · **p**^{2}]
and multiply it by this ratio. The problem is now to figure out what the ratio is.

The insight is to realize that the circumference of the base of the cone and curved part of the disk-minus-sector are exactly the same length. This is because the curved edge of the unrolled cone side defined the base of the cone before it was unrolled.

In the diagram below, the two red curves have identical length. (*Note:* the two
images are not to scale.)

Since we know the circumference of the base of the cone [2 · pi · **r**],
the curved part of the disk is in ratio (2 · pi · **r**) / (2 · pi · **p**)
= **r**/**p** to the entire disk. Multiplying this ratio by the
area of the full disk gives (**r**/**p**) · pi · **p**^{2}
= pi · **r** · **p**.

This is the area of the side.

Adding it all up gives

A_{cone} = pi · **r**^{2} + pi · **r**
· **p**

Factoring out terms:

A_{cone} = pi · **r** · (**r** + **p**)

Substituting for **p** gives the final formula:

A_{cone} = pi · **r** · (**r** + (**r**^{2}
+ **h**^{2})^{1/2})

*Et voilą!*. Here is the formula. No calculus, just one insight.

Let's do a quick sanity check on it. We would expect that as **h**
increases, so does the surface area. Does it? As **h** increases, so does **p**,
so the surface does increase. Good.

Now, what happens as **h** decreases to zero? The cone gets flatter and
flatter, with the tip approaching the base. So the surface area should approach two times
the area of the base. If we plug in zero for **h**, we get pi · **r**
· (**r** + **r**) = 2 · pi · **r**^{2}.
Looks like we have a winner.

Now that all that's over, I have to confess that I wrote the page not only because I like this derivation but because I can never remember the formula.

- The circumference of a circle of radius
**r**is 2 · pi ·**r** - The area of a disk of radius is
**r**is pi ·**r**^{2} - Given a right triangle with sides
**a**,**b**, and hypotenuse**c**,**a**^{2}+**b**^{2}=**c**^{2}[The Pythagorean theorem]

All contents ©1999 Mark L. Irons