Standard formulas for the area of a circle, and the surface area and volume of a sphere, involve the constant π. Most of us remember that π is about 3.1416. That’s pretty close, but not exact. Even closer, π is about 3.1415926536. Now that’s really close, but still not exact. π has no exact numerical value – it’s been computed out to more than a million decimal places, but no matter how many we compute, there are always millions more. So there’s really no point in asking for the exact value of π. We can come as close as we want – or that we can afford (in terms of computer time).

Another way of looking at it is to realize that there is no pair of integers, a and b, such that a/b = π. There are a few that come close. The first is 22/7 = 3.142857…. (the decimal part repeats). Then comes 355/113 = 3.141592920353982…. (that one goes on for at least 30 decimal places with no repeat in sight)

This gives us a hint that there may be a way of calculating approximate areas and volumes that are “close enough”.

Start with the standard formula:

Circle area:** A = πr ^{2}**

It’s not always convenient to measure the radius directly. We can measure the diameter and divide by 2, but suppose we could work with the diameter directly:

A = πr^{2}

r = d/2

A = π(d/2)^{2}

A = πd^{2}/4

which we rewrite as

A = (π/4)d^{2}

We can look up the value for π/4; it’s almost exactly 0.7854.

So our area formula becomes

A ~= 0.7854d^{2}

We use ~= to mean “approximately equal.

A very good approximation is

A ~= 0.8d^{2}

this is within slightly less than 2%

It’s easy to multiply something by .8: double it three times, then divide by 10.

If d=12 inches, then A = 0.8*12 = 9.6 in^2.

Let’s see how that works out:

Suppose you have a garden with a 10-foot diameter circular area you want to plant with flowers. You’ll need the area. The actual area is 78.53982 square feet. Using the “very good approximation”, we get 78.54 square feet.

The difference is 0.00018 square feet, or about 1/4 square inch. And, it’s always a little lager than the exact value, so you never have to worry about running out.

We can also get the area if we know the circumference. The circumference of a circle is not so easy to measure – unless you’re working with a cylinder (like a tin can) and you need to get the volume. But the formula will be useful later on, so let’s develop it here.

A = πr^{2}

c = πd

c = 2πr

r = c/2π

A = π(c^{2}/4π^{2})

**A = c ^{2}/4π**

**Moving right along to spheres**

First, the formulas:

**A _{s} = 4πr^{2}**

You may recognize the “πr^{2}” part as the area of a circle with radius r. The surface area of a sphere is exactly 4 times the area of the circle whose center is the center of the sphere (sometimes known as a great circle). An old Greek guy figured that out, before anybody knew what π was.

Rewriting that to use the diameter:

r = d/2

A_{s} = 4π(d/2)^{2}

A_{s} = 4πd^{2}/4

Which simplifies to

A_{s} = πd^{2}

That’s nice, but no help from the world of approximations. But note that the area of a sphere is 4 times the area of a great circle. We start over:

**A _{s} = 4A_{c}**

Now, we just go back to our circle area approximate formula:

A_{c} ~= 0.8d^{2}

Plug that into the sphere area formula:

A_{s} ~= 4*0.8d^{2}

A_{s} ~= 3.2d^{2}

Finally, let’s find the area from the circumference. That’s a lot easier to measure, on a sphere. (Think of a basketball.)

A_{s} = 4A_{c}

A_{s} = 4(c^{2}/4π)

A_{s} = c^{2}/π

Calculating, 1/π = 0.3183, which we can round off to 0.3, or

A_{s} ~= .3c^{2}

This gives a result that is just 6% too big.

Finally, the volume of a sphere:

**V _{s} = (4/3) * πr^{3}**

From the diameter:

r = d/2

V_{s} = (4/3) * π(d/2)^{3}

V_{s} = (4/3) * π(d^{3}/8

V_{s} = (π/6) * d^{3}

Calculate π/6 = 0.5326 Using 0.5 is too far off, so we’ll need to use 0.53:

V_{s} ~= 0.53d^{3}

Now let’s step back and ask if it’s possible to calculate the area of a circle without using π.

Start with the basic formula

**A = πr ^{2}**

Rewrite that as

A = πr*r

From

r = d/2

replace one of the r’s:

A = π(d/2)*r

A = πd*r/2

But πd = c, so

A = c*r/2

The trick is that π is contained in c – because c = πr.