# Calculate the Volume of a Sphere from Its Circumference

The formula for the volume of a sphere is traditionally given as a function of the radius. However, in practical applications, it is not always easy to determine the diameter or radius of a round object.

The circumference of a ball of sphere is much easier to measure. Simply wrap a measuring tape around the widest part of the sphere (the equator) and record the distance all the way around. Since the circumference and radius are related by a simple formula, you can then determine the radius of the spherical object, which allows you to calculate the volume. Here is the formula with some examples.

See also: How to Calculate the Surface Area of a Sphere from Its Circumference

## The Equation for Volume in Terms of the Circumference

First let's recall the formulas for volume (v) of a sphere and its circumference (c) in terms of the radius (r). These geometric equations are

v = (4/3)πr^{3}c = 2πr

If you solve the second equation for r, you get

c/(2π) = r

Now plug this expression into the volume equation as a substitution for r, in other words, replace r with the expression c/(2π). This gives you the equation

v = (4/3)π[c/(2π)]^{3}= (4/3)π[c^{3}/(8π^{3})]

= c^{3}/(6π^{2})

In other words, the volume of a sphere is the circumference cubed divided by the number 6π^{2}, which is approximately 59.2176.

## Example Calculation

**Problem:** Suppose an inflatable ball has a circumference of 60 cm. What is the volume of the ball in cubic centimeters (cc) and liters?

**Solution:** Since c = 60, we plug that value into the equation v = c^{3}/(6π^{2}) to find the volume. This gives us

v = 60^{3}/(6π^{2})

= 216000/59.2176

= 3647.5626 cc

= 3.648 liters

In practical terms, what this means is that if you deflate the ball with plans to reinflate it later, you will need to pump it up with about 3.648 liters of air.

## Another Example

**Problem:** Molly discovers that when the circumference of her ball increases by 2 inches, its volume increases by 100 cubic inches. What is the original circumference of the ball?

**Solution:** Let's say the original circumference is c and the original volume is v. This gives us the equation

v = c^{3}/(6π^{2})

When c increase by 2, v increases by 100. This gives us the new equation

v + 100 = (c + 2)^{3}/(6π^{2})

If we subtract the first equation from the second, we get a single equation that only involves the unknown variable c.

100 = (c + 2)^{3}/(6π^{2}) - c^{3}/(6π^{2})

With a little algebra, we can simplify this seeming cubic equation into a quadratic

600π^{2} = 6c^{2} + 12c + 8

Finally, using the quadratic formula or a polynomial solver, we get c ≈ 30.41 inches. To double check that this solution is correct, let's compute the volume of a sphere with a circumference of 30.41 inches and 32.41 inches and verify that the volumes differ by about 100.

(30.41)^{3}/59.2176 = 474.896

(32.41)^{3}/59.2176 = 574.892

## Another Example

**Problem:** Jan wants to make some round polymer clay beads with a circumference of 3 cm. If she wants to make 150 such beads, how much clay does she need?

**Solution:** Each bead has a volume given by

v = (27)/(6π^{2})

= 0.455945 cm^3

Since she needs 150 of them the total volume of clay she needs is (150)(0.455945) = 68.3918 cm^3.

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## 11 comments

Awesome! I like having a volume formula directly in terms of circumference instead solving for the radius, then plugging the radius into the usual volume formula. It's interesting that the circumference-to-volume formula has pi squared in the denominator, didn't expect that.

It really helped me with my math homework!!! Thanks, you're a lifesaver!!!

This is great, thank you. Also, what is volume of a cylinder from its circumference?

thanks, it's super helpful!

Can you help me solve this problem? A tall cyindrical bucket has a diameter of 12 inches and is filled halfway with water. If you drop a solid ball in the bucket and the ball has a diameter of 7 inches, how many inches will the water level rise? I think this is a volume problem, but I'm not sure what to do with the cylinder.

More practical to measure spherical volume this way!

Thanks! it helped a lot.

11