Surface Area of a Sphere Given the Circumference
In practical applications, it is easier to calculate the surface area of a sphere from the circumference rather than the radius. You can measure the circumference of a ball by wrapping a string or tape measure around its equator, but it's not so simple to measure the radius, since you would need to be inside the ball. Once you know the circumference C of any spherical object, you can plug this value into a formula to compute the square inches, square footage, or square centimeters of space of the surface. You can also calculate spherical volume from the circumference. Here is the surface area formula, its derivation, and several examples.
If a ball or sphere has a circumference of C, then its surface area is given by the surprisingly simple equation
Spherical Surface Area = (C^2)/pi
In other words, square the circumference measurement and divide that number by π, which is approximately 3.14159. Now we will see how this formula is derived.
The standard formula for the surface area of a sphere is given in terms of the radius R. That geometric formula is
Surface Area = 4*pi*R^2
The circumference and radius of a sphere are related by the equation C = 2*pi*R, which is equivalent to C/(2*pi) = R. Plugging this expression into the first formula gives you
Surface Area = 4*pi*[C/(2*pi)]^2
An large inflatable ball has a girth of 2.3 meters. What is its surface area in square centimeters?
For this problem, we first convert the circumference to 230 centimeters. Next, we square this value and divide by pi.
(230*230)/3.14159 = 16838.6 square centimeters of surface area, or approximately 1.68 square meters.
A soccer ball is inflated so that its circumference increases by 12.4%. By how much does its surface area increase?
Our original circumference is C and our original surface area is (C^2)/π. Our new circumference is 1.124C, and our new surface area is
[(1.124C)^2]/π = 1.263376(C^2)/π
This is an increase of 26.3376% over the original surface area. This figure is obtained by taking the factor 1.263376 and subtracting 1, then converting the remaining decimal to a percent.
Related Geometry Articles
Which is greater? The surface area of a sphere with a circumference of 6π, or the total surface area of two spheres whose circumferences are each 3π?
In the first case, the total surface area of the single ball is (6π)^2 ÷ π = (36π^2)/π = 36π.
In the second case, the total surface area of the two balls is 2*(3π)^2 ÷ π = 2*(9π^2)/π = 18π.
The single sphere has double the surface area of the two twin spheres.
The planet Mars has an average circumference of 21,344 kilometers. Using the formula, its approximate surface area in square km is (21344*21344)/3.14159265 ≈ 145,011,269 km^2. In comparison, the surface area of Earth is about 510,100,000 km^2.
Solving Backward: Circumference from Surface Area
Using the spherical geometry formula S = (C^2)/π you can calculate the circumference of a ball using the area of the surface; just solve the equation for C. This gives you C = sqrt(S*π). In other words, multiply the surface area by pi, then take the square root. For example, if a sphere has a surface area of 247.36 square centimeters, then the circumference is sqrt(247.36*3.4159) = 29.07 centimeters.
Related Rates Calculus Example
Related rates problems are something you learn in calculus to discover how a change in one quantity produces changes in a related quantity. Lets say you have a perfectly spherical snowball whose circumference is 15 inches. As the snowball is exposed to the sun, its surface area decreases by 0.5 square inches per minute, at a constant continuous rate. What is the instantaneous rate of change of the circumference when the surface area is 62 square inches?
First we have the equation
S(t) = [C(t)^2]/π
where S(t) and C(t) are the surface area and circumference of the snowball at t minutes. Taking the derivative of both sides gives us the related rates equation
dS/dt = (2/π) * C(t) * dC/dt
We want to solve this equation for dC/dt. At the moment when the snowball's surface area is 62 square inches, its circumference is sqrt(62π) = 13.9563 inches. Now using C(t) = 13.9563 and dS/dt = -0.5, we have
-0.5 = (2/π) * 13.9563 * dC/dt
-0.0563 = dC/dt
So, at the moment when the surface area is 62 square inches, the circumference is shrinking at a rate of 0.0563 inches per minute.