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# Surface Area and Volume

# Surface Area of Spheres

If you play basketball, eat oranges, or have eyeballs (which are sort of necessary to read this), you've most likely encountered spheres at some point in your life. They're such common everyday objects so learning how to find their surface area should prove useful...you know, in case you've ever wondered how much peel is actually on an orange.

Before we do that though, we need to talk about the Great Circle of Life.

Er...sort of. A **great circle** is a circle that has the same center and radius as its sphere. Imagine cutting a sphere exactly in half, creating two **hemispheres**. The circle we get is a great circle. Circles anywhere else on the sphere aren't that great. They're mediocre.

How does this help us find the surface area? Patience, young grasshopper.

The great circle (with area π*r*^{2}, obv) covers about a fourth of the sphere. Actually, it covers *exactly* a fourth of the sphere. That means our surface area for the entire sphere is four times the area of the great circle.

*SA* = 4π*r*^{2}

Coolness. We can get started on those surface areas. Hold your applause, please.

### Sample Problem

What is the surface area of this sphere?

This shouldn't be a problem. We'll start with our formula.

*SA* = 4π*r*^{2}

Our radius is 2 centimeters. That's all we need.

*SA* = 4π(2 cm)^{2}*SA* = 16π ≈ 50.3 cm^{2}

Could it get any simpler? Probably not. Could it get any harder? Unfortunately, always.

### Sample Problem

What is the surface area of this hemisphere?

This time, our formula is going to be a smidgeon more complicated. We're looking for the surface area of the *hemi*sphere, which is half the surface area of the sphere plus the area of the great circle.

*SA* = 2π*r*^{2} + π*r*^{2} = 3π*r*^{2}

Our diameter is 10 inches, so our radius is 5 inches.

*SA* = 3π*r*^{2}*SA* = 3π(5 in)^{2}*SA* = 75π ≈ 235.6 in^{2}

That's all there is to it. Well, let's throw in one more curveball to see if you really get it. (Get it? Because a curveball is a sphere, too!)

### Sample Problem

The small circle intersects the sphere where its center is 6.2 feet away from the center of the sphere. If the radius of the small circle is 5 feet, what is the surface area of the sphere?

Whoa, whoa, whoa. What in the name of Ke$ha is going on here? Let's make sense out of all of this.

The distance between the centers of the circle and the sphere is 6.2 feet, which makes a right angle with the $5 footlong radius of the circle. If we make it a right triangle and look at the hypotenuse, it's the light at the end of this spherical tunnel: the radius of the sphere.

The Pythagorean theorem, like the Energizer bunny, keeps going on and on and on…

*a*^{2} + *b*^{2} = *c*^{2}

Substitute in the values we know.

(6.2 ft)^{2} + (5 ft)^{2} = *c*^{2}

And solve for the sphere's radius.

*c* = *r* ≈ 7.96 ft

Once we have that, surface area's a breeze.

*SA* = 4π*r*^{2}*SA* = 4π(7.96 ft)^{2}*SA* = 796 ft^{2}