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Surface Area and Volume

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 πr2, 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 = 4πr2

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 = 4πr2

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

SA = 4π(2 cm)2
SA = 16π ≈ 50.3 cm2

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 hemisphere, which is half the surface area of the sphere plus the area of the great circle.

SA = 2πr2 + πr2 = 3πr2

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

SA = 3πr2
SA = 3π(5 in)2
SA = 75π ≈ 235.6 in2

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…

a2 + b2 = c2

Substitute in the values we know.

(6.2 ft)2 + (5 ft)2 = c2

And solve for the sphere's radius.

c = r ≈ 7.96 ft

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

SA = 4πr2
SA = 4π(7.96 ft)2
SA = 796 ft2

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