Derivation of Formula for Total Surface Area of the Sphere by Integration Video

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  • 0:04 Surface Area of a Sphere
  • 0:37 Radius, Angle, & Arclength
  • 2:38 Using Integration
  • 4:55 Lesson Summary
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Lesson Transcript
Instructor: Gerald Lemay

Gerald has taught engineering, math and science and has a doctorate in electrical engineering.

The total surface area of a sphere is found using an equation. In this lesson, we derive this equation using the arclength definition, radius relationships, and integral calculus.

The Surface Area of a Sphere

The idea of a surface area is easy to grasp when the object is flat like a square or a triangle. Even a three-dimensional object made up of flat surfaces has an easy to understand equation for the surface area. But what about a curved surface? What about a sphere?

A sphere is a special kind of three-dimensional object. If we know the radius, R, we can find other features of the sphere like its volume and surface area.

For example, the total surface area, A, is 4π times the square of R. In this lesson, we show how to derive this equation.

Radius, Angle, & Arclength

The radius, R, goes from the center of the sphere to it's outer surface. We could also draw another radius, r. This radius also touches the surface, but it starts perpendicular to a vertical line passing through the center of the sphere:

An angle, θ, provides a relationship between R and r:


We can displace r:

Use the definition of cosine (adjacent side over the hypotenuse):


Solving for r:


Great! We will use this r-R relationship later on. Now, lets delve into the idea of arclength.

We know once around a circle is 360o which is the same as 2π radians. We also know, the perimeter of a circle as being 2πR.

If θ is some arbitrary angle less than 2π, then θ / 2π is a fraction. The length along the arc (better known as the arclength), s, subtended by θ is a fraction of the total length 2πR. This fraction is given by s/2πR. These fractions are equal:


The 2π factors cancel from each side of the equation, leaving:


Solving for s:


We can take the derivative of both sides:


On the second line, we used the product rule which gives two terms. But dR is 0 because R is a constant. Meaning, θ dR is 0. Thus, we are left with only one term on the third line, giving us another very useful result:


Using Integration

With these results, we'll integrate to derive the total surface area equation. But first, lets revisit the circle traced out by r.

If we draw another circle immediately below it with the same radius we get the circle appearing here:


The separation between the two circles is an infinitely small arclength, ds:


We can open this band to get a rectangle:


The rectangle area, da, is a differential area equal to the length times the width. This rectangle has a length, 2πr, and a width, ds. Thus, the area, da, of this infinitely thin cylinder is:


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