1.5: Surface Area of Revolution

The area of a frustum is

$$A = 2\pi r(length). \nonumber$$

If we revolve a curve around the x-axis, we have that the surface area of revolution is given by

$$\text{Area} = 2\pi \int _a^b y \sqrt{1+\left( \dfrac{dy}{dx} \right)^2} dx. \nonumber$$

Example 1

Set up an integral that gives the surface area of revolution about the x axis of the curve

$$y = x^2 \nonumber$$

from 2 to 3.

Solution

We find

$$\left(\dfrac{dy}{dx} \right)^2=(2x)^2 = 4x^2. \nonumber$$

Now use the area formula:

$$A = 2\pi\int_2^3 x^2\sqrt{1+4x^2} dx. \nonumber$$

We will learn later how to work out this integral. However a computer gives that

$$A \approx 208.09. \nonumber$$

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.