1.5: Surface Area of Revolution

[ "article:topic", "frustum", "surface of revolution", "authorname:green" ]

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The area of a frustum is

\[ A = 2\pi r(length). \]

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

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

Example 1

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

\[ y = x^2\]

from 2 to 3.

Solution

We find

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

Now use the area formula:

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

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

\[A \approx 208.09.\]