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# 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

$A \approx 208.09.$

Larry Green (Lake Tahoe Community College)

• Integrated by Justin Marshall.