1.5: Surface Area of Revolution
- Page ID
- 504
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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.