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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.


    This page titled 1.5: Surface Area of Revolution is shared under a not declared license and was authored, remixed, and/or curated by Larry Green.

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