Skip to main content
\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)
Mathematics LibreTexts

1.5: Surface Area of Revolution

  • Page ID
    504
  • [ "article:topic", "frustum", "surface of revolution", "authorname:green" ]

    \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \)

    \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)

    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.