Determining the Surface Area of a Solid of Revolution

Key Questions

  • How do you find the surface area of the solid obtained by rotating about the y-axis the region bounded by y=x^2 on the interval 1<=x<=2 ?

    First of all, you are missing a bound. We will assume that the other bound is y=0 or the x-axis. The answer is (15pi)/2.

    The first step is to determine whether you are rotating along an axis that is parallel to the independent axis or the axis of the parameter (x in this case). And we are not, so this integration should be done with cylindrical shells.

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    Always draw a diagram to verify what is the parameter and what is the function.

    You should note that y is not always a parameter of x. For instance, x=y^2, x is now a parameter of y.

    The formula for cylindrical shells is:

    V=int_a^b2pirhdr
    h is represented by y, we have y=x^2 and y=0
    r is represented by x
    V=int_1^2 2 pi x (x^2-0) dx

    Now that the substitutions are done, we can solve:

    V=2 pi int_1^2x^3dx
    =2pi (x^4)/4|_1^2
    =2pi([2^4-1^4])/4
    =(15pi)/2

    Amory W. · · Sep 5 2014
  • How do you find the surface area of the solid obtained by rotating about the x-axis the region bounded by y=e^x on the interval 0<=x<=1 ?

    The answer is pi/2[e^2-1].

    Since you are only given a single function and we are rotating about the axis of the parameter, this requires the disk method. The disk method is:

    V=int_a^b Adx
    =int_a^b pi r^2dx
    =int_a^b pi [f(x)]^2dx

    We have the known values:

    f(x)=e^x
    a=0
    b=1

    And now we can substitute:

    V=int_0^1 pi (e^x)^2dx
    =pi int_0^1 e^(2x)dx
    =pi (e^(2x))/2|_0^1
    =pi/2[e^2-e^0]
    =pi/2[e^2-1]

    Amory W. · · Sep 16 2014
  • How do you find the surface area of a solid of revolution?

    If the solid is obtained by rotating the graph of y=f(x) from x=a to x=b, then the surface area S can be found by the integral

    S=2pi int_a^b f(x)sqrt{1+[f'(x)]^2}dx

    Wataru · · Sep 21 2014

Questions

Applications of Definite Integrals