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Math Help - Solid of revolution

  1. #1
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    Solid of revolution

    The area under the curve y= sin x, between x =0 and x = π
    Show that the volume of the solid formed by rotating this area through 360 about the x-axis is exactly (π^2)


    This is what I have after doing some of the work:
    V = π (-cos^2x) limits π and 0
    I am not sure how to take it any further to obtain the answer. Please help me out guys...
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  2. #2
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    Quote Originally Posted by mike789 View Post
    The area under the curve y= sin x, between x =0 and x = π
    Show that the volume of the solid formed by rotating this area through 360 about the x-axis is exactly (π^2)


    This is what I have after doing some of the work:
    V = π (-cos^2x) limits π and 0
    I am not sure how to take it any further to obtain the answer. Please help me out guys...
    your antiderivative is incorrect ... use the power reduction identity that follows.

    \displaystyle V = \pi \int_0^\pi \sin^2{x} \, dx

    \displaystyle V = \int_0^\pi \frac{1-\cos(2x)}{2} \, dx

    now try it again.
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  3. #3
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    Quote Originally Posted by skeeter View Post
    your antiderivative is incorrect ... use the power reduction identity that follows.

    \displaystyle V = \pi \int_0^\pi \sin^2{x} \, dx

    \displaystyle V = \int_0^\pi \frac{1-\cos(2x)}{2} \, dx

    now try it again.
    Do i need to integrate and substitute the limits?
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  4. #4
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    Quote Originally Posted by mike789 View Post
    Do i need to integrate and substitute the limits?
    if that's what you mean by evaluating the definite integral using the Fundamental Theorem of Calculus, then yes.
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