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Math Help - Area Under a Curve

  1. #1
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    Area Under a Curve

    Also if some1 can help me in this question that would be great.

    Find the equation of the tangent to the parabola y=2x^2 at (1,2). Calculate its point of intersection with the x-axis and the volume of the solid formed when the area between the parabola, the tangent line and the xaxis is reveloved about the xaxis.

    Especially the bit about forming a solid of revolution.

    The tangent, has equn of
    y=4x-2
    THe pt of intersection is (1/2,0)

    But the area formed i found to be 2/15 pi, which is different to the answer. Much help would be appreciated.
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  2. #2
    MHF Contributor Calculus26's Avatar
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    I believe 2pi/15 is correct

    Draw a diagram and you'll see you need 2 integrals

    between x = 0 and 1/2 you have disks

    V= (pi)integral(4x^4dx) = pi/40

    between x =1/2 and 1 you have washers

    V = pi integral (4x^4 - (4x-2)^2)dx) =13pi/120

    Adding the 2 you get 16pi/120 = 2pi/15
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  3. #3
    MHF Contributor Calculus26's Avatar
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    See attachment for diagram and set up
    Attached Files Attached Files
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  4. #4
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    Quote Originally Posted by Calculus26 View Post
    See attachment for diagram and set up
    Thanks SO much. Especially for the diagram.
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