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Math Help - Help setting up integral to find surface area

  1. #1
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    Aug 2008
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    Help setting up integral to find surface area

    I need help with the following problem:

    Let C be the curve defined by y=2sin(3x) on the interval [0,Pi/2]. I need help setting up the integral required to find the surface area of the surface that results when C is revolved about the specified axis:

    a.) The x-axis
    b.) The y-axis

    I think for a you would use S=2Pi∫(from 0 to Pi/2) 2sin(3x)*sqrt(1+(2cos(3x) *3)^2)

    I think for b you would use S=2Pi∫(from 0 to Pi/2) x*sqrt(1+2cos(3x)*3)^2)

    Have I set these integrals up correctly or have I done something wrong? Thanks in advance to anyone who can help me.
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  2. #2
    Super Member
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    Aug 2008
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    For the equation x=t,;\ y(t)=2\sin(3t), the surface areas of the figures around the x and y axes respectively are:

    A_y=2\pi\int x ds

    A_y=2\pi\int y ds

    with ds=\sqrt{(x')^2+(y')^2}

    Then for f(x)=2\sin(3x);\; 0\leq x\leq \pi/2, in parametric form that's x(t)=t,\;y(t)=2\sin(3t);\;0\leq t\leq \pi/2 and thus:

    A_y=2\pi\int_0^{\pi/2} t\sqrt{1+(6\cos(3t))^2}dt

    A_x=2\pi\int_0^{\pi/2} 2\sin(3t)\sqrt{1+(6\cos(3t))^2}dt
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