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Math Help - Surface Area of a Revolved Parametric Equations

  1. #1
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    Smile Surface Area of a Revolved Parametric Equations

    I have worked this problem over but still can not get a reasonable answer. I would like some help, please.

    I am given the parametric equations: x=e^t and y=(1/3)e^3t. I am to revolve the resulting shape around the x-axis and calculate the surface area. I know that I must use
    2*pi*the radius*sqrt((dx/dt)^2 + (dy/dt)^2)
    in order to find the surface area. However, I'm not sure what my radius is and the integral is difficult because of the tricks it needs.

    I could really use some assistance ASAP.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ekeonee888 View Post
    I have worked this problem over but still can not get a reasonable answer. I would like some help, please.

    I am given the parametric equations: x=e^t and y=(1/3)e^3t. I am to revolve the resulting shape around the x-axis and calculate the surface area. I know that I must use
    2*pi*the radius*sqrt((dx/dt)^2 + (dy/dt)^2)
    in order to find the surface area. However, I'm not sure what my radius is and the integral is difficult because of the tricks it needs.

    I could really use some assistance ASAP.
    Do you have to do it parametrically? The y(x) formula is very simple.

    -Dan
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  3. #3
    Super Member wingless's Avatar
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    x=e^t

    y=\frac{1}{3}e^{3t}

    y=\frac{1}{3}x^3

    This will make it a lot easier.
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  4. #4
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    But one must put it in terms of t with the bounds 0<=t>=(ln3)/4
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  5. #5
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    If you are given bounds for t then you know the bounds of x, as x is defined as a function of t.
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  6. #6
    Super Member wingless's Avatar
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    As it's easier to work with respect to x, put the bounds for t in x=e^t to find what they correspond for x and then integrate using these bounds.
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Ekeonee888 View Post
    I have worked this problem over but still can not get a reasonable answer. I would like some help, please.

    I am given the parametric equations: x=e^t and y=(1/3)e^3t. I am to revolve the resulting shape around the x-axis and calculate the surface area. I know that I must use
    2*pi*the radius*sqrt((dx/dt)^2 + (dy/dt)^2)
    in order to find the surface area. However, I'm not sure what my radius is and the integral is difficult because of the tricks it needs.

    I could really use some assistance ASAP.
    \frac{dx}{dt} = e^t

    \frac{dy}{dt} = e^{3t}

    So
    A = 2 \pi \int_0^{ln(3)/4} e^t \sqrt{e^{2t} + e^{6t}}~dt

    = 2 \pi \int_0^{ln(3)/4} e^{2t} \sqrt{1 + e^{4t}}~dt

    Let u = e^{2t} \implies du = 2 e^{2t}~dt, so
    A = \pi \int_1^{3/2} \sqrt{1 + u^2}~du

    Can you finish this now?

    -Dan
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