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Math Help - area of a surface of revolution

  1. #1
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    area of a surface of revolution

    Find the area of the surface obtained by rotating the curve about the x-axis.?

    y = sqrt(2 + 2 x)

    3<=x<=11



    I know that the formula used for this is S =(integral from 3to11) 2*pi*y*sqrt(1+(y')^2)dy

    for y' i get y'=1/( 2*sqrt(2+2x) )

    once i plug in both y and y'. i get stuff because i'm not sure how it will all work out with the u-substitution and all.

    can someone help? thanks!
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  2. #2
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    Quote Originally Posted by break View Post
    Find the area of the surface obtained by rotating the curve about the x-axis.?

    y = sqrt(2 + 2 x)

    3<=x<=11



    I know that the formula used for this is S =(integral from 3to11) 2*pi*y*sqrt(1+(y')^2)dy

    for y' i get y'=1/( 2*sqrt(2+2x) )

    once i plug in both y and y'. i get stuff because i'm not sure how it will all work out with the u-substitution and all.

    can someone help? thanks!
    The surface area is the integral of circle circumferences, multiplied by \sqrt{(dy)^2+(dx)^2} rather than dx from x=3 to 11,
    having y as radius.

    \sqrt{(dy)^2+(dx)^2}=\sqrt{\left(\frac{dy}{dx}\rig  ht)^2+\left(\frac{dx}{dx}\right)^2}\ dx

    f(x)=(2+2x)^{0.5}

    \frac{d}{dx}\ f(x)=0.5(2+2x)^{-0.5}2=\frac{1}{\sqrt{2+2x}}

    \sqrt{1+\left(\frac{dy}{dx}\right)^2}=\sqrt{1+\fra  c{1}{2+2x}}

    =\sqrt{\frac{2+2x+1}{2+2x}}

    \int_{3}^{11}{(2{\pi}r)}\sqrt{(dy)^2+(dx)^2}=2{\pi  }\int_{3}^{11}{\sqrt{2+2x}}\sqrt{\frac{3+2x}{2+2x}  }dx

    =2{\pi}\int_{3}^{11}{\sqrt{3+2x}}\ dx

    u=3+2x

    du=2dx

    dx=\frac{du}{2}

    If you like you can change the limits and work exclusively with u,
    or leave the limits alone and switch back to x having evaluated the integral.

    {\pi}\int_{9}^{25}{u^{\frac{1}{2}}}\ du

    Or

    {\pi}\int_{x=3}^{x=11}{u^{\frac{1}{2}}}\ du

    ={\pi}\frac{2}{3}\left[(3+2x)^{\frac{3}{2}}\right] from x=3 to x=11

    ={\pi}\frac{2}{3}\left[\left(25\right)^{\frac{3}{2}}-9^{\frac{3}{2}}\right]
    Last edited by Archie Meade; May 21st 2010 at 05:13 PM.
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  3. #3
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    Quote Originally Posted by Archie Meade View Post
    The surface area is the integral of circle circumferences, from x=3 to 11,
    having y as radius.

    \int_{3}^{11}{(2{\pi}r)}dx=2{\pi}\int_{3}^{11}{\sq  rt{2+2x}}\ dx
    This is incorrect! Please see this: Surface of Revolution -- from Wolfram MathWorld
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  4. #4
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    Quote Originally Posted by break View Post
    Find the area of the surface obtained by rotating the curve about the x-axis.?

    y = sqrt(2 + 2 x)

    3<=x<=11



    I know that the formula used for this is S =(integral from 3to11) 2*pi*y*sqrt(1+(y')^2)dy

    for y' i get y'=1/( 2*sqrt(2+2x) )

    once i plug in both y and y'. i get stuff because i'm not sure how it will all work out with the u-substitution and all.

    can someone help? thanks!
    y=\sqrt{2+2x} = (2+2x)^{1/2}

    y' = \frac{1}{2} (2+2x)^{-1/2} \cdot 2 = \frac{1}{\sqrt{2+2x}}

    So:

    1+(y')^2 = 1+ \frac{1}{2+2x} = \frac{3+2x}{2+2x}

    So:

    y \cdot \sqrt{1+(y')^2} = \sqrt{2+2x} \cdot \sqrt{\frac{3+2x}{2+2x}} = \sqrt{3+2x}

    So the integral to solve is

    2 \pi \int_3^{11} \sqrt{3+2x} \, dx
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