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Surface area of a polar equation

  1. May 1, 2008 #1
    Hello, the problem I'm working on is to find and set up the integral whose value is the area of the surface obtained by rotating the curve about the x-axis, then another integral to find the surface area by rotating about the y-axis. I do not need to evaluate these integrals, just set them up. (I'm sorry I'm not using prper variable and notation signs in some parts, i'm worried about them not showing up correctly.)

    r = 1 + sin(4*Θ)
    where 0<= Θ <= 2pi

    I understand that I can find surface area of parametric equations using
    S = ʃ 2(pi)y sqrt([dx/dt]² + [dy/dt]²) dt
    (a->b)

    I'm also familiar with:
    x = r cos(Θ)
    y = r sin(Θ)
    r = sqrt(x² + y²)
    Θ = tan(y/x)

    And lastly, with all my efforts, basically all I did was write a bunch of stuff down and hope something jumped out at me.

    I drew a triangle and labeled it...(you can laugh at my attempt to draw it, you can also tell where theta is supposed to go)

    .
    |\
    y | \ r = 1 + sin(4Θ) = sqrt(x² + y²)
    | \
    ----`
    x

    and so sin(Θ) = y / (1 + sin(4Θ))
    and sin(Θ) = y / (sqrt(x² + y²))

    1 + sin(4Θ) = sqrt(x² + y²)

    but it turns out I don't really know what I'm doing, I have no direction. Any help is greatly appreciated, thanks a lot!
     
  2. jcsd
  3. May 1, 2008 #2

    rock.freak667

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    The surface area obtained by rotating the curve about the x-axis is given by

    [tex]\int 2\pi y ds[/tex]

    Your formula is correct, but your parameter is [itex]\theta[/itex] and not t.

    So if you substitute in it..you could easily get a formula for it.
     
  4. May 1, 2008 #3
    Wait...I'm sorry, I don't know what you mean. Could you elaborate a little further? I think my question is substitute what for what?
     
    Last edited: May 1, 2008
  5. May 1, 2008 #4

    rock.freak667

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    my bad if I wasn't too clear.

    Let [itex]x=rcos\theta;y=rsin\theta[/itex]

    The surface area of revolution is given by

    [tex]\int 2\pi y ds[/tex]

    So we need to find ds (arc length)

    in parametric form:

    [tex]ds = \sqrt{\left (\frac{dx}{d\theta}\right ) ^2 + \left (\frac{dy}{d\theta}\right ) ^2} d\theta[/tex]

    [tex]\frac{dx}{d\theta}=\frac{dr}{d\theta}cos\theta}-rsin\theta[/tex]

    [tex]
    \frac{dy}{d\theta}=\frac{dr}{d\theta}sin\theta}+rcos\theta[/tex]

    [tex]\left (\frac{dx}{d\theta}\right ) ^2 + \left (\frac{dy}{d\theta}\right ) ^2

    = (\frac{dr}{d\theta}cos\theta}-rsin\theta)^2 + (\frac{dr}{d\theta}sin\theta}+rcos\theta)^2[/tex]

    [tex](\frac{dr}{d\theta})^2cos^2\theta -2r\frac{dr}{d\theta}cos\theta sin\theta +r^2sin^2\theta + (\frac{dr}{d\theta})^2 sin^2\theta+ 2r\frac{dr}{d\theta}cos\theta sin\theta + r^2cos^2\theta[/tex]

    (long and tedious) simplifies to

    [tex]r^2+ \left( \frac{dr}{d\theta}\right )^2[/tex]

    But end story is

    [tex] ds= \sqrt{r^2 +\left( \frac{dr}{d\theta} \right )^2} d\theta[/tex]
     
  6. May 1, 2008 #5
    Yuck. Haha, thanks so much for the help, saved the day for me. I think I understand now.

    But let me ask one more question, in the original formula
    [tex]\int 2\pi y ds[/tex]
    what would I do with that y variable? Leave it as a variable? Wouldn't that leave the result still with y, r, and theta?

    And what if I wanted to rotate about the y axis instead of the x axis? Is the formula the same, except with x instead of y?

    It appears that I'm getting tired from doing these problems all day...
     
    Last edited: May 1, 2008
  7. May 1, 2008 #6

    rock.freak667

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    y=rsin[itex]\theta[/itex]

    if around the y-axis then yes.
     
  8. May 1, 2008 #7
    Oh, right, that makes sense. Thanks again for all your help friend, you've been a great help.
     
  9. May 1, 2008 #8

    rock.freak667

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    No problem.
     
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