Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Finding Surface area of a Parametric Curve

  1. Oct 1, 2005 #1
    Can someone please help me with this question?

    x = 1-sint, y = 2+cost, rotate about y = 2

    Find the surface area of the parametric curve.

    I don't know how to do it with y=2, I only know how if the question askes for rotating about the x-axis.
    The answer to the question is 2(pi)^2.
  2. jcsd
  3. Oct 1, 2005 #2
    Since you are rotating about y=2, that makes each of your y-values 2 less, so the equations become:

    x = 1 - sin(t)
    y = cos(t)

    Now I think you can do the rest:

    [tex]\text{SA}=2\pi\int_{0}^{2\pi}y\,\sqrt{\left(\frac{dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt[/tex]
    Last edited: Oct 2, 2005
  4. Oct 1, 2005 #3
    I forgot to mention that the interval is t between (-pi/2, pi/2), how do i get the answer 2pi^2 with

    [tex]\text{SA}=2\pi\int_{-\pi/2}^{\pi/2}y\,\sqrt{\left(\frac{ dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt[/tex] ??

    so i have... dx = -cost
    dy = -sint

    here's what i've done..but couldn't get the answer

    [tex]\text{SA}=2\pi\int_{-\pi/2}^{\pi/2}y\,\sqrt{(sint)^2+(cost)^2}\, dt[/tex]

    [tex] =2\pi\int_{-\pi/2}^{\pi/2}y\,\sqrt{1}\, dt[/tex]

    [tex] =2\pi\int_{-\pi/2}^{\pi/2}cost\, dt[/tex]

    [tex] =2\pi\sin(\pi/2) - 2\pi\sin(-\pi/2)[/tex]

    [tex] =4\pi[/tex]
    Last edited: Oct 1, 2005
  5. Oct 2, 2005 #4
    but i got the formula of surface area from my textbook, which is,

    [tex]\text{SA}=2\pi\int_{}^{}y\,\sqrt{\left(\frac{ dx}{dt}\right)^2+\left(\frac{dx}{dt}\right)^2}\, dt[/tex]
  6. Oct 2, 2005 #5
    Yes, right. I fixed it sorry about that. Anyways, using this you should come up with your answer (I myself am coming up with -4π). Are you sure it's 2π2?

    The only way I get 2π2 is by the following:


    ...but you clearly don't have a cos2(t)...
    Last edited: Oct 2, 2005
  7. Oct 2, 2005 #6
    yes..the answer is supposed to be 2pi^2, unless it is a typo in the textbook
  8. Oct 2, 2005 #7
    It happens but I doubt that. Does anyone have any ideas why the work above isn't giving the correct answer?
  9. Apr 11, 2011 #8
    i think you were suppose to multiplied by x(t) not y(t)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook