1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Surface area help please

  1. May 2, 2005 #1
    Surface area help please....

    How do you find a surface area of a function around a vertical line?

    For example: surface of [tex]\ y=x^3[/tex] between [tex]\ 0<= x <= 3[/tex], rotating around [tex]\ x=4[/tex]???

    I tried the formula for finding surface area, but I confuse with that vertical line x=4.... what should I do?
    Last edited: May 2, 2005
  2. jcsd
  3. May 2, 2005 #2
    How would you set up the integral if it asked you to rotate it around the y-axis and what change can you make to account for this difference?
  4. May 2, 2005 #3
    If around the y-axis, then the integral should be

    [tex]\int 2\pi*y^{1/3}* \sqrt{1+ {1/9}*y^{-4/3}}*dy[/tex]

    right? From then, what change should I make to adjust with that x=3 vertical line?
  5. May 2, 2005 #4
    Well, I just need to know how to adjust that integral with the vertical line at x=3... the formula is easy, but the interval of integral should be ...?
  6. May 2, 2005 #5


    User Avatar
    Science Advisor
    Homework Helper

    Your integral for around the axis x = 0 contains the radical term that represents the line segment along the curve from y to y + dy. It also contains a term that represents the distance of that line segment from the axis of rotation related to computing the circumference of the circle around the axis. If you move to a new axis, the thing that changes is the circumference of the circle around the axis of rotation. How far are you from the new axis of rotation for each value of y?
  7. May 2, 2005 #6
    Im sorry... You lost me... I guess what u try to say is that I substituted y with (y+3) in the integral formula?
  8. May 2, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    The integral you wrote for rotation around x = 0 was

    [tex]\int 2\pi*y^{1/3}* \sqrt{1+ {1/9}*y^{-4/3}}*dy[/tex]

    The part of the integrand that is


    is the circumference of a circle around the x = 0 axis of rotation corresponding to any particular value of y = x^3. The radical term times dy is the length of the curve y = x^3 between a particular value of y and y + dy. Isn't that how you found the integrand?

    What you must change is not the length of the curve between y and y + dy. You must change the circumference of the circle around the axis for that value of y. The new radius of the circle would be (4 - x) expressed in terms of y. Do you see why? Draw the curve and the axis of rotation.
  9. May 3, 2005 #8
    Thank you very much... now I got the problem...^_^
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook