1. Not finding help here? Sign up for a free 30min 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 of a curve

  1. Feb 27, 2013 #1
    1. The problem statement, all variables and given/known data
    The given curve is rotated about the y-axis. Find the area of the resulting surface.
    [tex] y = 3 - x^2 [/tex]
    [itex] 0 \leq x \leq 4 [/itex]



    2. Relevant equations
    [tex] A_{y} =2\pi\int_a^b x \sqrt{1+\left(\frac{dx}{dy}\right)^2} dy [/tex]


    3. The attempt at a solution
    now we need to write x in terms of y so we get:
    [tex] x = \pm \sqrt{3-y} [/tex]
    and if we differentiate the above, we get:

    [tex] \frac{dx}{dy} = -\frac{1}{\sqrt{3-y}} [/tex]

    Our range also changes
    when x = 0, then y = 3
    and x = 4, then y = -13

    so then our integral is:

    [tex] A_{y} =2\pi\int_{-13}^{3} 2\sqrt{3-x} \sqrt{1+ \left(-\frac{1}{\sqrt{3-y}}\right)^2} dy [/tex]


    So i multiplied the above by 2 instead of plus or minus
    and in addition to that, do we need to consider the dx/dy term? Does that need to be multiplied as well?

    Thanks for your help!
     
  2. jcsd
  3. Feb 27, 2013 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    For one thing, I think your derivative is a bit off. For another thing if x=sqrt(3-y) why did change it to sqrt(3-x) in your integral? And you don't really have to worry about the +/- bit. x in your integral is really a radius. The sign doesn't matter. Draw a picture.
     
    Last edited: Feb 27, 2013
  4. Feb 27, 2013 #3

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    It might have been a little easier to work with ds =√(1+(dy/dx)2).dx instead, but it should work either way.
    I don't understand why you multiplied by 2 at the end. The choice between ±√ for the value of x is just that, a choice. One will give you a negative result, one positive. You know the answer is to be positive, so just take the positive value at the end.
     
  5. Feb 27, 2013 #4
    Thanks for your replies. Let me fix what you have mentioned:

    The derivative is off:

    [tex] x = \pm \sqrt{3-y} [/tex]

    [tex] \frac{dx}{dy} = -\frac{1}{2 \sqrt{3-y}} [/tex]

    And fixing that and the 2 in the integral:
    [tex] A_{y} =2\pi\int_{-13}^{3} \sqrt{3-y} \sqrt{1+ \left(\frac{1}{2 \sqrt{3-y}}\right)^2} dy [/tex]

    How does this look now?
     
  6. Feb 27, 2013 #5

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    Looks good now.
     
  7. Feb 27, 2013 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    Much better. Now if you can try to work it out without Wolfram, that would be better yet. It's really not terribly hard.
     
  8. Feb 27, 2013 #7
    since they are both under root we can multiply the 3-y into the big root. this will yield under the root [itex] -y +\frac{13}{4} [/itex] which we can integrate by guessing or by u-substitution.

    Thank you!!
     
  9. Feb 22, 2015 #8
    Can someone explain the range change to me?
     
  10. Feb 23, 2015 #9

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The original range was stated as bounds on x. Because the integral was constructed as with respect to y, the solver had to find the range for y. As I remarked at the time, the whole problem would have been easier using x as the variable of integration.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Surface area of a curve
Loading...