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 functions without definite integrals

  1. Sep 27, 2008 #1
    1. The problem statement, all variables and given/known data
    The curve is rotated about the y axis, find the area of the resulting surface.

    y=(1/4)X2-.5ln|x| 1<_X<_2



    2. Relevant equations
    S=2(pi)(f(x))[tex]\sqrt{}1+f'(x)^2[/tex]



    3. The attempt at a solution
    Alright I'm not entirely sure where to even begin. Since I'm rotating about the Y-axis I know that I need to solve the limits wrt y, which would be 1/4<_y<_0.65345
    Then I need to solve the equation so X is isolated, then take the derivative of that.
    I'm not entirely sure where to even begin, I know that I could take everything to the e^ to get rid of the natural log of x, but then everything is to the e^.
     
    Last edited: Sep 27, 2008
  2. jcsd
  3. Sep 27, 2008 #2

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    You can't explicitly write a formula for x(y). But you don't have to. You have to integrate 2*pi*x(y)*sqrt(1+x'(y)^2)*dy. dy=(x/2-1/(2x))*dx. Use that to find dx/dy and convert the integration over dy to an integration over dx. You can manipulate the expression inside the square root into a perfect square.
     
  4. Sep 27, 2008 #3
    Thanks for the help, I don't completely follow you though.

    You can't explicitly write a formula for x(y). But you don't have to. You have to integrate 2*pi*x(y)*sqrt(1+x'(y)^2)*dy. dy=(x/2-1/(2x))*dx. Use that to find dx/dy and convert the integration over dy to an integration over dx. You can manipulate the expression inside the square root into a perfect square.


    1/(x/2-1/(2x))=dx/dy, now that's x'(y), and then take the integral of this function to get X? If I take the integral of that I get 2ln|x|+x^2 = y, but I don't really think this is right because it's almost where I started from, and there aren't any y's in the equation and x isn't isolated. x(y) should equal the integral of dx/dy though? Thanks a lot.
     
  5. Sep 27, 2008 #4

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    No, don't integrate dx/dy. You want to integrate x*sqrt(1+(dx/dy)^2)*dy. That's x*sqrt(1+(dx/dy)^2)*(x/2-1/(2x))*dx. You can write it completely as an integral over x without ever knowing an explicit form for x as a function of y. You are basically just changing the variable of integration from y to x.
     
  6. Sep 27, 2008 #5
    Alright I'm still stuck.

    so dx/dy= 1/((x/2-1/(2x)) which is the same as 2/x - 2x, so (2/x-2x)(2/x-2x)= 4/x2-8+4x2, take a 4 out and you get 1/x2-2x2+1, get a common denominator and you get (x4-2x2+1)/(x2), factors to (x2-1)(x2-1)/(x2) + 1. I know this is going to be really obvious, but I can not see the perfect square factorization of this.

    Thanks a lot for the help, the book we have doesn't offer very good explanations.
     
  7. Sep 27, 2008 #6

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    1/(x/2-1/(2x)) is NOT the same as 2/x-2x. Now you are just being sloppy with algebra. Try again.
     
  8. Sep 27, 2008 #7
    Okay, just to check in, the solution to the square root should be (x2+1)/(x2-1) (providing my algebra is correct, which it most likely isn't). Then I multiply that solution by (x2/2 - 1/2) (by multiplying the x into dy). Then I get two integrals, and then I long divide and get a really long integral and another long one? (Just checking so I don't waste an hour going down the wrong track).
     
  9. Sep 27, 2008 #8

    Dick

    User Avatar
    Science Advisor
    Homework Helper

    I think your algebra is correct this time. (x^2/2-1/2)=(x^2-1)/2. Looks to me like you've got some cancellation there when you multiply by (x^2+1)/(x^2-1). May not be nearly as hard as you are thinking.
     
  10. Sep 27, 2008 #9
    Ohhh good call! I got a final answer of 12pi, which is simple enough to make me think it's the correct one! Thanks Dick!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Surface area of functions without definite integrals
Loading...