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Surface area of revolution about y

  1. Apr 15, 2015 #1
    1. The problem statement, all variables and given/known data

    Find the surface area obtained by rotating the curve
    [itex]y = x^2/4 - ln(x)/2 [/itex]

    [itex]1 \leq x \leq 2[/itex]

    2. Relevant equations
    [itex]2π \int f(x)\ \sqrt{1+(f'x)^2} dx [/itex]

    3. The attempt at a solution
    I can't seem to isolate for x in terms of y. I raised both sides to e and seperated the exponents and rearranged. Giving me [itex] e^x^2/x^2 = e^(4y) [/itex] but I'm stuck here.

    Thanks
     
  2. jcsd
  3. Apr 15, 2015 #2

    SammyS

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    Are you revolving this about the y-axis or about the x-axis?
     
  4. Apr 15, 2015 #3
    Sorry y-axis. Could've sworn i wrote it
     
  5. Apr 15, 2015 #4

    SammyS

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    Yes, it's in the title. Just wanted to make sure. -- It is always best to include all information in the Original Post as well.

    About the y-axis: should that be x rather than f(x) in the integral.

    ##\displaystyle\ 2π \int_1^2 x\, \sqrt{1+(f'(x))^2} dx\ ##


    Also, why do you want to find x in terns of y ?
     
    Last edited: Apr 16, 2015
  6. Apr 15, 2015 #5

    Mark44

    Staff: Mentor

    The LaTeX interpreter choked on your equation, and I can't make sense of it either. What is the left side supposed to be? With LaTeX, if an exponent is more than one character, you need braces. IOW, this -- ##e^2## (that is, e^2) is OK, but ##e^2x## (using e^2x) doesn't render the same as e^{2x} would.

    Yes. All the information should be in the post, not just in the title.
     
  7. Apr 15, 2015 #6
    I see now. That x in your equation i'm used to thinking of as f(y), just like when you rotate the curve about the x-axis it's a y; i also see that as an f(x).

    So for each radius which is each x value on the curve i thought I had to rewrite it as a function of y integrate over the y interval...just like how y=f(x) i thought x had to be f(y)...didnt realize you can just stick the actual x and y in by themselves, much more convenient.

    Thanks a ton
     
  8. Apr 15, 2015 #7
    Does this apply to volume as well or just surface area, because for volumes i've always rearranged for x as a function of y if it asks to rotate a curve about the y axis.
     
  9. Apr 16, 2015 #8
    I would suspect it's just surface area because of that invariant arc length term
     
  10. Apr 16, 2015 #9
    ya...because each point on that arc length is mapped to a certain x value regardless of whether in terms of y or x...so you can just use that x and the x interval instead of flipping it and using the y interval in terms just because it's about y...like rotation of height only sideways............anyways im rambling

    thank you
     
  11. Apr 16, 2015 #10

    Mark44

    Staff: Mentor

    Not necessarily. This is true if the function happens to be one-to-one on the interval in question (your function is one-to one on the interval).

    A big problem is that what you show as your relevant equation isn't relevant. Since the curve is being revolved around the y-axis, the radius is NOT f(x), but is instead x. The relevant formula here would be ##2\pi\int_1^2 x \sqrt{1 + (f'(x))^2}dx##. Did you draw a sketch of the curve and another of the surface of revolution? Doing that will help you derive the formula or at least use the right one.
     
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