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Area Of The Surface Of Revolution

  1. Sep 8, 2007 #1
    I have tried this question (http://img524.imageshack.us/img524/9539/scan0001pa1.png) a number of times and always use the formula

    S = 2*pi*int( y*sqrt( (dx/dt)^2+(dy/dt)^2 ) dt

    i always get S=6*pi*a^2[1/5*(sin t)^5] 0<t<pi, and because sin 0 = 0 and sin pi = 0 the answer i get is 0. If you were to replace the y in the equation with x then i get the correct answer but that would not be the formula for rotation about the x-axis.

    Any help would be much appreciated.
     
    Last edited: Sep 8, 2007
  2. jcsd
  3. Sep 8, 2007 #2

    HallsofIvy

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    x= a cos3(t), y= a sin3(t) so x'= 3a cos2(t)sin(t) and y'= a sin2(t)cos(t). x'2+ y'2= 9a2 cos4(t)sin2(t)+ 9a2 sin4(t)cos2(t). Factoring out 9a2 cos2(t)sin2(t) leaves cos2(t)+ sin2(t)= 1.
    [tex]\sqrt{x'^2+ y'^2}= \sqrt{9a^2 cos^2(t)sin^2(t)}= 3a cos(t)sin(t)[/tex]
    [tex]y\sqrt{x'^2+ y'^2= 3a^2 cos(t)sin^2(t)[/tex]
    After substitution, I get
    [tex]6\pi a^2\int_{-\pi}^{pi}(sin^2(t)cos(t)) dt[/tex]

    You are right- the substitution u= sin(t) gives u= 0 at both points.

    Aha! I think I see the problem. cos(t), for t> [itex]\pi/2[/itex], is negative. The square root of cos2(t) is NOT cos(t) for t> [itex]\pi/2[/itex]! More generally, [itex]\sqrt{x^2}= |x|[/itex], not x.

    Use symmetry. integrate from 0 to [itex]\pi/2[/itex]. That will give you a positive value. Because the integral from 0 to [itex]\pi[/itex] is 0, the integral from [itex]\pi/2[/itex] to [itex]pi[/itex] must be the negative of that (you can, if you wish, integrate and see). The true value of the integral from 0 to [itex]\pi[/itex], using absolute value, must be twice the integral from 0 to [itex]\pi/2[/itex]. And, of course, the actual surface area, the integral from [itex]-\pi[/itex] to [itex]\pi[/itex], is 4 times the integral from 0 to [itex]\pi/2[/itex].
     
  4. Sep 8, 2007 #3

    arildno

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    Hmm..we should use absolute value signs here!

    Thereby, we get the integral:
    [tex]6a^{2}\pi\int_{0}^{\pi}\sin^{4}(t)|\cos(t)|dt[/tex]
    where I have utilized that sin(t) is positive on the entire interval.

    EDIT:
    Okay, Halls found the flaw as well..:smile:
     
  5. Sep 8, 2007 #4

    HallsofIvy

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    I just type faster!
     
  6. Sep 8, 2007 #5
    Thanks guys, i totally understand it now :smile:
     
  7. Sep 8, 2007 #6

    HallsofIvy

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    I'm glad YOU do!
     
  8. Sep 9, 2007 #7
    and you dont...? how come?
     
    Last edited: Sep 9, 2007
  9. Sep 9, 2007 #8

    arildno

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    I think Halls is flabbergasted as to why he forgot the absolute value sign in the first place!

    Besides, in his expression, he mistakenly used y=asin(t), rather than the correct third power of sine in the y-expression.
     
    Last edited: Sep 9, 2007
  10. Sep 9, 2007 #9
    ... and that would keep on giving him the wrong answer therefore not understanding it?
     
  11. Sep 9, 2007 #10

    arildno

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    I think he was merely angered at himself for making a dumb mistake, that's all.
     
  12. Sep 9, 2007 #11

    HallsofIvy

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    Unfortunately, that happens a lot!
     
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