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Homework Help: Polar Arc length

  1. Apr 6, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the length pf the curve over the given interval.
    [tex] r=1+\sin\theta [/tex]
    [tex] 0\preceq\theta\preceq\2\pi [/tex]

    3. The attempt at a solution
    Ok I set it up as:
    [tex] 2\pi[/tex]
    [tex]\int\sqrt((1+\sin\theta)^2+cos^2\theta) [/tex]

    and by simplifying and integrating, I get

    [tex]2\pi [/tex]
    [tex] -2\sqrt2[\sqrt(1-\sin\theta)] [/tex]

    [tex] -2\sqrt2[(1-0)-(1-0)] =0 [/tex]

    and obviously it is wrong,

    I check the solution it has the same everything but the range ,
    it obviously broke down the whole length into 2 times 1 piece from [tex] \pi/2 to 3\pi/2 [/tex] and the answer is 8

    My question is why I get zero within my range, and why broke it down into the range above?
    Last edited: Apr 6, 2008
  2. jcsd
  3. Apr 6, 2008 #2


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    You might wish to elaborate about how you got that integral. It's not what I get. You want to do the integral using a double angle formula to write the integrand as a perfect square. It also may help to shift the limits of integration by using sin(x+pi/2)=cos(x). You should also remember that sqrt(x^2)=abs(x) for any expression x. So to integrate something like that you'll need to break into regions where 'x' is positive.
  4. Apr 7, 2008 #3
    There is some mistake on my first post but anyway I integrate it like this: (ignore the limits for the moment)

    [tex] \int\sqrt((1+\sin\theta)^2+cos^2\theta) [/tex]

    = [tex] \sqrt2 \int\sqrt((1+\sin\theta) [/tex]

    = [tex] \sqrt2 \int \frac{\cos\theta}{\sqrt(1-\sin\theta)} [/tex]
    (multiply up and down by [tex] \sqrt(1-\sin\theta) [/tex]

    = [tex] 2\sqrt2 [ \sqrt((1-\sin\theta)][/tex] (by substitution)

    which if I use the limit [tex] \pi/2 [/tex] to [tex] 3\pi/2 [/tex] and times 2
    I get the answer, by I still don't understand what decision it based on setting the limits:uhh:,
    I also tried from 0 to pi and times 2 and it came out to be zero :yuck:, also when I trace on the calculator ( [tex] \pi/2 [/tex] to [tex] 3\pi/2 [/tex])the x's are all -ve :uhh: clue?

    So yes why the selected limits :confused: Also it will be great if you can demonstrate me a different way of integrating it and what do you mean by to shift the limits of integration by using sin(x+pi/2)=cos(x) :smile:

    Thanks a million~
    Last edited: Apr 7, 2008
  5. Apr 7, 2008 #4
    Ok wait I am up to something, it is because of the behaviour of the sin curve ?

    for not getting zero I have to use that certain limit? :rofl:

    http://hk.geocities.com/ymtsang2606/sin.jpg [Broken]
    Last edited by a moderator: May 3, 2017
  6. Apr 7, 2008 #5


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    That's a clever trick. But it's hiding something from you. sqrt(cos(theta)^2)=abs(cos(theta)). As theta goes from 0 to pi, cos changes sign. Integrate from 0 to pi/2, then from pi/2 to pi and add the absolute values.
  7. Apr 7, 2008 #6
    How do u get to sqrt(cos(theta)^2)?
  8. Apr 7, 2008 #7


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    sqrt(1-sin(theta)^2)=sqrt(cos(theta)^2). Isn't that what you did?
  9. Apr 7, 2008 #8
    [tex] \int\sqrt((1+\sin\theta)^2+cos^2\theta) [/tex]

    = [tex] \sqrt2 \int\sqrt((1+\sin\theta) [/tex] (then I multiply up and down by [tex] \sqrt(1-\sin\theta) [/tex] ) <-- do u mean the sqrt cos^2theta on the top after multiplying?

    = [tex] \sqrt2 \int \frac{\cos\theta}{\sqrt(1-\sin\theta)} [/tex]

    = [tex] 2\sqrt2[\sqrt((1-\sin\theta)] [/tex]
  10. Apr 7, 2008 #9


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    Yes, I mean the cos(theta)^2 on the top under the square root.
  11. Apr 7, 2008 #10

    Ok I get it now, so when you integrate you must make sure that the signs of the function would not change through out the limit range , cause otherwise they will offset each other, and given this case :[tex] \sqrt2 \int \frac{\cos\theta}{\sqrt(1-\sin\theta)} [/tex]

    the sqrtcos^2theta on the numerator change signs with the range 0 to pi so you have to break it down in to two parts and integrate, while sin doen't change signs in the denominator within 0 to pi so it doen't matter.
  12. Apr 7, 2008 #11

    There's a superfluous integral sign in that post (the last line, where you've already performed the integration)... sorry to nitpick, but it might be confusing for those who are first signing on.
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