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Homework Help: Arclength of polar cardiod problem

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data
    My textbook sets up the integral, but does not solve, claiming that it's "trivial to solve manually or by using a CAS". I put the integral into my TI-89, and sure enough, there is a solution, and that solution happens to be "8". However...

    2. Relevant equations

    The actual problem is to find the total arclength of r = 1+sin(x)
    (It's polar, I just let theta = x, since theta is a pain to LaTeX)

    3. The attempt at a solution
    Here is my attempt at solving this integral manually:

    [tex]s = \int_{0}^{2π}\sqrt{(1+sin^{2}(x))+cos(x)^{2}}dx[/tex]


    [tex]\int_{0}^{2π}\sqrt{2+2sin(x)} \frac{\sqrt{2-2sin(x)}}{\sqrt{2-2sin(x)}}dx[/tex]




    [itex]u = 1-sin(x)[/itex]
    [itex]du = -cos(x)dx[/itex]


    I stopped here, because it's clear to me that the integral from one to one will be zero...

    ...and now the whole thing equals zero... I have tried doing the integral with substitutions other than [itex]u=1-sin(x)[/itex] to no avail. I tried to devise a way so I could take a limit (much like some improper integral problems are done), no luck. I can't seem to figure this out without using a CAS! It is one thing if I just got to a point where I didn't know how to proceed I could deal with, but I come up with zero every time!

    This is my best attempt, I have done it 11 different ways now..
    Last edited: Nov 8, 2011
  2. jcsd
  3. Nov 8, 2011 #2


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    You did a pretty dangerous thing there. You said sqrt(cos(x)^2)=cos(x). That's not right. cos(x) is negative for some values in [0,2pi]. It's |cos(x)| isn't it?
  4. Nov 8, 2011 #3
    Yeah, that makes sense, good catch. I still have no clue how to proceed though, especially now with the abs in the integral :cry:

  5. Nov 9, 2011 #4


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    Split the region of integration up into parts where cos(x) has a definite sign. Isn't that what you usually do with abs problems? It's premature to cry over it.
  6. Nov 9, 2011 #5

    Since, cos(x) will be positive in that domain.



    Like that? Forgive me, I have never encountered an integral with an abs yet.


    [tex]\sqrt{2}\int_{\frac{-π}{2}}^{\frac{π}{2}}\frac{cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{π}{2}}^{\frac{3π}{2}}\frac{-cos(x)}{\sqrt{1-sin(x)}}dx[/tex]
  7. Nov 9, 2011 #6


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    No, you want to integrate from 0 to 2pi. Integrate separately over the regions [0,pi/2], [pi/2,3pi/2] and [3pi/2,2pi] and add them up with appropriate signs.
  8. Nov 9, 2011 #7
    Thanks for the help.

    \sqrt{2}\int_{0}^{\frac{π}{2}}\frac{cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{π}{2}}^{\frac{3π}{2}}\frac{-cos(x)}{\sqrt{1-sin(x)}}dx + \sqrt{2}\int_{\frac{3π}{2}}^{2π}\frac{cos(x)}{\sqrt{1-sin(x)}}dx

    (No clue why that sqrt is not showing.)

    I see that you are splitting it up for the regions in which cos(x) would be positive or negative, but I am not sure how to sign the cos(x) in the functions. If integrating through the range of pi/2 to 3pi/2, cos(x) would turn out negative anyway? I am not seeing what effect this would have exactly?

    Here is how I am imagining the situation:

    Suppose you had [itex]\int_{0}^{2π}|sin(x)|dx[/itex].

    Since, that integral would be zero if there were no abs value, to handle the abs value case I would split the integral up like so:

    [tex]\int_{0}^{π}sin(x)dx + \int_{π}^{2π}-sin(x)dx[/tex]

    In this way, when sin(x) is negative, the negative that I placed into the split region would force the sin(x) to act positive?
  9. Nov 9, 2011 #8


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    That's it exactly. Now evaluate those integrals and add them. I get 8.
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