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Area Polar

  1. Nov 11, 2013 #1
    1. The problem statement, all variables and given/known data
    Question: Sketch the curve and find the area that encloses it.
    r = 1- sin(theta)


    2. Relevant equations

    I sketched it

    3. The attempt at a solution

    Now
    A = (1/2) ∫ (1-sinθ)^2 dθ [-pi/2, pi/2] ...Explain the limits please. Could it be 3pi/2 instead of -pi/2? I will result in a neg sign.

    I just did the integral I got A = (1/2)θ + cos(θ) + (1/2)θ - (1/4)sin(2θ) evaluate at the limits I gave and I got (3∏/4) then I multiply by 2 for symmetry. If you look at the graph you will understand what I'm getting at.
    Did I do this OK? Thanks, J
     
  2. jcsd
  3. Nov 11, 2013 #2

    LCKurtz

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    You need to learn how to post images. How can we explain your limits when we don't know what you are looking at? What values of ##\theta## did you use for your graph?
     
  4. Nov 11, 2013 #3
    Yeah I know. Umm what do you mean what values of theta? I just graphed it on wolfram. They don't actually give you values. Do you mean like 0 <= theta < = pi/6
    Or something like this? You know wolfram alpha? If you just copy the original equation into there it will draw it out for you. I don't know how to do images.
     
  5. Nov 11, 2013 #4
    Disregard this post.
     
  6. Nov 11, 2013 #5

    LCKurtz

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    If you graphed it by hand you would make a table of ##\theta## and ##r## values and plot the corresponding ##(r,\theta)## values. You would know what ##\theta## values were needed to produce the whole graph. You need to know them to know what limits to use. Perhaps you should try graphing it by hand and don't use Wolfram until you understand what's going on.

    So you should learn. Click the Site Info tab at the top of this page and open the FAQ (frequently asked questions) and read about it.
     
  7. Nov 11, 2013 #6
    I used this for my limits [-pi/2, pi/2]. I know what limits. But my question was if I use [-pi/2, pi/2] or [ 3pi/2, pi/2] which preferred? One gives the negative of the other. Also I wanted to know if my set up was OK.
     
  8. Nov 11, 2013 #7

    umm?
     
  9. Nov 11, 2013 #8
    First, I'm going to let you go HERE so that you can learn to actually make things legible.

    Second, I think the problem is ill-posed. There are infinitely many areas that enclose the given graph of the curve. Instead, a better problem would be to find the area that the curve encloses. This would be written "Sketch the curve and find the area that it encloses."

    Now, let's look at your integral. I'm assuming you wrote ##\displaystyle \frac{1}{2}\int\limits_{[0,2\pi]}(1-\sin\theta)^2\mathrm{d}\theta##, where we use the standard abuses of notation, correct?
     
  10. Nov 11, 2013 #9

    LCKurtz

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    [ 3pi/2, pi/2] is not in the positive direction. Why are you bothering with using symmetry, which, by the way, you should prove if you are going to use it?
     
  11. Nov 11, 2013 #10
    No I wrote
    ## \displaystyle \frac{1}{2}\int\limits_{[-pi /2, pi/2]}(1-\sin\theta)^2\mathrm{d}\theta ##
    Not sure why it written under the integral like that.


    Because this sucker is symmetric. Limits always go in positive direction?
     
  12. Nov 11, 2013 #11

    LCKurtz

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    I fixed your Latex. Quote it to see how I fixed your integral limits.

    Yes, looking at the graph it appears to be symmetric. But you should prove it mathematically if you intend to use it. And it is unnecessary in this problem.
     
  13. Nov 11, 2013 #12
    xxxxxxx means ?
     
  14. Nov 11, 2013 #13

    LCKurtz

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    Nothing. I accidentally posted before I was done. I have finished editing it now.
     
  15. Nov 11, 2013 #14

    Mark44

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    If you reverse the order of the limits, you get the opposite value when you integrate.

    IOW,
    $$\int_a^b f(x)dx = -\int_b^a f(x)dx$$
     
  16. Nov 11, 2013 #15

    Mark44

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    I don't see this in the thread. Possibly it was there but whoever posted it edited it. If you're quoting someone, use the Quote button so we have some context for the question.
     
  17. Nov 11, 2013 #16
    So I should just do [ 0, 2pi ]?
     
  18. Nov 11, 2013 #17

    Mark44

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    That should work.
     
  19. Nov 11, 2013 #18
    If you graph it you would be able to tell what you need to integrate with the correct bounds. Since usually in polar you'll be asked to find the total area or the area of one "petal" of the polar curve, for example r=sin2θ.
     
  20. Nov 11, 2013 #19
    You're off by a factor of 2. Do you see why?

    It's just preference of notation. I like to remember that I'm integrating over chains is all. Plus, it makes substitution easier because you can write ##\displaystyle \int\limits_{u([a,b])}##. Plus, application of the generalized Stokes' Theorem is easier to see. Plus, as a friend I tutor in calculus puts it, "It makes it look like a pretty bad*** integral." The list goes on. :tongue:

    The only downside I know of to putting the limits underneath the integral is that some people will think you mean the "unsigned area bounded by the curve." In cases where that may lead to potential confusion, I just make it clear from either context or explicit wording (id est "THIS IS NOT THE UNSIGNED AREA"). Otherwise, in case you couldn't tell, I think my notation for integrals is fabulous.
     
  21. Nov 11, 2013 #20
    Yeah I see.
     
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