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Area in polar coordinate using multiple integral

  1. Aug 12, 2015 #1
    The question is to find the area of a disk, r ≤ 2a×cos(θ)
    as in the figure "example-just an illustration"

    yvLb1.png


    I used two methods, each gave different wrong answers

    - integrate 2a×cos(θ) dθ dr - from θ=0 to θ=π/2 and from r=2a to r=2a×cos(θ) ; then I simply multiplied the answer by 2.

    - integrate 2a×cos(θ) dθ dr - from θ=0 to θ=2π and from r=2a to r=2a×cos(θ)

    I was hopping to get the right answer using the first method, but surprisingly it didn't work, integration seems to be fine I even used wolfram to check it.

    -----------------

    I have another small question. I had a question about finding the volume of flipped cone "its point in the origin and its shape is in the positive side of z-axis", its height is the same as its base radius so that, r=h z=h, thus r=z
    "example-just an illustration"

    http://01.edu-cdn.com/files/static/mcgrawhillprof/9780071624756/INVERTED_CONE_WATER_TANK_PROBLEM_01.GIF [Broken]

    thing is, when I put the limits say : r=0 ⇒ h //// z=0 ⇒ r //// θ=0 ⇒ 2π . It give wrong answer

    but r=0 ⇒ z //// z=0 ⇒ h //// θ=0 ⇒ 2π . give the right one

    that's really confusing!!
     
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Aug 12, 2015 #2

    Ray Vickson

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    In polar coordinates the area element is ##dA = r \, dr \, d \theta##. To get the area inside the curve ##r = 2a \cos(\theta)## we need to integrate ##dA## for ##r =0 \to 2 a \cos(\theta)## and for ##\theta = 0 \to \pi/2##, then multiply the result by 2 (or integrate for ##\theta = -\pi/2 \to \pi/2##).

    The boundary curve ##r = 2a \cos(\theta)## has a simple form when expressed in cartesian coordinates---looking at the plot suggests what it should be. From the resulting form one can easily determine the interior area with essentially no work.
     
    Last edited by a moderator: May 7, 2017
  4. Aug 12, 2015 #3
    we begin at ##θ=0##, why should ##r=0## ?, logicaly ##r=2acosθ## give us ##r=2a## !
    the thing is, ##r## is the distance between the point and the origin and ##θ## is the angle between r and x-axis !

    I don't know how did you do it but! the limits you suggest were wrong "both limit from ##-π/2## to ##π/2## and ##0## to ##π/2## multiplied by ## 2##" give an answer of## = 2πa^2##, which is double the area we expect.

    Thank you.
     
  5. Aug 12, 2015 #4

    Ray Vickson

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    Look at the picture! For a given value of ##\theta##, ##r## starts at ##r = 0## and ends at ##r = 2a \cos(\theta)##.

    My limits are not wrong, and when I do the integration I get an area of exactly ##2 \times \pi a^2/2 = \pi a^2## in the ##0 \to \pi/2## case and ##\pi a^2## directly in the ##-\pi/2 \to \pi/2## case. Yes, I have done the integrations, and that is what I get!

    Please re-read what I said in post #2. There I told you exactly what integrations I performed, and they are not at all the same as the ones you did.
     
  6. Aug 13, 2015 #5
    but the computer gave me solutions different than yours
    look

    http://im50.gulfup.com/rtRHjt.png [Broken]

    http://im50.gulfup.com/lL8Idy.png [Broken]
     
    Last edited by a moderator: May 7, 2017
  7. Aug 13, 2015 #6

    Ray Vickson

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    Those are not the correct integrals; they do not represent the area you want. Again: re-read (carefully!) what I wrote; or else, look up the topic in a book or via Google.
     
    Last edited by a moderator: May 7, 2017
  8. Aug 13, 2015 #7
    Got it!, thank you. But can I ask another question ?

    why can't I substitute ##r=2a cos(θ)## in the integral?, I did it because it seems reasonable in an integral. I mean since r depend on θ, then I can't integrate ##2a cos(θ)## which is in ##r## with respect to r because it contain a constant!. All that doesn't matter, still the question is why I can't substitute with the value of r ?

    is it because the disk is represented by ##r ≤ 2a cos(θ)## ??
     
    Last edited: Aug 13, 2015
  9. Aug 13, 2015 #8

    Ray Vickson

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    Basically, yes.

    I did suggest you look up the topic in Google. Have you done that? You should; there are several nice articles with derivations, diagrams, worked examples, etc. All your questions will be answered there.
     
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