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Math Help - double integral in polar coordinates:sketch graph

  1. #1
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    double integral in polar coordinates:sketch graph

    I want to ask how can we sketch the region in polar coordinate?For example r=8sin4θ?Thanks.Any tips to sketch this type of graph?
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    Super Member Aryth's Avatar
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    This is an example for r  = 4cos\theta

    Notice how the straight lines come out of the center to form common angles (Remember when you had to study the unit circle?).

    The points are extended the value of r, and the r values are the circles that continue to get larger as you leave the origin.



    t = \theta On this graph.

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    Thanks for your reply.That a good way to sketch polar graph.But is there any intuitively method to determine the shape of the graph?And another question is how can we determine the limit of the inner and outer integral?Can you take r=2sin2θ as example?Thanks.
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    Super Member Aryth's Avatar
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    Well, for sin\theta We'll take a look.

    Here's plain r = sin\theta



    Here's r = sin3\theta



    Notice that the graph gets what we call "petals", which differ according to the coefficient of \theta

    Here's a few equations on the same graph that show what happens when multiply the whole statament:



    Notice all it does is stretch the graph vertically, Also notice that the cosine graph is a 90 degree rotation of the sine graph.

    Here's what happens when you add to or subtract the functions:







    General Statements

    1. r = a
    This equation is saying that no matter what angle we’ve got the distance from the origin must be a. If you think about it that is exactly the definition of a circle of radius a centered at the origin.

    2. r = 2acos\theta
    This is a circle of radius |a| and center (a, 0). Note that a might be negative and so the absolute value bars are required on the radius. They should not be used however on the center.

    3. r = 2bsin\theta
    This is a circle of radius |b| and center (0, b).

    4. r = 2acos\theta + 2bsin\theta
    This is a combination of the previous two and by completing the square twice it can be shown that this is a circle of radius \sqrt{a^2 + b^2} and center (a, b). In other words, this is the general equation of a circle that isn’t centered at the origin.

    5. Cardioids : r = a \pm acos\theta and r = a \pm asin\theta.
    These have a graph that is vaguely heart shaped and always contain the origin.

    6. Limacons with an inner loop : r = a \pm bcos\theta and r = a \pm bsin\theta with a < b.
    These will have an inner loop and will always contain the origin.

    7. Limacons without an inner loop : r = a \pm bcos\theta and r = a \pm bsin\theta with a > b.
    These do not have an inner loop and do not contain the origin.
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    Thanks for the information.Very useful,dude.By the way,i have a question.The graph of r=sinxθ where x=positive integer,how can we determine the petal will distributed on the quadrants?As your graph of r=sin3θ,can the two petals located on third and fourth quadrants?
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    Super Member Aryth's Avatar
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    For r = sin3\theta

    The petals will distributed in that way every time. It is a fixed graph.

    I did forget that General Statement though.

    8. r = sin(n\theta)
    For n = 2k (n must end up being greater than 1), there are 2n petals distributed evenly among the four quadrants. For even n, the graph is symmetric about the polar axis, the pole and the line \theta = \frac{\pi}{2}.

    For n = 2k + 1, there are n petals distributed in a particular order, though it depends on n. It has no polar symmetry. (It appears that n that are multiples of 3 have symmetry about the line \theta = \frac{\pi}{2} but don't quote me on this.)


    9. r = cos(n\theta)
    For n = 2k, there are 2n petals distributed in the same way as sin(n\theta) There are size differences. Sine petals are fatter and vertically shrunk compared to cosine. But think about it, an evenly distributed petal system of at least 4 petals rotated 90 degrees doesn't change the location of any of the petals. Same symmetry as well.

    For n = 2k + 1, there are also n petals distributed in a particular order dependent on n. But for cosine, these graphs are symmetric about the line \theta = \frac{\pi}{2}. So they have a small shift to the right compared to sine graphs.

    There are also:

    10. r = ab^{\theta}
    These are logarithmic spirals, fun to look at. I don't know that much about them though.

    11. r = a\theta + b
    These are Archimedes spirals. Don't know too much about them either.
    Last edited by Aryth; March 21st 2008 at 12:04 PM. Reason: Put the "don't quote me on this" on the wrong function...
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    Really appreciate on your help. Does help me a lot.
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  8. #8
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    Another question here,bro.r=8sin4θ,one petal.does the limit 0<=θ<=pi/4 and 0<=r<=8sin4θ correct?If not, how should I find the limit?
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  9. #9
    Super Member Aryth's Avatar
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    I'm not sure what you're asking.

    Are you asking how to find the limits of one pedal?
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  10. #10
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    I want to find the area of the region bounded by the graph r=8 sin 4θ, one petal. And I want to ask does my inner integral is integral from 0 to 8sin4θ w.r.t. r and outer integral is from 0 to pi/4 w.r.t θ correct?If not can you teach me how to find the limit of integral when we duel with this type of question?Thanks.
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