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Homework Help: Polar integral questions

  1. Jan 12, 2008 #1
    the first one is

    find the area around this formulas

    1. r=a(1+cos(teta))
    i know the formula
    my problem is,
    how am i suppose to know what is the range of the angles of the integral

    where there is a point (a/2,0)
    inside the area.

    how is that point sopposed to change the way i am solving it
    why shouldnt i solve it the normal way as if the point wasnt mentioned at all.
  2. jcsd
  3. Jan 12, 2008 #2
    1. You need to look at how the plot is going to look like. For values of theta= 0..2Pi plot r. e.g. If r= 1, (here there is no theta) but the plot is a circle of rad=1.

    Polar plots are difficuit to plot so I would suggest polarplot function in Maple

    2. After looking at the plot, I see that point (a/2,0) is point (a/2,0) on the x-y axis.

    See attached.

    Attached Files:

    Last edited: Jan 12, 2008
  4. Jan 12, 2008 #3
    what shaded area
    i see only a formula no shades

    about the second question
    i was given a formula
    generally i whouls solve it with a given formula
    (if i knew what angles to put in the integral)

    but they add another info that puzzles me
    "where there is a point (a/2,0)
    inside the area"

    what does it meen??
  5. Jan 13, 2008 #4
    i know how polar works but..

    but i still dont see in what way i am suppose to know the rangle
    only by looking at the formula
  6. Jan 13, 2008 #5
    You're right. You cannot have any clue of the range by just looking at the formula. You have to plot it. That's the only way. Plotting polar graphs is hard, so I suggest use of a software such as Matlab or Maple which will give you some insight initially.
  7. Jan 13, 2008 #6


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    Science Advisor

    You don't really need to do a very accurate graph. Since sine and cosine are periodic with period [itex]2\pi[/itex] You can "check" submultiples of that.

    The first one, [itex]r= a(1+ cos(\theta)), is particularly easy since r is never negative (assuming a> 0). Since a distance is always positive, the graph when r< 0 is interpreted as "going the other way", [itex]\theta+ \pi[/itex] and that can confuse things. When [itex]\theta= 0[/itex], r= 2a. When [itex]\theta= [itex]\pi/2[/itex], r= a, when [itex]\theta= \pi[/itex], r= 0, when [itex]\theta= 3\pi/2[/itex], r= a, and when [itex]\theta= 2\pi[/itex], r= 2a again. After that, it traces the graph over again. In order to go around the graph a single time, [itex]\theta[/itex] can go from 0 to [itex]2\pi[/itex].

    For the second one, [itex]r=a(cos(\theta)+sin(\theta))[/itex], you need to be more careful.
  8. Jan 13, 2008 #7
    about the second one i was tald that there is two common feilds

    and that we need to find the are of the feild for which the given point exists

    so i will try to buil them by entering point of 0 pi/2 pi p*1.5 2pi
    is this the universal way to plot a graph??
  9. Jan 14, 2008 #8


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    Science Advisor

    and that we need to find the are of the feild for which the given point exists[/quote]
    Did you write the equation correctly then?

    If [itex]r= a(cos(\theta)+ sin(\theta))= acos(\theta)+ asin(\theta)[/itex], then, multiplying both sides by r, [itex]r^2= arcos(\theta)+ arsin(\theta)[/itex] or
    [itex]x^2+ y^2= ax+ ay[/itex] so [itex]x^2- ax+ y^2- y^2= 0[/itex]

    Completing the square, [itex]x^2- ax+ a^2/4+ y^2- ay+ a^2/4= a^2/2[/itex] so [itex](x-a/2)^2+ (y-a/2)^2= a^2/2[/itex].

    The graph is a circle with center at (a/2, a/2) and radius [itex]a/\sqrt{2}[/itex], not two different "fields".
  10. Jan 14, 2008 #9
    (a/2,0) in (r,theta) is (a,a/2) on the x-y axis
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