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Homework Help: Area of overlapping polar coordinates

  1. May 14, 2012 #1
    1. The problem statement, all variables and given/known data
    find the over lapping area of the following equations


    2. Relevant equations

    area =1/2 ∫ f(x)^2 dx

    3. The attempt at a solution

    first off I started by finding the intersecting angle by:

    and the peak is at pi/2

    so I started with a = 2×1/2 ∫ (1+sin(x))^2 dx from pi/6 to pi/2

    so (3/2)x-2cos(x) -1/4 sin(x) from pi/6 to pi/2
    comes out to 2pi/4 + (7√3)/8

    then I have the other segment

    a= 2×1/2 ∫ (3sin(x))^2 dx from 0 to pi/6

    for which I get 3pi/4 - (9√3)/8

    which totals up to 7pi/4 -(2√3)/8

    and according to my professor's key the answer is pi

    needless to say I am lost.
  2. jcsd
  3. May 14, 2012 #2


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    Gold Member

    You should always plot the graph to easily understand and deduce the limits. I have attached the graph and the overlapped area is shaded in blue.

    You could use the formula for area of sector of a circle, but the problem will become longer than needed.

    Calculating the half area on the right of the line ∏/2 using polar coordinates:
    [tex]Half\; area\;=\int^{x=\pi /2}_{x=0} \int^{r=1+\sin x}_{r=0} rdrdx-
    \int^{x=\pi /6}_{x=0} \int^{r=1+\sin x}_{r=3\sin x} rdrdx
    [/tex] Then, multiply by 2.

    Now, using the method involving the formula for area of sector of a circle:

    First, find the area from x = 0 to x=∏/2 inside the cardioid.
    [tex]\frac{1}{2}\int^{x=\pi /2}_{x=0} (1+\sin x)^2dx[/tex]
    Then, you need to find the small area wedged between both graphs, in the interval x=0 and x=∏/6
    [tex]\frac{1}{2}\int^{x=\pi /6}_{x=0} (1+\sin x)^2dx-\frac{1}{2}\int^{x=\pi /6}_{x=0} (3\sin x)^2dx[/tex]
    Finally, subtract area of wedge from cardioid, to get the half area, then multiply by 2 to get the total overlapping area.

    There is still another way (a shorter version) of doing this problem using the area of sector formula:

    First, find the area of sector inside cardioid from x=∏/6 to x=∏/2:
    [tex]\frac{1}{2}\int^{x=\pi /2}_{x=\pi / 6} (1+\sin x)^2dx[/tex]
    Then, find the area inside circle from x=0 to x=∏/6:
    [tex]\frac{1}{2}\int^{x=\pi /6}_{x=0} (3sin x)^2dx[/tex]
    Add them up and multiply by 2.
    You could also write its equivalent in terms of polar coordinates. :smile:

    Attached Files:

    Last edited: May 14, 2012
  4. May 14, 2012 #3
    Thanks but that is actually doing. I think I am messing up the integration somewhere. It is just hard to type this all out on a tablet.
  5. May 14, 2012 #4


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    [tex]2\times \frac{1}{2}\int^{x=\pi /2}_{\pi /6} (1+\sin x)^2 \,.dx=\int^{x=\pi /2}_{\pi /6} (1+\sin x)^2 \,.dx=\int^{x=\pi /2}_{\pi /6} (1+2\sin x+\sin^2x) \,.dx=\frac{\pi}{2}+\frac{9\sqrt 3}{8}
    [tex]\int^{x=\pi /6}_{x=0} (3\sin x)^2\,.dx=\int^{x=\pi /6}_{x=0} 9\sin^2x\,.dx=\frac{3\pi}{4}-\frac{9\sqrt 3}{8}[/tex]
    Adding them up, the total area is: [itex]\frac{5\pi}{4}[/itex]
    Last edited: May 14, 2012
  6. May 14, 2012 #5
    Thanks, so I did make at least one mistake :-D
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