Results 1 to 4 of 4

Math Help - Polar Equations and Areas

  1. #1
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19

    Polar Equations and Areas

    I'm having a difficult time applying the formula \frac{1}{2}\int_{\alpha}^{\beta}r^2d\theta for areas under curves in polar form. My problem is deducing the limits of integration. For example, if I want to find the area of the interior for r=1-sin(\theta), I graph the function on my calc, and then I trace from the origin back to the y-axis. My calculator gives the angles in radians, which are \frac{\pi}{2} and \frac{3\pi}{2}, so I can integrate 2\left(\frac{1}{2}\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}}[1-sin(\theta)]^2d\theta\right) to find the area.

    What method can I use, in general, to find the limits of integration for problems like this? How can I determine on paper the interval through which the curve is traced?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Unknown008's Avatar
    Joined
    May 2010
    From
    Mauritius
    Posts
    1,260
    While I'm not sure to be understanding what you are doing, I feel like you are not familiar with trigonometric graphs.

    y = sin(x)

    Is a curve starting at 0, going up to 1 at 90 degrees, goes back to 0 at 180 degrees, going further down till -1 at 270 degrees and up again to 0 at 360 degrees.

    y =sin(x - Wolfram|Alpha)

    Since you put the negative sign in front of the sine, you reflect the graph through the x-axis, giving:

    y =-sin(x - Wolfram|Alpha)

    Now, there is 1 in front. This means that you 'raise' the curve one unit up. Zeros thus become 1, 1s become 2s and -1s becomes 0s, like this:

    y =-sin(x) +1 - Wolfram|Alpha

    Now, if you can sketch such graphs on paper, knowing where the maxima, minima, and where the curve cuts the axes, you'll be better off finding the limits.

    I hope it helped!
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    Joined
    Jun 2009
    From
    United States
    Posts
    676
    Thanks
    19
    Quote Originally Posted by Unknown008 View Post
    While I'm not sure to be understanding what you are doing, I feel like you are not familiar with trigonometric graphs.

    y = sin(x)

    Is a curve starting at 0, going up to 1 at 90 degrees, goes back to 0 at 180 degrees, going further down till -1 at 270 degrees and up again to 0 at 360 degrees.

    y =sin(x - Wolfram|Alpha)

    Since you put the negative sign in front of the sine, you reflect the graph through the x-axis, giving:

    y =-sin(x - Wolfram|Alpha)

    Now, there is 1 in front. This means that you 'raise' the curve one unit up. Zeros thus become 1, 1s become 2s and -1s becomes 0s, like this:

    y =-sin(x) +1 - Wolfram|Alpha

    Now, if you can sketch such graphs on paper, knowing where the maxima, minima, and where the curve cuts the axes, you'll be better off finding the limits.

    I hope it helped!
    These are polar equations. r=1-sin(\theta) in a polar coordinate system (where points are represented by a radial distance from the origin and an angle  (r,\theta)) is much different then the function f(x)=1-sin(x) on the x-y plane. The graph of the polar form is a closed heart-shaped structure.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member bondesan's Avatar
    Joined
    Jul 2010
    From
    Brazil
    Posts
    58
    I'll try to explain how I do this: Imagine that you have a clock on the origin with only one pointer. The pointer direction depends on each graph you deal, but it necessarily will rest, initially, in one of the cardinal points (north, south, west, east).

    So, imagine that this pointer will start rotating to some direction - think that it will sweep the area that you want to find. You "stop" either when the graph "ends" (in other words, when you swept the region you wanted) or when it completes a full turn (360 deg or 2\pi rad).

    If it completes a full turn, then the limits will be from 0 to 2\pi. In the cardioid r=1-\sin(\theta) you do a full turn, while, for example, if you look the polar function r = \sin(\theta) in the figure, the limits of the area integral will be from 0 to \pi, if you start from east to west.

    Polar Equations and Areas-circle1.jpg

    On the other hand, the limits of the function r=\cos(\theta) can be found by starting the "pointer" south and then going north:

    Polar Equations and Areas-circle.pg.jpg

    In other words, you go from -\pi/2 to \pi/2.

    I hope you understand this tip hehe! Good luck.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Areas with polar coordinates
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 22nd 2010, 03:56 PM
  2. Some more areas with polar stuff
    Posted in the Calculus Forum
    Replies: 1
    Last Post: April 22nd 2010, 01:27 PM
  3. Areas in Polar Coordinates
    Posted in the Calculus Forum
    Replies: 8
    Last Post: April 28th 2009, 11:44 AM
  4. Areas in polar coordinates
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 27th 2008, 11:03 PM
  5. Areas under and between polar curves
    Posted in the Calculus Forum
    Replies: 2
    Last Post: April 25th 2008, 05:40 PM

Search Tags


/mathhelpforum @mathhelpforum