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Homework Help: Area of a sector of a circle

  1. Jun 22, 2008 #1
    Prove the formula [tex] A = \frac{1}{2}r^{2}\theta[/tex] for the area of a sector of a circle with radius r and central angle [tex]\theta[/tex]. (Hint: Assume 0 < [tex]\theta[/tex] < [tex]\frac{\pi}{2}[/tex] and place the center of the circle at the origin so it has the equation [tex]x^{2} + y^{2} = r^{2}[/tex] . Then A is the sum of the area of the triangle POQ and the area of the region PQR in the figure.)


    So the area of the triangle is 1/2bh which comes to [tex]\frac{1}{2}r^{2}cos\theta sin\theta[/tex]

    Now, for the other region I used the integral [tex]\int\sqrt{r^{2} - x^{2}}dx[/tex]

    I make x = r sin[tex]\theta[/tex]

    I plug that in under the square root sign and get r cos[tex]\theta[/tex].

    I changed the limits of integration from r cos[tex]\theta[/tex] to r, to pi/4 to pi/2.

    Now since dx = r cos[tex]\theta[/tex]d[tex]\theta[/tex] the integral for the area of the second region is [tex]\int r^{2}cos^{2}\theta d\theta[/tex]. Now here is where the problem starts, I pull the r^2 out of the equation and use half angle theorem on the cos^2 theta. After his I end up with [tex]\frac{1}{2}r^{2}(\theta+\frac{1}{2}sin2\thetacos\theta)[/tex] with limits pi/4 to pi/2. Now I am assuming there should be a negative equivalent somewhere here to cross out the first area and leave me with just [tex]\frac{1}{2}r^{2}\theta[/tex]but I don't see how I can get it to work, especially after I finish integration and sub the limits in at which point I will lose the trig ratios. Can anyone help me finish this question up?

    Attached Files:

    Last edited: Jun 22, 2008
  2. jcsd
  3. Jun 22, 2008 #2
    there's a better way:

    Area of all / Area of sector = Total arc length / sector arc length

    use r.theta = arc length
  4. Jun 22, 2008 #3
    While I see that does work, I am more interested in solving the problem using trigonometric substitution.
  5. Jun 22, 2008 #4
    why divide it into two areas
    when you can use Polar co-ordinates?

    Using Jacobi transformations
    integrate from 0 to theta
  6. Jun 22, 2008 #5
    I was preoccupied watching movie..

    using your method
    I got
    0.5r^2.theta - 0.5 * int (0,theta) cos 2*t dt
    for second area...

    I think you took wrong limits

    starting from very beginning:
    second area:
    int (0, theta) int (r. cos thata -- > r) [r] .dr.d(theta)

    urs different

    \int r^{2}cos^{2}\theta d\theta

    I have
    \int r^{2}sin^{2}\theta d\theta
  7. Jun 23, 2008 #6


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    Homework Helper

    Hi Calcotron! :smile:

    You've used θ to mean two different things.

    First, it's a fixed value, then it's a variable of integration.

    Don't be stingy … use another letter! :smile:

    (and be careful about the limits of integration!)
  8. Jun 23, 2008 #7
    Yeah, I always use theta for that part and I wasn't thinking here. Ok, so if I make the substitution x = r sin u the limits change to [tex] sin^{-1}cos\theta[/tex] to [tex] \frac{\pi}{2}[/tex]correct?

    I end up with [tex] \frac{1}{2}r^{2}[\frac{\pi}{2} - (sin^{-1}cos\theta + \frac{1}{4}sin(2sin^{-1}(cos\theta))][/tex]

    I assume the sin and arcsin cross out but I still cannot see how to get this into a form so that it will cancel out the area of the first triangle that I found. Can anyone see where I made a mistake?
  9. Jun 23, 2008 #8
    Well, if I replace [tex]sin^{-1}cos\theta[/tex] with [tex] \frac{\pi}{2} - \theta[/tex] in both spots there and then an angle sum identity on the last term, and then the double angle sin identity I get the right answer. Can anyone confirm this is correct?
    Last edited: Jun 23, 2008
  10. Jun 24, 2008 #9


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    go for the obvious …

    Hi Calcotron! :smile:
    uhh? :confused:

    Why x = rsinu?

    Why deliberately give yourself limits like arcsin(cos)??

    Try again with (the far more obvious?) x = rcosu. :smile:
  11. Jun 24, 2008 #10
    that works too
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