
#1
Jun2208, 04:57 PM

P: 17

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? 



#2
Jun2208, 06:09 PM

P: 1,295

there's a better way:
Area of all / Area of sector = Total arc length / sector arc length use r.theta = arc length 



#3
Jun2208, 06:32 PM

P: 17

While I see that does work, I am more interested in solving the problem using trigonometric substitution.




#4
Jun2208, 07:55 PM

P: 1,295

Area of a sector of a circle
why divide it into two areas
when you can use Polar coordinates? Using Jacobi transformations integrate from 0 to theta 



#5
Jun2208, 11:07 PM

P: 1,295

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 for [tex] \int r^{2}cos^{2}\theta d\theta [/tex] I have [tex] \int r^{2}sin^{2}\theta d\theta [/tex] 



#6
Jun2308, 09:17 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

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! (and be careful about the limits of integration!) 



#7
Jun2308, 10:03 PM

P: 17

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? 



#8
Jun2308, 11:05 PM

P: 17

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?




#9
Jun2408, 09:21 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

Hi Calcotron!
Why x = rsinu? Why deliberately give yourself limits like arcsin(cos)?? Try again with (the far more obvious?) x = rcosu. 



#10
Jun2408, 07:36 PM

P: 17

that works too



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