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Area of a sector of a circle

by Calcotron
Tags: circle, sector
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Calcotron
#1
Jun22-08, 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?
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rootX
#2
Jun22-08, 06:09 PM
rootX's Avatar
P: 1,294
there's a better way:

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

use r.theta = arc length
Calcotron
#3
Jun22-08, 06:32 PM
P: 17
While I see that does work, I am more interested in solving the problem using trigonometric substitution.

rootX
#4
Jun22-08, 07:55 PM
rootX's Avatar
P: 1,294
Area of a sector of a circle

why divide it into two areas
when you can use Polar co-ordinates?

Using Jacobi transformations
integrate from 0 to theta
rootX
#5
Jun22-08, 11:07 PM
rootX's Avatar
P: 1,294
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]
tiny-tim
#6
Jun23-08, 09:17 AM
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P: 26,148
Quote Quote by Calcotron View Post
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].
Hi Calcotron!

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!)
Calcotron
#7
Jun23-08, 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?
Calcotron
#8
Jun23-08, 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?
tiny-tim
#9
Jun24-08, 09:21 AM
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P: 26,148
Hi Calcotron!
Quote Quote by Calcotron View Post
Now, for the other region I used the integral [tex]\int\sqrt{r^{2} - x^{2}}dx[/tex]
Quote Quote by Calcotron View Post
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?
uhh?

Why x = rsinu?

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

Try again with (the far more obvious?) x = rcosu.
Calcotron
#10
Jun24-08, 07:36 PM
P: 17
that works too


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