What is the formula for the area of a sector of a circle?

[Calcotron]

Prove the formula $$A = \frac{1}{2}r^{2}\theta$$ for the area of a sector of a circle with radius r and central angle $$\theta$$. (Hint: Assume 0 < $$\theta$$ < $$\frac{\pi}{2}$$ and place the center of the circle at the origin so it has the equation $$x^{2} + y^{2} = r^{2}$$ . 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 $$\frac{1}{2}r^{2}cos\theta sin\theta$$

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

I make x = r sin$$\theta$$

I plug that in under the square root sign and get r cos$$\theta$$.

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

Now since dx = r cos$$\theta$$d$$\theta$$ the integral for the area of the second region is $$\int r^{2}cos^{2}\theta d\theta$$. 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 $$\frac{1}{2}r^{2}(\theta+\frac{1}{2}sin2\thetacos\theta)$$ 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 $$\frac{1}{2}r^{2}\theta$$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?

 

[rootX]

there's a better way:

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

use r.theta = arc length

 

[Calcotron]

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

 

[rootX]

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

Using Jacobi transformations
integrate from 0 to theta

 

[rootX]

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

I have
$$\int r^{2}sin^{2}\theta d\theta$$

 

[tiny-tim]

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

I make x = r sin$$\theta$$

I plug that in under the square root sign and get r cos$$\theta$$.

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

Now since dx = r cos$$\theta$$d$$\theta$$ the integral for the area of the second region is $$\int r^{2}cos^{2}\theta d\theta$$.

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]

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 $$sin^{-1}cos\theta$$ to $$\frac{\pi}{2}$$correct?

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

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]

Well, if I replace $$sin^{-1}cos\theta$$ with $$\frac{\pi}{2} - \theta$$ 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]

go for the obvious …

Hi Calcotron!

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

Ok, so if I make the substitution x = r sin u the limits change to $$sin^{-1}cos\theta$$ to $$\frac{\pi}{2}$$correct?

uhh?

Why x = rsinu?

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

Try again with (the far more obvious?) x = rcosu.

 

[Calcotron]

that works too