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

In summary: I got0.5r^2.theta - 0.5 * int (0,theta) cos 2*t dtfor second area...In summary, the area of the sector is 0.5r^2.theta - 0.5*int(0,theta)cos2*t dt.
  • #1
Calcotron
17
0
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.)

07_trigonometric_functions-516.gif


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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  • #2
there's a better way:

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

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

Using Jacobi transformations
integrate from 0 to theta
 
  • #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

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

I have
[tex]
\int r^{2}sin^{2}\theta d\theta
[/tex]
 
  • #6
Calcotron said:
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! :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!)
 
  • #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?
 
  • #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?
 
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  • #9
go for the obvious …

Hi Calcotron! :smile:
Calcotron said:
Now, for the other region I used the integral [tex]\int\sqrt{r^{2} - x^{2}}dx[/tex]
Calcotron said:
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? :confused:

Why x = rsinu?

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

Try again with (the far more obvious?) x = rcosu. :smile:
 
  • #10
that works too
 

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

What is a sector of a circle?

A sector of a circle is a portion of the circular region enclosed by two radii and an arc. It is a wedge-shaped section that resembles a slice of pizza or a piece of pie.

How do you calculate the area of a sector of a circle?

To calculate the area of a sector of a circle, you can use the following formula: \[A = \frac{\theta}{360} \times \pi r^2\] Where: - \(A\) represents the area of the sector. - \(\theta\) is the central angle of the sector in degrees. - \(r\) is the radius of the circle.

What is the central angle in the formula for the area of a sector?

The central angle, represented by \(\theta\), is the angle formed at the center of the circle by the two radii that define the sector. It determines the size of the sector and is measured in degrees. The central angle varies depending on the fraction of the circle's circumference the sector represents.

Can you provide an example of calculating the area of a sector of a circle?

Sure! Let's say you have a circle with a radius (\(r\)) of 6 inches, and you want to find the area of a sector that subtends a central angle (\(\theta\)) of 60 degrees. Using the formula, you can calculate it as follows: \[A = \frac{60}{360} \times \pi \times (6^2) = \frac{1}{6} \times \pi \times 36 \approx 18.85 \text{ square inches}\] So, the area of the sector is approximately 18.85 square inches.

What units should I use for the radius when calculating the area of a sector?

The units for the radius should match the units used for the area. If you're using inches for the radius, the area will be in square inches. Similarly, if you use meters for the radius, the area will be in square meters. Consistency in units is essential for accurate calculations.

When would you need to find the area of a sector in practical applications?

Finding the area of a sector is useful in various real-world scenarios. For example, it can be applied in geography to calculate the area of a portion of a city on a circular map, in physics to determine the angular displacement covered by an object in motion, or in geometry to solve problems involving circular regions or pie charts representing data.

Are there any special cases or considerations when calculating sector areas?

One special case to consider is when the central angle (\(\theta\)) equals 360 degrees, which means the sector encompasses the entire circle. In this case, the sector's area will be the same as the area of the entire circle (\(A = \pi r^2\)). Additionally, make sure to use consistent units for all measurements when calculating sector areas.

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