Finding the area of a sector of a cardioid

[Mampac]

Good day,
I'm given a cardioid radius r = 8 + 8 sinϕ and i have to find its SECTOR.

I know the formula for obtaining the area of this cardioid, and I can obtain it, but I need to subtract from it the not-highlighted part.

So far I had a thought about:

1) moving the cardioid 4 points down the y-axis and compute the area enclosed by the cardioid and the x-axis, but I realized I don't have the parameters (such that the cardioid is represented as a parametric function of x and y). That's why I was wondering: can I get the parameters x, y from the radius? Because I have solved such a problem but it provided me with the definitions of x and y.

2) computing the area of cardioid using Reimann sums... I remember stuff about subtracting bottom function from the top function: in this case, the bottom function could be 4. but isn't this polar space? I'm not allowed to do it, am I? Moreover, the area functions are different.

Any thoughts on how to proceed? It seems like I know all the formulae but can't see how to advance in this problem.

 

[Mampac]

UPDATE: i just figured out how to find the sector: in the cardioid formula, i have to specify the limits as the intersections of the cardioid with the line y = 4. that is, if I parametrize, I get:

4 = r * sinϕ
4 = (8 + 8 sinϕ) * sinϕ

I get a quadratic equation, but am failing to find the roots of ϕ.

 

[nasi112]

[MATH]y = 4[/MATH]
you can convert it to Polar
[MATH]r = \frac{4}{\sin \Phi}[/MATH]
then find when cardioid insect the line 4

[MATH]\frac{4}{\sin\Phi} = 8 + 8 \sin \Phi[/MATH]
after finding the intersection, i will call it [MATH]\Phi[/MATH]
then you want to find

[MATH]2\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 8 + 8 \sin \theta \ d\theta - 2\int_{\frac{-\pi}{2}}^{\Phi} 8 + 8 \sin \theta \ d\theta \ - \ [/MATH] 2Triangle

Triangle [MATH] = 0.5(4)(r \cos\Phi)[/MATH]

 

[nasi112]

I get a quadratic equation, but am failing to find the roots of ϕ.

[MATH] 8\sin^2 \Phi + 8\sin\Phi - 4 = 0 [/MATH]
let [MATH]u = \sin \Phi[/MATH]
then

[MATH]8u^2 + 8u - 4 = 0[/MATH]
[MATH]u = \frac{\sqrt{3} - 1}{2}[/MATH], we have taken only the positive value of the root

then

[MATH]\sin\Phi = \frac{\sqrt{3} - 1}{2}[/MATH]
[MATH]\Phi = \sin^{-1}(\frac{\sqrt{3} - 1}{2})[/MATH]

 

[Mampac]

[MATH]y = 4[/MATH]
you can convert it to Polar
[MATH]r = \frac{4}{\sin \Phi}[/MATH]
then find when cardioid insect the line 4

[MATH]\frac{4}{\sin\Phi} = 8 + 8 \sin \Phi[/MATH]
after finding the intersection, i will call it [MATH]\Phi[/MATH]
then you want to find

[MATH]2\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} 8 + 8 \sin \theta \ d\theta - 2\int_{\frac{-\pi}{2}}^{\Phi} 8 + 8 \sin \theta \ d\theta \ - \ [/MATH] 2Triangle

Triangle [MATH] = 0.5(4)(r \cos\Phi)[/MATH]

could you explain your choice of integration limits? and why there are two integrals instead of one?
isn't area of a cardioid equal to one half of integral from [MATH]ϕ[/MATH]₀ to [MATH]ϕ[/MATH]₁ of r squared? why do you have 2 as a coefficient and r to the power of one? and where do the [MATH]{\frac{\pi}{2}}[/MATH] and [MATH]{\frac{-\pi}{2}}[/MATH] come from?

 

[lex]

@Mampac
Yes you are right.

[MATH]\frac{1}{2}\left(8\right)^2\int_{\Phi}^{\frac{\pi}{2}}{\left(1+sin\theta\right)^2d\theta}-\frac{1}{2}\left(4\right)\left(8\right)\left(1+sin\Phi\right)cos\Phi[/MATH]
where [MATH] \Phi={sin}^{-1}\left(\frac{\sqrt3-1}{2}\right)[/MATH]

This will give you the required area on the right hand side of the shape. As it is symmetrical, you can just double this answer.

 

[nasi112]

could you explain your choice of integration limits? and why there are two integrals instead of one?
isn't area of a cardioid equal to one half of integral from [MATH]ϕ[/MATH]₀ to [MATH]ϕ[/MATH]₁ of r squared? why do you have 2 as a coefficient and r to the power of one? and where do the [MATH]{\frac{\pi}{2}}[/MATH] and [MATH]{\frac{-\pi}{2}}[/MATH] come from?

Thank you a lot for alerting me. You are correct about the area. I will correct my integral now.

[MATH]2\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2}(8 + 8\sin\theta)^2 \ d\theta[/MATH]
this integral will find the whole area of the cardioid.

you can also use this integral

[MATH]\int_{0}^{2\pi} \frac{1}{2}(8 + 8\sin\theta)^2 \ d\theta[/MATH]
So, it does not matter what limits you use as long as you make a complete revolution.

Now, look at the picture.



I will set the integral to find the yellow area.

[MATH]2\int_{\frac{-\pi}{2}}^{\sin^{-1}(\frac{\sqrt{3} - 1}{2})} \frac{1}{2}(8 + 8\sin\theta)^2 \ d\theta[/MATH]
Now, look at the picture again.



I will set the integral to find the green area.

[MATH]2\int_{\sin^{-1}(\frac{\sqrt{3} - 1}{2})}^{\frac{\pi}{2}} \frac{1}{2}(\frac{4}{\sin\theta})^2\ d\theta[/MATH]
You can also find the green area by using the triangle area.

[MATH]2[/MATH]triangle area [MATH]= 2\frac{1}{2}(4)r\cos\theta = 2\frac{1}{2}(4)\frac{4\cos\theta}{\sin\theta} = 16\cot\theta = \frac{16}{\tan\theta}[/MATH]
So, the green area is [MATH]\frac{16}{\tan(\sin^{-1}(\frac{\sqrt{3} - 1}{2}))}[/MATH]
Look at the picture.



Now, we want to find the red area.

Red Area = Whole Cardioid Area - Yellow Area - Green Area

Or by the integral, the required area is

[MATH]2\int_{\frac{-\pi}{2}}^{\frac{\pi}{2}} \frac{1}{2}(8 + 8\sin\theta)^2 \ d\theta \ - \ 2\int_{\frac{-\pi}{2}}^{\sin^{-1}(\frac{\sqrt{3} - 1}{2})} \frac{1}{2}(8 + 8\sin\theta)^2 \ d\theta \ - \ 2\int_{\sin^{-1}(\frac{\sqrt{3} - 1}{2})}^{\frac{\pi}{2}} \frac{1}{2}(\frac{4}{\sin\theta})^2\ d\theta[/MATH]