Area in polar coordinates

[MathsLearner]

The problem summary statement is, When recording live a microphone is used with cardioid pickup pattern. suppose the microphone is placed 4m from the front of the stage and the boundary is given by cardioid r = 8 + 8[math] \sin\theta[/math], where r is measured in meters and the microphone at pole. What is the area of the optimal pickup range of the microphone.

My attempt is i am finding the area of the blue region.
The total area is
A = [math]\int_0^{2\pi} \frac {r^2} 2 d\theta[/math]A=[math]\int_0^{2\pi} 32(1+\sin\theta)^2 d\theta[/math] = 301.59
But i am confused how much area i need to subtract. That is which is the point i have to take at [math]0\deg[/math]? What is the angle at which cardioid intersects with the stage. Please advise.

 

[Deleted member 4993]

The problem summary statement is, When recording live a microphone is used with cardioid pickup pattern. suppose the microphone is placed 4m from the front of the stage and the boundary is given by cardioid r = 8 + 8[math] \sin\theta[/math], where r is measured in meters and the microphone at pole. What is the area of the optimal pickup range of the microphone.
View attachment 20017
My attempt is i am finding the area of the blue region.
The total area is
A = [math]\int_0^{2\pi} \frac {r^2} 2 d\theta[/math]A=[math]\int_0^{2\pi} 32(1+\sin\theta)^2 d\theta[/math] = 301.59
But i am confused how much area i need to subtract. That is which is the point i have to take at [math]0\deg[/math]? What is the angle at which cardioid intersects with the stage. Please advise.

First you need to locate the origin and the line \(\displaystyle \theta = 0\)

At \(\displaystyle \theta = 0 \ \ \to \ \ \) r = ?

At \(\displaystyle \theta = \frac{\pi}{2} \ \ \to \ \ \) r = ?

At \(\displaystyle \theta = \pi \ \ \to \ \ \) r = ?

Plot the function using wolframalpha.com

You should be able to calculate the answers to your questions.

 

[Dr.Peterson]

The problem summary statement is, When recording live a microphone is used with cardioid pickup pattern. suppose the microphone is placed 4m from the front of the stage and the boundary is given by cardioid r = 8 + 8[math] \sin\theta[/math], where r is measured in meters and the microphone at pole. What is the area of the optimal pickup range of the microphone.
View attachment 20017
My attempt is i am finding the area of the blue region.
The total area is
A = [math]\int_0^{2\pi} \frac {r^2} 2 d\theta[/math]A=[math]\int_0^{2\pi} 32(1+\sin\theta)^2 d\theta[/math] = 301.59
But i am confused how much area i need to subtract. That is which is the point i have to take at [math]0\deg[/math]? What is the angle at which cardioid intersects with the stage. Please advise.

I would start by writing an equation for the line marking the front of the stage, and put that in polar coordinates. Then find the intersections of that with the cardioid.

 

[Sonal7]

I would start by writing an equation for the line marking the front of the stage, and put that in polar coordinates. Then find the intersections of that with the cardioid.

Can you just not find theta when r is 4?

[MATH]8 + 8 sin \theta =4 [/MATH]
Yes i got it, you cant theta is -pi/6 so you need the points on intersection first.

 

[Dr.Peterson]

Can you just not find theta when r is 4?

[MATH]8 + 8 sin \theta =4 [/MATH]
Yes i got it, you cant theta is -pi/6 so you need the points on intersection first.

How did you get that? It isn't right; rather than r=4, you need y=4, right?

 

[Sonal7]

How did you get that? It isn't right; rather than r=4, you need y=4, right?

Yes we need to find the points of intersection. It is easily done using a graphical calculator or desmos, then one just need to work out the angle.
But i m tying to work this out a normal calculator.

[MATH]y=4\\ r= 8+ 8 sin \theta\\[/MATH]
substitute in [MATH]r^2= x^2+y^2[/MATH]
[MATH] y^2+x^2=(8+8 sin\theta)^2 \\ 16+ x^2 = (8+ 8 sin \theta)^2\\ [/MATH]

 

[Dr.Peterson]

That doesn't really help, because now you have three variables!

Do you know how you express y in terms of r and theta? That's fundamental in discussing polar coordinates.

 

[Sonal7]

That doesn't really help, because now you have three variables!

Do you know how you express y in terms of r and theta? That's fundamental in discussing polar coordinates.

I was trying to work out x as that would let me use an inverse than to work an angle I need, initial line to the the point of the intersection. ah ok!

[MATH]y= r sin \theta[/MATH]
We still dont know theta but I got the equation we need:

[MATH]4= r sin \theta\\ r=4/sin\theta\\ 4/sin\theta=8+8sin\theta[/MATH]
so [MATH]\theta =21.47 degrees \\or\\ 0.375 rad [/MATH]

 

[Sonal7]

i have an ans.

 

[Dr.Peterson]

I was trying to work out x as that would let me use an inverse than to work an angle I need, initial line to the the point of the intersection. ah ok!

[MATH]y= r sin \theta[/MATH]
We still dont know theta but I got the equation we need:

[MATH]4= r sin \theta\\ r=4/sin\theta\\ 4/sin\theta=8+8sin\theta[/MATH]
so [MATH]\theta =21.47 degrees \\or\\ 0.375 rad [/MATH]

That is the correct angle; I got it just a little differently, as [MATH]r\sin(\theta) = (8+8\sin(\theta))\sin(\theta)[/MATH], so that [MATH]8\sin(\theta) +8\sin^2(\theta) = 4[/MATH]. You end up with the same quadratic equation.

 

[Sonal7]

That is the correct angle; I got it just a little differently, as [MATH]r\sin(\theta) = (8+8\sin(\theta))\sin(\theta)[/MATH], so that [MATH]8\sin(\theta) +8\sin^2(\theta) = 4[/MATH]. You end up with the same quadratic equation.

yes I did the same as you, but I didn't wish to spoil it for the OP. I solved it. The point of intersection is what desmos is giving me (i think)

 

[Dr.Peterson]

Sorry, I was thinking you were the OP.

 

[Sonal7]

The OP seemed to lose interest, and i had questions on the same topic- i had a good practise with this one.

 

[MathsLearner]

Ok, I understood the equation of the line and arrived at the angle as 21.47 degrees. But still i think i did not get the proper answer.

[math]A =\int_{21.47}^{180+21.47}32(1+\sin\theta)^2d\theta \\ A = 32\int_{21.47}^{201.47}(1+\sin\theta)^2d\theta [/math]I used online tool to find the integral, but the answer is strange it is some 8513, which i did not expect. What is the mistake?

 

[Dr.Peterson]

The calculations must be done using radians.

Also, your upper bound is wrong.

And you are not finding only the shaded area. Doing it the way you are, you will have to subtract the area of a triangle.

 

[Sonal7]

Ok, I understood the equation of the line and arrived at the angle as 21.47 degrees. But still i think i did not get the proper answer.

[math]A =\int_{21.47}^{180+21.47}32(1+\sin\theta)^2d\theta \\ A = 32\int_{21.47}^{201.47}(1+\sin\theta)^2d\theta [/math]I used online tool to find the integral, but the answer is strange it is some 8513, which i did not expect. What is the mistake?

Also you might need to decompose the area, I had to work out an area of a triangle.. You might need to work out an angle in the triangle..
Yup always in rads when doing integration. thanks for posting this Q.

 

[MathsLearner]

The calculations must be done using radians.

Sorry for this i converted it into radians 0.374.

Also, your upper bound is wrong.
And you are not finding only the shaded area. Doing it the way you are, you will have to subtract the area of a triangle.

I am confused about this.
[math] A= 32\int_0^{0.374}(1+\sin\theta)^2d\theta; \\ A= 16.99; [/math]Two areas so twice of this [math]= 16.99*2 = 34 [/math]The area of blue region is [math]301-34=267[/math]. I don't think the answer is yet correct. But i am not understanding from where the triangle is coming, i need to think more about this.

 

[Sonal7]

Sorry for this i converted it into radians 0.374.


I am confused about this.
[math] A= 32\int_0^{0.374}(1+\sin\theta)^2d\theta; \\ A= 16.99; [/math]Two areas so twice of this [math]= 16.99*2 = 34 [/math]The area of blue region is [math]301-34=267[/math]. I don't think the answer is yet correct. But i am not understanding from where the triangle is coming, i need to think more about this.

Think about which area the integrand will give you. The formula is essentially area of a sector but the theta has so small (this is how integration works), its been ignored. so if you make a sweep of tiny sectors from the pole what do you get? So whats the area( its the not the white area enclosed by the circoid).

 

[Dr.Peterson]

Here are the other sides of the triangle, in red:

Your integral needs to go from one red line to the other; your error in the first version, besides being in degrees, was that you added rather than subtracting. The way you are doing it now, subtracting the two side pieces (below the red lines) from the whole cardioid, is also valid, but the next paragraph applies either way.

The area you get will then include both the blue area and the isosceles triangle below it. You need to subtract the area of that triangle. (You could instead take this into account in your integrand, but that would probably be harder for you.)

 

[MathsLearner]

Ok, I am getting cleared now, i had misunderstandings about the concepts.


Sorry i don't have a proper tool to draw exactly, but i have shown
a. the origin,
b. the Degress 21.47 and 180-21.47 = 158.3 Deg
c. The Isosceles triangle ABC.

Just i want to make clear that according to the problem i need to calculate the area of the blue region? For me it is "Yes".
Next the calculations
[math] \int_{0.375}^{5.908}32(1+\sin\theta)^2 d\theta; =276.49 [/math]Now the area of the isosceles triangle
[math] r\sin(21.47) = 4; \ r = 4/\sin(21.47) = 10.92 \\ x axis = 10.92*\cos(21.47) = 10.17[/math]Area of ABC triangle = [math]4 * 10.17 = 40.68 [/math]
The area of the Blue region [math]= 276.49 - 40.68 = 235.8 [/math] Is it correct? The answer does not seem to match.