Area in polar coordinates

[Sonal7]

I

Ok, I am getting cleared now, i had misunderstandings about the concepts.
View attachment 20052

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.

YOu should integrate from A to B then double it. Origin is actually called the POLE. In an exam the terminology might throw you off, as happened to me. They said origin i the exam and I had only known it to be called 0,0. I got the same area for the triangle. Then you need to add the triangle to the lower bits and subtract to the value you gave us for the whole area.

 

[Sonal7]

Ok, I am getting cleared now, i had misunderstandings about the concepts.
View attachment 20052

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.
[/QUOT

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.

This step is wrong, you have to integrate from 0 to 0.375 (I think you rounded off incorrectly) and then 3/2pi to 2pi. Please do some basic problems as so that basics does not throw you off. Then add those two together and double it, that give you the 'wings', add that to the triangle and then subtract by the full area.

 

[MathsLearner]

This step is wrong, you have to integrate from 0 to 0.375 (I think you rounded off incorrectly) and then 3/2pi to 2pi. Please do some basic problems as so that basics does not throw you off. Then add those two together and double it, that give you the 'wings', add that to the triangle and then subtract by the full area.

But i am not doing method, i have changed to method as shown in post 20. Can you advise the mistakes in that method? what is wrong.

Then you need to add the triangle to the lower bits

What is lower bits? I don't understand the terminology. It confuses me more.

 

[Sonal7]

But i am not doing method, i have changed to method as shown in post 20. Can you advise the mistakes in that method? what is wrong.


What is lower bits? I don't understand the terminology. It confuses me more.

Sorry I am going mad. You did integrate from 0 to 0.375 and then *2, and you have correctly subtracted that from the full area minus the triangle. The reason the area for 0 to 0.375 is not the full area that looks like wings ( attached to the triangle) , is that this integrad is giving you the area from the INITIAL LINE in small sectors to the point of intersection. But the initial line is above the lowest point of these wings and thus you have cut off the lower bits. This bit can be calculated by adding 2* area given by the integrad from 3/2 pi to 2 pi. I think you dont understand the basics. Let me see if I can send you a link

 

[Sonal7]

I was doing to draw on the diagram but you have already identified correctly what the initial line is. You know where you drew a horizontal line on your diagram. That is correct. That is where the initial line is so 0 to 0.375 is the area above the initial line. You also need the area below the horizontal line which unfortunately you forgot to consider!

The initial line is where the angle is 0. on your diagram you drew a horizontal over it. so you calculated the area from 0 to 0.375 - from the initial line to the point of intersection, so what about the area below this? You cant just ignore it.

 

[LarlSangan42]

So you are saying he forgot to double the integral from 0 to 0.375 , because it has another symmetrical area he needs to substract too ?

 

[Dr.Peterson]

Whom are you talking to? Sonal7 in post #24 said, "You did integrate from 0 to 0.375 and then *2 ". That is not the issue.

I just realized I never saw post #20, because of going to the second page. The error there is in the upper limit, which was converted to radians incorrectly.

 

[MathsLearner]

I just realized I never saw post #20, because of going to the second page. The error there is in the upper limit, which was converted to radians incorrectly.

Thank you for pointing out the mistake, the final result is
[math] \int_{0.375}^{2.76}32(1+\sin\theta)^2 d\theta = 244.41 [/math]Subtract the area of triangle = [math] 40.68 [/math]. The final result is
[math] 244.41 - 40.68 = 203.72 [/math]