What is the area?

enter image source here

2 Answers
Dec 9, 2017

There is no real solution to this question.

Explanation:

The integral is not valid, if you look at what happens when theta approaches 0, you will see the function goes large, thus the area between [0,1] is kind of infinity.
---by @miles-a

I'm sorry. I don't know the steps of solving this. But you can try the formula mentioned by @alvin-l-2 as the following.

Dec 9, 2017

The calculation is hard. I used http://www.wolframalpha.com/ and obtained the result(about 0.50754).

Explanation:

This is the graph of r=1/(1-cos3theta) (by Wolframalpha).
http://www.wolframalpha.com/input/?i=polar+r%3D1%2F(1-cos(3theta))http://www.wolframalpha.com/input/?i=polar+r%3D1%2F(1-cos(3theta))

Then, how to interpret the graph?
[1] The graph crosses x=0 at theta=pi/2. There,
r=1/(1-cos((3pi)/2))=1.

[2] When does the graph cross x=1?
To solve this, apply the formula x=rcostheta.
1=rcostheta
r=1/costheta
1/(1-cos3theta)=1/costheta
1-cos3theta=costheta
cos3theta+costheta=1

Use the formula cos3theta=4cos^3theta-3costheta.
4cos^3theta-2costheta-1=0

The real solution for this is:
costheta≒0.88465
theta≒0.48505 [rad] ≒27.79 [deg]

In cartesian form, the graph crosses x=0 at (0,1) and crosses x=1 at (1,tan(0.48505))=(1,0.52705).

Now we are ready to calculate the area.
The area bounded by polar curves is S=1/2int_alpha^beta{f(theta)}^2dtheta.
In this problem,
S=1/2int_0.48505^(pi/2)(1/(1-cos3theta))^2dtheta=0.24401.

However, this integration is done in polar coordination. You must add the area of the right triangle with vertices(in cartesian form) (0,0), (1,0) and (1,0.52705).

The answer is 0.24401+1/2*1*0.52705=0.50754.

By the way, calculating areas bounded by polar curves is so difficult. I don't think it is necessary for us to do these types of calculation without a calculator.

I posted a question https://socratic.org/questions/calculating-areas-bounded-by-polar-curves-looks-extremely-difficult-do-americans two months ago but nobody has answered this. What are your opinions?