# What is the area?

## 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 $r = \frac{1}{1 - \cos 3 \theta}$ (by Wolframalpha). Then, how to interpret the graph?
 The graph crosses $x = 0$ at $\theta = \frac{\pi}{2}$. There,
$r = \frac{1}{1 - \cos \left(\frac{3 \pi}{2}\right)} = 1$.

 When does the graph cross $x = 1$?
To solve this, apply the formula $x = r \cos \theta$.
$1 = r \cos \theta$
$r = \frac{1}{\cos} \theta$
$\frac{1}{1 - \cos 3 \theta} = \frac{1}{\cos} \theta$
$1 - \cos 3 \theta = \cos \theta$
$\cos 3 \theta + \cos \theta = 1$

Use the formula $\cos 3 \theta = 4 {\cos}^{3} \theta - 3 \cos \theta$.
$4 {\cos}^{3} \theta - 2 \cos \theta - 1 = 0$

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 $\left(0 , 1\right)$ and crosses $x = 1$ at $\left(1 , \tan \left(0.48505\right)\right) = \left(1 , 0.52705\right)$.

Now we are ready to calculate the area.
The area bounded by polar curves is $S = \frac{1}{2} {\int}_{\alpha}^{\beta} {\left\{f \left(\theta\right)\right\}}^{2} d$$\theta$.
In this problem,
$S = \frac{1}{2} {\int}_{0.48505}^{\frac{\pi}{2}} {\left(\frac{1}{1 - \cos 3 \theta}\right)}^{2} d$$\theta = 0.24401$.

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

The answer is $0.24401 + \frac{1}{2} \cdot 1 \cdot 0.52705 = 0.50754$.

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?