# Area enclosed between half lines in polar space

I don't know if the anwser to my question is obvious because I cannot find any explanation anywhere on google.

Question

The blue region $$R$$ is bounded by the curve C with equation $$r^{2} = a^{2}cos(2\theta)$$ $$0 \leqslant \theta \leqslant \frac{\pi}{4}$$, the line $$\theta = \frac{\pi}{2}$$ and the line $$l$$ which is parallel to the initial line. The point $$P(\frac{a}{\sqrt{2}},\frac{\pi}{6})$$

Show the area of the blue region $$R$$ is $$\frac{a^{2}}{16}(3\sqrt{3} - 4)$$.

I tried solving this by finding the area of the rectangle up to $$P$$ then taking the area of the triangle up to $$P$$ away as well as the area of C enclosed by the green half line $$\theta = \frac{\pi}{6}$$ and $$\theta = \frac{\pi}{2}$$ which is where I made a mistake but do not understand why it does not work for $$\theta = \frac{\pi}{2}$$.

I've graphed the curve and shaded the regions as seen in the picture:

I know the general method of answering questions like these but what I am really asking is

Given a polar curve , if I choose two half lines (in this example the green one with equation $$\theta = \frac{\pi}{6}$$ and another half line that does enclose the curve BUT is greater than $$\theta = \frac{\pi}{4}$$ i.e. it is any half line that encloses the curve but isn't the closest one to it which is the pink one $$\theta = \frac{\pi}{4}$$ , why do I get a different area than the actual area?

Is the curve not enclosed by any half line that is greater than $$\frac{\pi}{4}$$ and $$\frac{\pi}{6}$$ ?

How do I determine the other half line to enclose the curve that is correct then?

In the past I've come across several questions where I could avoid this situation but now since my exams are approaching fast I feel I need to understand this properly.

The portion of R above the $\pi/4$ line (let us call it R1) is half of a square with side $\pi/(2\sqrt(2))$, so that its area is $\pi^2/16$.

The area of the remaining part of R (let us call it R2) can be calculated as the difference between the area of the triangle delimitated by the lines $\pi/4$, $\pi/6$, and L, and that of the small portion of C above the $\pi/6$ line.

The triangle has base equal to $\pi (\sqrt{3}-1)/(2\sqrt{2})$ and height equal to $\pi/(2\sqrt{2})$, so that its area is $\pi^2 (\sqrt{3}-1)/16)$.

The area of the small portion of C above the $\pi/6$ line can be calculated by integrating $\frac{1}{2} a^2 cos(2\theta)$ between $\pi/4$ and $\pi/6$ (in polar coordinates, integration of $f(\theta)$ is performed by standard integration of $(f(\theta))^2/2$). The indefinite integral is $\frac{1}{2} a^2 sen(\theta) cos(\theta)$, which calculated over the above mentioned interval yields $\pi^2 (4-2\sqrt{3}/16)$.

Thus the area of R2 is:

$\pi^2 (\sqrt{3}-1)/16 - \pi^2 (4-2\sqrt{3})/16)=\pi^2 (3\sqrt{3} - 5)/16$.

Summing R1 and R2 we get:

$\pi^2/16 + \pi^2 (3\sqrt{3} - 5)/16=\pi^2 (3\sqrt{3} - 4)/16$.

Clearly we could also have obtained the same result more directly as the difference between the area of the triangle delimitated by the $\pi/6$ line, the y-axis and L, and that of the portion of C above the $\pi/6$ line.

• Why do I get an area different to the actual one if I choose the half line to be greater than $\frac{\pi}{4}$ – Nubcake May 30 '14 at 22:24
• When I first worked it out I used pi/2 as the second half line and I believe I did get a different area , I'll try this again now. – Nubcake May 30 '14 at 22:51
• You do not get a different area if you choose a half line greater than $\pi/4$. In this case, R1 decreases and R2 increases, but their sum does not change. – Anatoly May 30 '14 at 22:55
• I've ended up with $\frac{3\sqrt{3}a^{2}}{16}$ when I used pi/2. – Nubcake May 30 '14 at 22:59
• So I re did the working out and I came up with this $\frac{a^{2}}{16}(3\sqrt{3} - 4Sin(2\theta ))$ derived from $\frac{a^{2}\sqrt{3}}{16} - \frac{a^{2}}{4} \left[Sin(2\theta)]_\frac{\pi}{6}^\theta$ where $\theta$ is the equation of the half life. If the area as you say will be the same regardless of the value of the half line what is wrong with what I posted here? This provides the correct answer for $\theta = \frac{\pi}{6}$ but not for other values. – Nubcake May 30 '14 at 23:16