Region bounded by a Polar Curve

Question

For a National Board Exam:

Find the area of the region bounded by a polar curve $r^2 = a^2 \cos(2\theta)$

Answer = $a^2$.

So I cheated a bit and plotted the curve on wolfram so i could visualize things quickly: http://www.wolframalpha.com/input/?i=r%5E2+%3D+5%5E2+cos%282theta%29 . Saw this infinity symbol type shape so I decided to set the limits to 0 and pi/2 and just multiply it by 4 from there.

$${ A = \int^b_a \frac{1}{2} f(\theta)^2 d\theta }$$

Put this on my calculator, I get wrong answer but its in the choices...

$${ A = 4\int^{\pi /2}_{0} \frac{1}{2} a^2cos({2\theta}) d\theta = 2a^2}$$

What am I doing wrong?

Answer 1

Hint:

note that the part of the curve in the first quadrant is given by: $$ r=\sqrt{a^2 \cos 2 \theta} \quad for \quad 0\le \theta \le \frac{\pi}{4} $$ (for $\frac{\pi}{4}<\theta\le \frac{\pi}{2}$, $\cos 2\theta <0$)

so the area is: $$ A=\frac{1}{2}\int_0^{\frac{\pi}{4}}r^2(\theta)d \theta=\frac{1}{2}\int_0^{\frac{\pi}{4}}a^2 \cos (2\theta)d \theta $$ that you can easily calculate.