We have two circles $r = a \cos \theta$ and $ r = a \sin \theta$. We need to find the area between them using Polar Co-ordinates (Area 1).
This is what I did :
$$ Area = \int_{0}^{\frac{\pi}{2}} \left(\int_{a \sin \theta}^{a \cos \theta} r \ dr \right) \ d\theta = 0$$
Which shouldn't be the case, since area must be $> 0$ here.
What is wrong in my approach?
Please help.
The area of intersection is not given by $$0 \leq \theta \leq \frac{\pi}{2}, a \sin \theta \leq r \leq a \cos \theta$$ For the reasons geodude mentioned, the (net) area of this region is zero. Note that due to the phase shift between the sine and cosine curves, the sine circle boundary of the intersection region is drawn left to right when $0 \leq \theta \leq \frac{\pi}{4}$, and the cosine circle boundary is drawn right to left when $\frac{\pi}{4} \leq \theta \leq \frac{\pi}{2}$.
Therefore, one way to go about finding the are of the intersection region is to split it into two pieces with a line segment $L$ from $(0,0)$ to $(a,a)$. The total area $A$ is then given by
$$A = \underbrace{\int_0^{\frac{\pi}{4}} \int_0^{a \sin \theta} r \,dr \,d\theta}_{S} + \underbrace{\int_{\frac{\pi}{4}}^{\frac{\pi}{2}} \int_0^{a \cos \theta} r \,dr \,d\theta}_{C} $$
$S$ gives the area between $L$ and the green boundary from the sine circle, and $C$ given the area between $L$ and the red boundary from the cosine circle.
$$\begin{align} S &= \int_0^{\frac{\pi}{4}} \int_0^{a \sin \theta} r \,dr \,d\theta \\ &= \int_0^{\frac{\pi}{4}} \frac{1}{2} a^2 \sin^2 \theta \,d\theta \\ &= \left [ \frac{1}{4} a^2 \left ( \theta - \sin \theta \cos \theta \right )\right ]_0^{\frac{\pi}{4}} \\ &= \frac{1}{4}a^2 \left ( \frac{\pi}{4} - \frac{1}{2} \right ) \end{align} $$
$C$ is also equal to this (which makes sense since the region of interest is symmetrical about $L$), so
$$ A = \frac{1}{2}a^2 \left ( \frac{\pi}{4} - \frac{1}{2} \right ) $$