Though less common in JEE Main, polar area problems do appear. The area enclosed by r = f(theta) from theta1 to theta2 is A = $\frac{1}{2}$ * integral from theta1 to theta2 of [f(theta)]^2 d(theta). For a full cardioid r = a(1 + cos(theta)), the area is $\frac{1}{2}$ * integral from 0 to 2pi of $a^2$(1 + cos(theta))^2 d(theta) = $\frac{3*pi*a^2}{2}$. For the rose curve r = acos(2theta), each petal has area pi$a^2$/8, and with 4 petals the total is pi*$a^2$/2. The key challenge is determining the correct angular limits for the desired region.