Revising past exam papers and I am completely lost

(c) The area of this is twice the area of its upper half. Luckily, the symmetry axis is the x-axis, so you need only look for the part above the x-axis. Those parts of the curve, it's easily calculated, correspond to **phi ∈ $0,2pi/3$** (the "bigger" one) and **phi ∈ $pi,4pi/3$** (the "smaller" one). The formula for area bounded in polar coordinates for **phi ∈ $alpha, beta$** is **A=1/2\*int$alpha,beta$ r(phi)\^2 dphi**. To find this *half,* you compute this formula for **phi ∈ $0,2pi/3$**, and then *subtract* the one for **phi ∈ $pi,4pi/3$**. What you got is area of half the figure, so you **multiply it by 2.**