Return to Revisions

As described on this page, the area enclosed by a polar curve is given by

$$A = \int_\alpha^\beta \frac{r^2}{2} \mathrm{d}\theta$$

In your case this is,

Integrate[Sin[2 θ]^2/2, {θ, 0, π}]
N@%
(* π/4 *)
(* 0.785398 *)

You can get this same answer using Region functionality by first making a RegionPlot, converting it to a MeshRegion and finding the area,

RegionPlot[{Sqrt[x^2 + y^2] <= Sin[2 ArcTan[y/x]], 
   Sqrt[x^2 + y^2] <= Sin[2 ArcTan[-y/x]]}, {x, 0, 1}, {y, -1, 1}, 
  PlotPoints -> 60] // DiscretizeGraphics

Area@%
(* 0.785374 *)

But this is only good to a limited degree of precision.