Area
As described on this page, the area enclosed by a polar curve is given by
$$A = \int_\alpha^\beta \frac{r^2}{2} \mathrm{d}\theta$$$$A = \int_\alpha^\beta \frac{r(\theta)^2}{2} \mathrm{d}\theta$$
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.
Arc Length
Going back to that same site, the formula for the arc length is
$$L = \int \mathrm{d}s$$
where
$$\mathrm{d}s = \sqrt{r^2+\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}\right)^2}$$$$\mathrm{d}s = \sqrt{r(\theta)^2+\left(\frac{\mathrm{d}r(\theta)}{\mathrm{d}\theta}\right)^2}\mathrm{d}\theta$$
Using Integrate we get
Integrate[
Sqrt[Sin[2 θ]^2 + D[Sin[2 θ], θ]^2], {θ,
0, π}]
N@%
(* 4 EllipticE[3/4] *)
(* 4.84422 *)
Or, we can extract the Line object from the PolarPlot and find its length using ArcLength
Cases[
PolarPlot[Sin[2 θ], {θ, 0, π}, PlotRange -> All],
Line[_], Infinity] // First // ArcLength
(* 4.8441 *)
Again, this seems to be good for three decimal places.
