The graph shows the designated area as dark shading. To find out where the lemniscate and circle intersect we solve for θ: r²=8sinθ for r=2, sinθ=4/8=½, θ=π/6 and 5π/6. The intersection points are given by polar coordinates (2,π/6) and (2,5π/6).

If we consider any point P(r,θ) we can create a small arc which subtends an angle dθ at the origin a distance r from the origin, and the area of the resultant sector is ½r²dθ. The sum of such areas will give the area under a curve. In this case we have two curves, so the difference of two areas will give the area of the space between them.

∫[⅙π,⅚π]8sinθdθ-∫[⅙π,⅚π]4dθ gives us the required area. We can combine these:

∫[⅙π,⅚π](8sinθ-4)dθ=(-8cosθ-4θ)[⅙π,⅚π]=

4√3-10π/3-(-4√3-2π/3)=8√3-8π/3=5.4788 approx.