Consider the 4 leaf rose and a circle having polar equations: r=10cos(2Θ) and r=5 with 0≤Θ≤2π, respectively. Find the area of the region that lies inside the rose and outside the circle. hint: find the smallest positive value of Θ for which the two curves intersect; use symmetry.

Question

So far I have gotten; 1/2*int[-π/4,π/4](10cos^2(2Θ))dΘ
=5/2*int[-π/4,π/4](1+cos(4Θ))dΘ
I am not sure about my lower and upper limits. Will this set up get me the answer?

Answer 1

The curves intersect at π/6, not π/4. The leaves of the rosette return to (0,0) at multiples of π/4, but that's not where the circle intersects the rosette.

Using the 8-way symmetry, I'd just integrate over [0,π/6], so

a = 8∫[0,π/6] 1/2 (R^2-r^2) dθ
where R=10cos(2θ) and r=5, so
a = 8∫[0,π/6] 1/2 (100cos^2(2θ)-25) dθ = 25/3 (3√3+2π)