Answer:
Step-by-step explanation:
Given are two polar curves as [tex]r = 6 sin θ and r = 2+ 2 sin θ[/tex]
Let us find the point/s of intersection
Eliminate r to get
[tex]6 sin θ = 2+ 2 sin θ\\sin\theta = 0.5\\\theta = 30, 150[/tex]
[tex]r=6sin 30 = 3[/tex]
Thus we find that area outside II curve and inside first curve would be
[tex]\int {\frac{r^2}{2} } \, d\theta \\ =\int{\frac{36sin^2 \theta - (2+2sin\theta)^2 }{2} } \, d\theta[/tex] where lower limit is 30 degrees and 150 degrees.
