Question

Sketch the polar curves r = 6 sin θ and r = 2+ 2 sin θ, then set up an integral that represents the area inside r = 6 sin θ and outside r = 2 + 2 sin θ. Do not evaluate your integral.

Answer 1

Explanation:

Given are two polar curves as
`r = 6 sin θ and r = 2+ 2 sin θ`

Let us find the point/s of intersection

Eliminate r to get

`6 sin θ = 2+ 2 sin θ\\sin\theta = 0.5\\\theta = 30, 150`

`r=6sin 30 = 3`

Thus we find that area outside II curve and inside first curve would be

`\int {(r^2)/(2) } \, d\theta \\ =\int{(36sin^2 \theta - (2+2sin\theta)^2 )/(2) } \, d\theta` where lower limit is 30 degrees and 150 degrees.