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

Answer:

Step-by-step 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

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