# Find the area of the region outside r=10+10sinθ , but inside r=30sinθ.

Question: Find the area of the region outside r=10+10sinθ , but inside r=30sinθ. Please show work.

Solution:

Given that to find the area of the region outside
r = 10 + 10 sinθ and the inside r = 30 sinθ
Two curves intersected of
10 + 10 sinθ = 30 sinθ
20 sinθ = 10
sinθ = 1/2 => θ = π/6, 5π/6

$A=\frac{1}{2}\int_{\pi /6}^{5\pi/6}(r_1^2 - r_2^2)d\theta$
$=\frac{1}{2}\int_{\pi /6}^{5\pi/6}(30 \sin \theta)^2 -(10+10 \sin \theta)^2 d \theta$
$=\frac{1}{2}\int_{\pi /6}^{5\pi/6}[800 \sin^2\theta - 200 \sin \theta -100] d \theta$
$=756\pi /6 - 150\pi/6 = 100\pi = 314 \text{ units}$