Find the area of the region that lies inside the first curve and outside the second curve. r = 19 sin(θ), r = 10 − sin(θ)

A graph of the functions is attached. The area you need to calculate is colored with red.

The area can be calculated by the formula:
[tex] \int\limits^\alpha_\beta {(1/2) (r_{1}^{2} - r_{2}^{2}) } \, d\theta [/tex]

In order to find the limits of integration you need to calculate the two points at which the circumferences intersect:

[tex] \left \{ {{r=19sin\theta} \atop {r=10-sen\theta}} \right. [/tex]

19sinθ = 10 - senθ
20senθ = 10
senθ = 1/2
θ₁ = π/6 
θ₂ = 5π/6

Therefore:
[tex]\int\limits^ {5\pi/6}_{\pi/6} {(1/2) [(19sin\theta)^{2} - (10 - sin\theta)^{2}] } \, d\theta[/tex] =

[tex]\int\limits^ {5\pi/6}_{\pi/6} {(1/2) [361sin^{2}\theta - (100 - 20sin\theta + sin^{2}\theta)] } \, d\theta[/tex] =

[tex]\int\limits^ {5\pi/6}_{\pi/6} {(1/2) [360sin^{2}\theta + 20sin\theta - 100] } \, d\theta[/tex] =

[tex]10\int\limits^ {5\pi/6}_{\pi/6} {18sin^{2}\theta + sin\theta - 5 } \, d\theta[/tex] =

[tex]10[9\theta - (9/2)sin(2\theta) - cos\theta -5\theta]_{\pi/6}^{5\pi/6}[/tex] =

[tex][-45sin(2\theta) - 10cos\theta +40\theta]_{\pi/6}^{5\pi/6}[/tex] =

[45√3/2 + 5√3 + 100π/3] - [-45√3/2 - 5√3 + 20π/3] =

55√3 + 80π/3 =

≈179.04

Hence, the area of the region that lies inside the first curve and outside the second curve is around 179 square units.