Find the area inside the circle $r = 10 \sin \theta$ and above the line $r = 2 \csc \theta$. Function Plotter graph:



I think the formula is
$$A = \frac 1 2 \int_{\alpha}^{\beta} (\text{outer})^2 - (\text{inner})^2 d\theta$$
where $\alpha, \beta$ are where they intersect in $[0, 2\pi]$.
This is what I got based on that
$$A = \frac 1 2 \int_{x}^{\pi-x} (10 \sin \theta)^2 - (2 \csc \theta)^2 d\theta$$
where $x= \sin^{-1}(\frac {1}{\sqrt{5}} )$
Is that right?

We can do this all with classical geometry.
The region in question is a circle or radius 5 less a section of a circle with angle $2\cos^{-1} \frac 35$ plus 2 $3-4-5$ right triangles.
$25\pi - 25\cos^{-1} \frac 35 + 12$
Lets see if calculus gives us the same answer.
limits of integration.
$2\csc \theta = 10\sin \theta\\
\frac 15 = \sin^2 \theta\\
\sin\theta = \pm \frac 1{\sqrt 5}$
$\frac 12\int_{\sin^{-1} \frac 1{\sqrt 5}}^{\pi - \sin^{-1} \frac 1{\sqrt 5}} (10\sin\theta)^2 - (2\csc\theta )^2\ d\theta\\
\frac 12\int_{\sin^{-1} \frac 1{\sqrt 5}}^{\pi - \sin^{-1} \frac 1{\sqrt 5}}  50(1-\cos2\theta) - 4\csc^2\theta\ d\theta\\
25 (\theta -\sin\theta\cos\theta) + 2\cot\theta |_{\sin^{-1} \frac 1{\sqrt 5}}^{\pi - \sin^{-1} \frac 1{\sqrt 5}}\\
25 \pi -50\sin^{-1}\frac {1}{\sqrt 5} +20 - 8 
$
Does $25 \cos^{-1} \frac 35 = 50 \sin^{-1} \frac 1{\sqrt5}?$
Indeed it does!
Set up in Cartesian
inside:
$x^2 + y^2 = 10 y$
above $y = 2$
$\int_2^{10} \sqrt {10y - y^2} dy$ 
