Area inside polar curve Find the area of the region inside 
$r=7\sin\theta$ but outside $r=1$.
I have tried finding the area of both using $A=\frac{1}{2}\int_\alpha^\beta r^2 d\theta$, tried arc length...but I can't find the answer.
 A: Draw a picture, when possible! Don't try arc length, you're finding polar area. Your first integral is the one to use!
Our functions are $r_1=7\sin(\theta)$ and $r_2=1$ (why I number them will be clear I hope later)
So we see we should be integrating from where $7\sin(\theta)=1$ to $\pi/2$ ($\alpha$ and $\beta$, respectively) and then by symmetry we just multiply our integral's result by 2. State this last part and remember to do it!
So we should have an integral like $\int_{\alpha}^{\beta} (outside - inside) d\theta$, better said $\int_{\alpha}^{\beta} r_1^2 - r_2^2 d\theta$
I've left actually writing out the integral (do that!), finding the limits, and doing the calculation. Hopefully there is enough help here, though.

A: \begin{align*}
  A &= \frac{1}{2}
       \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
       (r^{2}-1) \, d\theta \\
    &= \frac{1}{2} \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
       (49\sin^{2}-1) \, d\theta \\
    &= \frac{1}{2} \int_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}}
       \left[ 49\left( \frac{1-\cos 2\theta}{2} \right)-1 \right] \, d\theta \\
    &= \left[
         \frac{47}{4}\theta-\frac{49}{8} \sin 2\theta
       \right]_{\sin^{-1} \frac{1}{7}}^{\pi-\sin^{-1} \frac{1}{7}} \\
    &= \frac{47\pi}{4}-\frac{47\pi}{2}\sin^{-1} \frac{1}{7}+
       \frac{49}{4} \sin \left( 2\sin^{-1} \frac{1}{7} \right) \\
    &= \frac{47\pi}{4}-\frac{47\pi}{2}\sin^{-1} \frac{1}{7}+
       \frac{49}{2} \times \frac{1}{7} \times \sqrt{\frac{48}{49}} \\
    &= 2\sqrt{3}+\frac{47\pi}{2}-\frac{47}{2}\sin^{-1} \frac{1}{7} \\
    &= 2\sqrt{3}+\frac{47}{2}\cos^{-1} \frac{1}{7}
\end{align*}
