Find area outside circle $r=7$ and inside cardioid $r=6+7\sin \theta$ So this is the question I have problem dealing with.
I know that firstly I need to equate $7$ and $6 + 7\sin \theta$ to get the intersection. 
And then I am supposed to apply the formula..
But I am not sure how to find out the bounds and how to subtract the areas to get the answer
 A: I think you need to look at the graphs to get an idea of it. Wolfram can help here. Once there, you notice that there are two places where the circle and the cardioid intersect. That may be calculated by doing:
$$
f(\theta)=7\ ;\ g(\theta)=6+7\sin(\theta) \\
f(\theta) = g(\theta) \\
\theta_0 \approx .14334
$$
Then, by symmetry, the area that is both inside the cardioid and outside the circle is the same from $\theta_0$ to $\frac{\pi}{2}$ as from $\frac{\pi}{2}$ to the corresponding angle $\pi - \theta_0$. Doing the integral of the difference of the functions:
$$
2\frac{1}{2} \int_{\theta_0}^{\frac{\pi}{2}}(g(\theta)^2 - f(\theta)^2)\ \mathrm{d}\theta = \\
\int_{\theta_0}^{\frac{\pi}{2}} (g(\theta)^2 - f(\theta)^2)\ \mathrm{d}\theta = \\
\int_{\theta_0}^{\frac{\pi}{2}} ((6+7\sin(\theta))^2 - 7^2)\ \mathrm{d}\theta = \\
\int_{\theta_0}^{\frac{\pi}{2}} (36 + 84 \sin(\theta) + 49 \sin(\theta)^2 - 49)\ \mathrm{d} \theta \\
\int_{\theta_0}^{\frac{\pi}{2}} (36 + 84 \sin(\theta) - \frac{49}{2} \cos(2\theta) + \frac{49}{2} - 49)\ \mathrm{d} \theta
$$
By evaluating then we get an area of approximately $103.18$ units.
Hope it helps!
A: The area is given by
$$\frac12\int_{\sin^{-1}\frac17}^{\pi-\sin^{-1}\frac17}
[(6+7\sin \theta )^2-7^2]d\theta=50\sqrt3+\frac{23}2\cos^{-1}\frac17
$$
