<s>[INST] Find the area between a cardioid and a straight line with polar equations I have to find the area between a line and a cardioid, given by $\rho_1 = 8 + 8 \sin(\theta)$ and $\rho_2 = 4/\hspace{-0.5mm}\sin(\theta)$.
First I found when both are positive, which is always for the first one, and only from $0$ to $2\pi$ for the second one. Then I graph them both and now I have to find the intersections, but I get $8+\sin(\theta) = 4/\hspace{-0.5mm}\sin(θ)$, which is $\sin^2(\theta) + 1 = 1/2$, and I have no idea how to solve that.
Then, for the area, I know it's the top part of the graphic, so let's say they intersect in $\theta=a$ and $\theta=b$, so I have $$\int_{a}^{b} (8+8\sin\theta)^2 - (4/\hspace{-0.5mm}\sin\theta)^2 d\theta$$ for the cardioid is over the line. Is that correct? I'm not supposed to solve it, just to propose it.
And that's all, I hope you can help me. Thank you!
 [/INST] The general formula for the area between two polar curves is $$\frac{1}{2}\int_{\theta_1}^{\theta_2}\left(r_1^2-r_2^2\right)\mathrm{d}\theta,$$ where $r_1$ is the "outer" curve (the cardioid in this case) and $r_2$ is the "inner" curve (the line in this case). You were mostly on the right track, aside from the $1/2$ out front.
Finding the points of intersection of the two polar curves can often be, as you found, the trickiest part of setting up the integral. In this case, since $r_1=8+8\sin\theta$ and $r_2=4/\hspace{-0.5mm}\sin\theta$, we have
\begin{align}
8+8\sin\theta&=4/\hspace{-0.5mm}\sin\theta \\
\hspace{2.5cm}8\sin\theta+8\sin^2\theta&=4 \\
2\sin^2\theta+2\sin\theta-1&=0 \\
2x^2+2x-1&=0 &&\text{Let $x=\sin\theta$} \\
\end{align}
\begin{align}
x&=\frac{-2\pm\sqrt{2^2-4(2)(-1)}}{2(2)} \\
x&=\frac{-2\pm\sqrt{12}}{4} \\
x&=-\frac{1}{2}\pm\frac{\sqrt{3}}{2} \\
\sin\theta&=-\frac{1}{2}\pm\frac{\sqrt{3}}{2}
\end{align}
We can see from this result that $\theta_1=\arcsin\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\right)$. Unfortunately, since the domain of the $\arcsin$ function is only $[-1,1]$, $-\frac{1}{2}-\frac{\sqrt{3}}{2}$ is outside the domain of the $\arcsin$ function. We can correct for this by taking $\theta_2=\pi-\theta_1$, since it's clear from graphing the two functions that that's the angle we want. Therefore, the final form of the integral is
$$\frac{1}{2}\int_{\arcsin\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\right)}^{\pi-\arcsin\left(-\frac{1}{2}+\frac{\sqrt{3}}{2}\right)}\left[\Big(8+8\sin\theta\Big)^2-\left(\frac{4}{\sin\theta}\right)^2\right]\mathrm{d}\theta$$
Numerical evaluation shows that this integral is approximately equal to $204.16$.
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