What is the area bound by two shifted sine waves in polar coordinates? I need to calculate the area bound by the curve $(\sin(\theta), \sin(\theta -\phi)), \theta \in [0, 2\pi]$ and have the area in terms of $\phi$.
So far I have been saying $x=\sin(\theta)$ and $y = \sin(\theta -\phi)$ then $r^2 = \sin^2(\theta)+\sin^2(\theta - \phi)$.
I then do a polar double integral of them form
$$A=\int_0 ^{2\pi}\int_0^{\sqrt{\sin^2(\theta)+\sin^2(\theta - \phi)}} r dr d\theta$$
$$A=\frac12 \int^{2\pi} _0\left[r^2\right]^{\sqrt{\sin^2(\theta)+\sin^2(\theta - \phi)}}_0 d\theta = \frac12 \int^{2\pi}_0\sin^2(\theta)+\sin^2(\theta - \phi) d\theta$$
$$A = \frac12 \int _0 ^{2\pi} \left(\frac{1 - \cos(2\theta)}2 +\frac{1 - \cos(2(\theta-\phi))}2\right) d\theta$$
$$A=\frac12\left[\frac{2\theta-\sin(2\theta)}4 + \frac{2\theta-\sin(2(\theta-\phi))}4\right]_0 ^{2\pi}$$
$$A=\pi$$
After a few attempts I keep getting the answer to be $\pi$, when obviously the area should be a function of $\phi$. Am I formulating the problem incorrectly or do I keep making a mistake during the integration?
For reference, this is the area I am trying to determine, as a function of $\phi$ https://www.desmos.com/calculator/qeq6toiqki.
Thanks in advance!
 A: The error is essentially the fact that $t$ is not your polar angle as you are treating it in your integration. The process will go something like this.
We want to convert $(\sin t,\ \sin(t-\phi))$ to polar coordinates. So let $(r,\theta)$ be the required variables. Then $$\sin t=r\cos\theta$$ $$\sin(t-\phi)=r\sin\theta$$
From this, we calculate $$r^2=\sin^2t+\sin^2(t-\phi)$$ $$\theta=\tan^{-1}\left(\frac{\sin(t-\phi)}{\sin t}\right)$$ and hence $$\mathrm d\theta=\frac{\mathrm d\left(\frac{\sin(t-\phi)}{\sin t}\right)}{1+\left(\frac{\sin(t-\phi)}{\sin t}\right)^2}=\frac{\sin t\cos(t-\phi)-\sin(t-\phi)\cos(t)}{\sin^2t+\sin^2(t-\phi)}\mathrm dt=\frac1{r^2}\sin(\phi)\mathrm dt$$
Now, the area of a curve in polar coordinates (where the limits in our case come from $t\in[0,2\pi]$) is $$A=\frac12\int_{\theta_1}^{\theta_2} r^2\ \mathrm d\theta=\frac12\int_0^{2\pi}r^2\frac1{r^2}\sin\phi\ \mathrm dt=\pi\sin\phi$$
where $\theta_1$ and $\theta_2$ are the bounds for $\theta$ that I was too lazy to calculate (one can just put $\theta\in [0,2\pi]$ by say, rotating the diagram, but that's some more work which is unnecessary).

One can completely bypass this conversion by using Green's Theorem. By Green's Theorem in 2 dimensions, letting $\mathrm d$ be the exterior derivative, we have
$$\int_{R}\mathrm d(P,Q)=\int_{\partial R}(P,Q)$$ or more concretely
$$\int_{R} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\ \mathrm dA=\int_{\partial R}P\ \mathrm dx+Q\ \mathrm dy$$
Let $(P,Q)=(0,x)$. Then
$$\int_R1\ \mathrm dA=\int_{\partial R}x\ \mathrm dy$$
But the LHS is just the area of the region $R$, and the RHS integration is over the boundary. Hence, plugging in our equations, we have:
\begin{align*}\text{Area}&=\int_{0}^{2\pi}\sin t\ \mathrm d(\sin(t-\phi))\\&=\int_{0}^{2\pi}\sin t\cos(t-\phi)\ \mathrm dt\\&=\frac12\cos\phi\int_0^{2\pi}\sin(2t)\ \mathrm dt+\frac12\sin\phi\int_{0}^{2\pi}[1-\cos(2t)\ \mathrm dt]\\&=\frac12 2\pi\sin\phi&\\&=\pi\sin\phi
\end{align*}
A: If you don't like integration, here's some trigonometry for you. We first rotate the whole figure by $\pi/4$
\begin{align*}\frac1{\sqrt2}\begin{bmatrix}1&& 1\\-1&&1\end{bmatrix}\begin{bmatrix}\sin t\\ \sin(t-\phi)\end{bmatrix}&=\frac1{\sqrt2}\begin{bmatrix}\sin t+\sin(t-\phi)\\ \sin(t-\phi)-\sin t\end{bmatrix}\\&=\frac{1}{\sqrt 2}\begin{bmatrix}2\sin\left(t-\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)\\ -2\cos\left(t-\frac{\phi}{2}\right)\sin\left(\frac{\phi}{2}\right)\end{bmatrix}\\&=\sqrt2\begin{bmatrix}\sin\left(t-\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)\\ -\cos\left(t-\frac{\phi}{2}\right)\sin\left(\frac{\phi}{2}\right)\end{bmatrix}\end{align*}
Now note that this is a standard ellipse
$$\frac{\left(\sqrt2\sin\left(t-\frac{\phi}{2}\right)\cos\left(\frac{\phi}{2}\right)\right)^2}{\left(\sqrt2\cos\left(\frac{\phi}{2}\right)\right)^2}+\frac{\left(-\sqrt2\cos\left(t-\frac{\phi}{2}\right)\sin\left(\frac{\phi}{2}\right)\right)^2}{\left(\sqrt2\sin\left(\frac{\phi}{2}\right)^2\right)}$$ $$=\sin^2\left(t-\frac{\phi}{2}\right)+\cos^2\left(t-\frac{\phi}{2}\right)=1$$
with the identification that $a=\sqrt2\cos\left(\frac{\phi}{2}\right)$ and $b=\sqrt2\sin\left(\frac{\phi}{2}\right)$. Well, since we just rotated the figure, the area is still the same and hence $$\text{Area}=\pi ab=2\pi\cos\left(\frac{\phi}{2}\right)\sin\left(\frac{\phi}{2}\right)=\pi\sin\phi$$
