Here is a proof that the uniform distribution over $[0, 2\pi]$ (i.e. the normalized Lebesgue measure) maximizes the integral, based on variational argument. The idea in this proof is based on my other answer for the similar question.
On the space $\mathcal{M}$ of signed Borel measures on $[0, 2\pi]$, define $\langle \cdot, \cdot \rangle$ by
$$ \forall \mu, \nu \in \mathcal{M}, \quad \langle \mu, \nu \rangle = \int_{[0,2\pi]^2} \lvert \sin(x-y)\rvert \, \mu(\mathrm{d}x)\nu(\mathrm{d}y). $$
Obviously $\langle \cdot, \cdot \rangle$ is a symmetric bilinear form on $\mathcal{M}$. Using this, OP's question is rephrased as the problem of finding maximizers of the functional $I(\mu) = \langle \mu, \mu \rangle$ over the set of all probability measures on $[0, 2\pi]$. In this regard, we have the following characterization:
Proposition. Write $\mathcal{M}_1$ for the set of all $\mu \in \mathcal{M}$ satisfying $\mu([0,2\pi]) = 1$. Then, for $\mu \in \mathcal{M}_1$, the followings are equivalent:
- $\mu$ maximizes the functional $I$.
- $s \mapsto \langle \mu, \delta_s \rangle$ is constant for $s \in [0, 2\pi]$.
Assuming this proposition, we see that the normalized Lebesgue measure $\mu = \frac{1}{2\pi} \operatorname{Leb}|_{[0,2\pi]}$ satisfies the second condition, and so, it maximizes $I$ not only among probability measures but on all of $\mathcal{M}_1$. (On the other hand, I have not proved the uniqueness of the maximizer. I guess that the maximizer is unique, based on the idea that the map $s \mapsto \langle \mu, \delta_s \rangle$ behaves like Fourier transform, and so, it should have inverse transform.)
As for the proof, we resolve an easier part first. This is a typical application of the variational argument.
Proof of $(1)\Rightarrow(2)$. Assume that $\mu \in \mathcal{M}_1$ maximizes the functional $I$. Then for any $s, t \in [0, 2\pi]$, the map $ \epsilon \mapsto I(\mu + \epsilon \delta_s - \epsilon \delta_t) $ attains the maximum at $\epsilon = 0$. Differentiating with respect to $\epsilon$ and plugging $\epsilon = 0$ proves the desired implication. $\square$
So we turn to proving the opposite direction.
Proof of $(2)\Rightarrow(1)$. We begin by citing the following Fourier series.
$$ \lvert \sin x \rvert = \frac{2}{\pi} - \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{\cos (2nx)}{4n^2 - 1}. $$
For instance, see this answer, for the proof. To make use of this, write $\hat{\mu}(\xi) = \int_{[0,2\pi]} e^{i \xi x} \, \mu(\mathrm{d}x)$ for the Fourier transform of $\mu$. Now, since this series converges uniformly, we can plug this formula to $\langle \mu, \nu \rangle$ and interchange the order of the integration and summation. This yields
$$ \langle \mu, \nu \rangle
= \frac{2}{\pi} \hat{\mu}(0)\overline{\hat{\nu}(0)} - \frac{4}{\pi} \sum_{n=1}^{\infty} \frac{1}{4n^2 - 1}\operatorname{Re}\left( \hat{\mu}(2n)\overline{\hat{\nu}(2n)} \right). $$
From this, we immediately observe that
$$ \text{if} \quad \mu([0,2\pi]) = 0, \qquad \text{then} \quad \langle \mu, \mu \rangle
= -\frac{4}{\pi} \sum_{n=1}^{\infty} \frac{1}{4n^2 - 1}\left|\hat{\mu}(2n)\right|^2 \leq 0. $$
Now we are ready for the final blow. Let $\mu \in \mathcal{M}_1$ be such that $\langle \mu, \delta_s \rangle = \int_{[0,2\pi]} \lvert \sin(x-s)\rvert \, \mu(\mathrm{d}x)$ does not depend on $s \in [0,2\pi]$. If $c$ denotes this constant value, then $ \langle \mu, \nu \rangle = c $ holds for any $\nu \in \mathcal{M}_1$. So,
$$ \forall \nu \in \mathcal{M}_1, \qquad
I(\nu)
= I(\mu) + 2\underbrace{\langle \mu, \nu-\mu \rangle}_{c-c = 0} + \underbrace{I(\nu - \mu)}_{\leq 0} \leq I(\mu). $$
Therefore $\mu$ maximizes $I$ over $\mathcal{M}_1$ as desired. $\square$
Best Answer
Given the area and perimeter given by @LeeDavidChung, the expected area is
$$\int _0^1\int _0^1x y (1-\alpha (1-2 x) (1-2 y))dydx=\frac{9-\alpha }{36}$$
The maximum ($1/4$) occurs when $\alpha=0$.
The expected value of the perimeter is
$$\int _0^1\int _0^12 \left(\sqrt{x^2+y^2}+x\right) (1-\alpha (1-2 x) (1-2 y))dydx=\frac{1}{15} \left(-7 \sqrt{2} \alpha +6 \alpha +5 (\alpha +2) \sinh ^{-1}(1)+10 \sqrt{2}+15\right)$$
When $\alpha=0$ the expected value of the perimeter is $\frac{1}{3} \left(2 \sqrt{2}+3+2 \sinh ^{-1}(1)\right)$.