Probability - Expected Value and Variance of the Area of a Rectangle Suppose $D$ is $\mathbb{B}_1(0)\subseteq\mathbb{R}^2$. Choose points $P$,$V$ (uniformly and independently) on $\partial D$ and a point $Q$ on $D$ (again, uniformly and independently on the other two). Construct a rectangle $R$ this way: its sides are either parallel or perpendicular to the segment $\bar{0V}$, and the segment $\bar{QP}$ is one of the diagonals.
How can I find $\mathbb{E}(A_R)$ and $\mathbb{V}ar(A_R)$, where $A_R$ denotes the area of $R$?
If it is of any help, you may suppose you already know the value of the probability that $R\subseteq D$.
 A: I'm assuming that by $\mathbb B_1(0)$ you mean the unit disk.
The only function of $V$ is to uniformly randomly choose a direction for the sides of the rectangle; so we're looking for the expectation and variance of the area of a rectangle with opposite corners given by a point $P$ uniformly chosen on the unit circle and a point $Q$ uniformly chosen in the unit disk, with sides parallel or perpendicular to a uniformly randomly chosen direction.
Given $P$ and $Q$ that define the diagonal $PQ$, the rectangle consists of two congruent right triangles on either side of $PQ$, whose other corners lie on the circle with diameter $PQ$. The altitude of these triangles over $PQ$ is $\cos2\phi$, with $0\le\phi\le\frac\pi2$ the angle uniformly randomly determined by $V$. The average of the cosine is $\frac2\pi$, so the mean area of these rectangles is $\frac2\pi\left|PQ\right|^2$.
The mean square distance between $P$ and $Q$ is the sum of their mean square distances from the centre. $P$ lies on the unit circle and thus has constant square distance $1$ from the centre, and $Q$ ranges over the unit disk and thus has mean square distance
$$
\frac{\int_0^1 2\pi rr^2\mathrm dr}{\int_0^1 2\pi r\mathrm dr}=\frac{\frac14}{\frac12}=\frac12
$$ 
from the centre. Thus the average area of the rectangle is
$$\mathsf E(A)=\frac2\pi\cdot\frac32=\frac3\pi\approx0.955\;.$$
To find the variance, we need the mean square area of the rectangle. Given $PQ$, the mean square area of the rectangle is $\frac12\left|PQ\right|^4$, since the average of the squared cosine is $\frac12$. To find the mean of $\left|PQ\right|^4$, we actually need to do an integration, since this isn't just the sum of the means of $|OP|^4$ and $|OQ|^4$, as in the quadratic case. With $P=(\sin\theta,\cos\theta)$ and (without loss of generality) $Q=(0,r)$, we obtain
\begin{eqnarray*}
\mathsf E\left(|PQ|^4\right)
&=&
\frac{\int_0^{2\pi}\mathrm d\theta\int_0^1 2\pi r\mathrm dr \left(\sin^2\theta + (r - \cos\theta)^2\right)^2}{\int_0^{2\pi}\mathrm d\theta\int_0^1 2\pi r\mathrm dr}
\\
&=&
\frac{\int_0^{2\pi}\mathrm d\theta\int_0^1\mathrm 2\pi rdr \left(r^2-2r\cos\theta+1\right)^2}{\int_0^{2\pi}\mathrm d\theta\int_0^1\mathrm 2\pi rdr}
\\
&=&
\frac{\int_0^1r\mathrm dr \left(\left(r^2+1\right)^2+2r^2\right)}{\int_0^1r\mathrm dr}
\\
&=&
\frac{\frac16+\frac24+\frac12+\frac24}{\frac12}
\\
&=&
\frac{10}3\;.
\end{eqnarray*}
Thus the variance of the area of the rectangle is
\begin{eqnarray*}
\mathsf{Var}(A)
&=&
\mathsf E\left(A^2\right)-\mathsf E(A)^2
\\
&=&
\frac12\cdot\frac{10}3-\left(\frac3\pi\right)^2
\\
&=&
\frac53-\frac9{\pi^2}
\\
&\approx&
0.755\;.
\end{eqnarray*}
