Variance of an unbiased estimator $L = \frac\pi4\sqrt{X_1X_2}$ Find variance of unbiased estimator L, where $L = \dfrac\pi4\sqrt{X_1X_2}$.
$f(x) = \dfrac1\theta e^{\frac{-x}{\theta}}$, $x>0$,
$X_1$ and $X_2$ are independent, and exponentially distributed. 
Since $L$ is unbiased so I know $E[L] = \theta$, right?
Also, $\operatorname{Var}[x] = E[x^2] - E[x]^2$. But I'm struggling with finding $E[x^2]$. Any help will be appreciated. Thank you.
 A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\begin{align}
\mathbb{E}\bracks{L} & \equiv
\int_{0}^{\infty}{\expo{-x_{1}/\theta} \over \theta}
\int_{0}^{\infty}{\expo{-x_{2}/\theta} \over \theta}
\pars{{\pi \over 4}\root{x_{1}x_{2}}}\dd x_{1}\,\dd x_{2}
\\[5mm] & =
{\pi \over 4}\pars{\root{\theta}
\int_{0}^{\infty}\expo{-x_{1}/\theta}
\,\root{x_{1} \over \theta}\,{\dd x_{1} \over \theta}}
\\[2mm] & \phantom{===\,}
\pars{\root{\theta}
\int_{0}^{\infty}\expo{-x_{2}/\theta}\,\root{x_{2} \over \theta}
\,{\dd x_{2} \over \theta}}
\\[5mm] & =
{\pi \over 4}\
\underbrace{\pars{\int_{0}^{\infty}x^{1/2}\expo{-x}\dd x}^{2}}
_{\ds{=\ \Gamma^{2}\pars{3/2}\ =\ {\pi/4}}}\ \theta\ =\
\bbx{{\pi^{2} \over 16}\,\theta}
\\[1cm]
\mathbb{E}\bracks{L^{2}} & \equiv
\int_{0}^{\infty}{\expo{-x_{1}/\theta} \over \theta}
\int_{0}^{\infty}{\expo{-x_{2}/\theta} \over \theta}
\pars{{\pi \over 4}\root{x_{1}x_{2}}}^{2}\dd x_{1}\,\dd x_{2}
\\[5mm] & =
{\pi^{2} \over 16}\pars{\theta\int_{0}^{\infty}
\expo{-x_{1}/\theta}\,{x_{1} \over \theta}
\,{\dd x_{1} \over \theta}}
\pars{\theta\int_{0}^{\infty}
\expo{-x_{2}/\theta}\,{x_{1} \over \theta}
\,{\dd x_{2} \over \theta}}
\\[5mm] & =
\bbx{{\pi^{2} \over 16}\,\theta^{2}}
\\[1cm]
\mbox{Var}\pars{L} & =
\bbx{{\pi^{2} \over 16}\pars{1 - {\pi^{2} \over 16}}\theta^{2}}
\end{align}
