Determine the value of $\theta$. Given two random variables $X_1$ and $X_2$ with pdf
$$
f\left(x_i\right)= \begin{cases}\frac{1}{2 \theta}, & -\theta<x_i<\theta \\ 0, & x_i \text { otherwise }\end{cases}
$$
If it is known that $X_1$ and $X_2$ are independent and $\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$, determine the value of $\theta$.
My working:
If the random variable X comes from continuous data with pdf
$$\operatorname{Var}\left(X_1 X_2\right)=\frac{64}{9}$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\mu)^2f(x) dx$$
for $\mu$ is the expected value. For example
$$\mu=\int xf(x) dx$$
if subtitute $f(x)=\frac{1}{2 \theta}$, thus
$$\mu=\int x\frac{1}{2 \theta} dx$$
$$\mu=\frac{x^2}{4 \theta}+C$$
So
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\mu)^2f(x) dx$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x-\frac{x^2}{4 \theta})^2\frac{1}{2 \theta} dx$$
$$\operatorname{Var}\left(X_1 X_2\right)=\int(x^2-\frac{x^3}{2 \theta}+\frac{x^4}{16 \theta})\frac{1}{2 \theta} dx$$
$$\frac{64}{9}=(\frac{x^3}{3}-\frac{2x^4}{\theta}+\frac{5x^5}{16 \theta})\frac{1}{2 \theta}$$
$$\frac{64}{9}=(\frac{x^3}{6 \theta}-\frac{x^4}{\theta^2}+\frac{5x^5}{32 \theta})$$
How the next step and please correct if i'm wrong for my work. Thank u
 A: Note that $EX_1 = EX_2 = 0$, hence
$$
\begin{align}
Var (X_1 X_2) &= E(X_1X_2)^2 - \left(E\left(X_1X_2\right)\right)^2 \\
&= EX_1^2 EX_2^2 - (EX_1 EX_2)^2 \\
&=(EX_1^2)^2 \\
&= \left(\int_{-\theta}^\theta\frac{x^2}{2\theta}  dx\right)^2 \\
&= \left(\frac{\theta^2}{3}\right)^2 \\
\end{align}
$$
Solve equatioin $\left(\frac{\theta^2}{3}\right)^2 = \frac{64}9$,
derive $\theta = 2\sqrt2$
A: As shown in xzm's answer, the variance can be obtained by using the independence of $X_1$ and $X_2$. We can also compute the distribution of $X_1X_2$. This approach might be closer to that of the OP.

Compute the Distribution of the Product
The distribution of $(X_1,X_2)$ is uniform on $[-\theta,\theta]\times[-\theta,\theta]$, but the distribution of $X_1X_2$ is not uniform on $\left[-\theta^2,\theta^2\right]$.
In the case for $\lambda\ge0$,

The shaded area is
$$
2\theta^2+2\lambda-2\lambda\log\left(\frac{|\lambda|}{\theta^2}\right)\tag1
$$
In the case for $\lambda\lt0$:

The shaded area is also given by $(1)$. Since the area of the whole space is $4\theta^2$, we get
$$
P(X_1X_2\le\lambda)=\frac12+\frac\lambda{2\theta^2}-\frac\lambda{2\theta^2}\log\left(\frac{|\lambda|}{\theta^2}\right)\tag2
$$
$(2)$ gives a density of
$$
-\frac1{2\theta^2}\log\left(\frac{|\lambda|}{\theta^2}\right)\tag3
$$

Compute the Variance
The distribution is an even function, so the mean of $X_1X_2$ is $0$:
$$
\begin{align}
E(X_1X_2)
&=-\int_{-\theta^2}^{\theta^2}\frac\lambda{2\theta^2}\log\left(\frac{|\lambda|}{\theta^2}\right)\,\mathrm{d}\lambda\tag{4a}\\
&=0\tag{4b}
\end{align}
$$
and the mean of $(X_1X_2)^2$ is
$$
\begin{align}
\mathrm{E}\!\left((X_1X_2)^2\right)
&=-\int_{-\theta^2}^{\theta^2}\frac{\lambda^2}{2\theta^2}\log\left(\frac{|\lambda|}{\theta^2}\right)\,\mathrm{d}\lambda\tag{5a}\\
&=-\theta^4\int_0^1t^2\log(t)\,\mathrm{d}t\tag{5b}\\
&=\frac{\theta^4}9\tag{5c}
\end{align}
$$
Thus, the variance is
$$
\begin{align}
\mathrm{Var}(X_1X_2)
&=\mathrm{E}\!\left((X_1X_2)^2\right)-\mathrm{E}(X_1X_2)^2\tag{6a}\\
&=\frac{\theta^4}9\tag{6b}
\end{align}
$$

Answer to the Question
Since the formula for the variance matches xzm's answer, the value of $\theta$ also matches:
$$
\frac{\theta^4}9=\frac{64}9\implies\theta=2\sqrt2\tag7
$$
