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If $Y_1$ is a standard normal random variable, then $Y_1^2$ will follow a Chi-square distribution with 1 degree of freedom

It can be observed that $Y_1$ has a range of $-1 \leq Y_1 \leq 1$ . Similarly, $Y_2$ is $0 \leq Y_2 \leq 1$ . To form the triangle shape, an additional constraint is needed: $|Y_1| + Y_2 \leq 1$ .

\begin{equation*} f(y_1,y_2) = \begin{cases} 1 & -1 \leq Y_1 \leq 1, 0 \leq Y_2 \leq 1, |Y_1| + Y_2 \leq 1 \\ 0 & \textnormal{otherwise} \end{cases} \end{equation*}

A split integral will be needed to find $E(Y_1Y_2)$ due to the special constraint.

\begin{align*} E(Y_1Y_2) &= \int_{-1}^{0}\int_{0}^{1+y_1}y_1y_2dy_2dy_1 + \int_{0}^{1}\int_{0}^{1-y_1}y_1y_2dy_2dy_1 \\ &= \int_{-1}^{0}\dfrac{y_1(1+y_1)^2}{2}dy_1 + \int_{0}^{1}\dfrac{y_1(1-y_1)^2}{2}dy_1 \\ &= \dfrac{1}{2}\int_{-1}^{0}y_1(1+y_1)^2dy_1 + \dfrac{1}{2} \int_{0}^{1}y_1(1-y_1)^2dy_1 \\ &= \dfrac{1}{2}\int_{-1}^{0}\left(y_1+y_1^3+2y_1^2\right)dy_1 + \dfrac{1}{2}\int_{0}^{1}\left(y_1+y_1^3-2y_1^2\right)dy_1 \textnormal{ (expanded)} \\ &= \dfrac{1}{2}\left[ \dfrac{y_1^2}{2} + \dfrac{y_1^4}{4} + \dfrac{2y_1^3}{3} \right]_{-1}^0 + \dfrac{1}{2}\left[ \dfrac{y_1^2}{2} + \dfrac{y_1^4}{4} - \dfrac{2y_1^3}{3} \right]_0^1 \\ &= -\cancel{\dfrac{1}{2}\left[ \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{2}{3} \right]} + \cancel{\dfrac{1}{2}\left[ \dfrac{1}{2} + \dfrac{1}{4} - \dfrac{2}{3} \right]} \\ &= 0 \end{align*}

The uniqueness property of mgfs allows us to show that $U$ follows a Chi-square distribution with 2 degrees of freedom.