Distribution density of the sum of two independent, uniform ditributioned random variables Random variable $ξ_1$ and $ξ_2$ are independent and have uniform distribution on interval [-a;a]. What is the disribution density of random variable $\eta = 2ξ_1 + ξ_2^2$? Obviously, we should use
convolution $f_{\eta}(y) = \int\limits_{-\infty}^{\infty} f_{2ξ_1}(x) \cdot f_{ξ_2^2}(y - x)dx$. I found $f_{2ξ_1}(x) = \frac{1}{2c - (-2c)} = \frac{1}{4c}, -2c \leq x \leq 2c$. $f_{ξ_2^2}(x) = \frac{1}{c^2 - 0^2} = \frac{1}{c^2}, 0 \leq x \leq c^2$. And as I know, now we need to set intervals of $y$ and then we will find intervals of $x$ and solve system of equations at different intervals of $y$. But I can't understand how to set these intervals to solve system of equations and how to find disribution density of $\eta = 2ξ_1 + ξ_2^2$ at the finish. Can someone explain me it, please?
 A: I will refer to $~ξ_1$ as $X$, $ξ_2$ as $Y$ and $\eta$ as $Z$.
$X, Y$ are independent and we have $X, Y \sim U(-a, a)$ and we need to find density of $Z = 2X + Y^2$. We can first find $F_z(z)$ and then differentiate to find the density.
We first note that $-2a \leq z \leq 2a + a^2$.
$i$) For $-2a \leq z \lt a^2 - 2a$, $-a \leq x \leq \frac{z}{2}, |y| \leq \sqrt {z - 2x}~$
$ii$) For $~a^2-2a \leq z \lt 2a$, $-a \leq x \leq \frac{z-a^2}{2}, |y| \leq a$ and for $\frac{z-a^2}{2} \leq x \leq \frac{z}{2}, |y| \leq \sqrt{z-2x}$
$iii$) For $~2a \leq z \leq a^2 + 2a$, $-a \leq x \leq \frac{z-a^2}{2}, |y| \leq a$ and for $\frac{z-a^2}{2} \leq x \leq a, |y| \leq \sqrt{z-2x}$
The above becomes easier to understand from the  below diagram showing curve $z = 2x + y^2$ for three different values of $z$ -
The blue curve is for a value of $z$ in $(i)$
The red curve is for a value of $z$ in $(ii)$
The black curve for a value of $z$ in $(iii)$

For $(i)$,
$ \displaystyle F_Z(z) = 2 \int_{-a}^{z/2} \int_0^{\sqrt{z-2x}} \frac{1}{4a^2} ~dy ~ dx $
For $(ii)$,
$ \displaystyle F_Z(z) = \frac{1}{4a^2} \cdot \left(\frac{z-a^2}{2} + a\right) \cdot 2a + $
$$ \displaystyle 2 \int_{(z-a^2)/2}^{z/2} \int_0^{\sqrt{z-2x}} \frac{1}{4a^2} ~dy ~ dx $$
For $(iii)$,
$ \displaystyle F_Z(z) = \frac{1}{4a^2} \cdot \left(\frac{z-a^2}{2} + a\right) \cdot 2a + $
$$ \displaystyle 2 \int_{(z-a^2)/2}^{a} \int_0^{\sqrt{z-2x}} \frac{1}{4a^2} ~dy ~ dx $$
You could also write the integral in the order $dx ~ dy$, which may be easier to evaluate.
$(i) ~ F_Z(z) = \displaystyle 2 \int_0^{\sqrt{z+2a}} \int_{-a}^{(z - y^2)/2} \frac{1}{4a^2} ~dx~dy$
$(ii)~F_Z(z) =  \displaystyle 2 \int_0^a \int_{-a}^{(z - y^2)/2} \frac{1}{4a^2} ~dx~dy$
$(iii) ~ F_Z(z) = \displaystyle 1 - 2 \int_{\sqrt{z-2a}}^a \int_{(z - y^2)/2}^a \frac{1}{4a^2} ~dx~dy$
Can you take it from here?
