Question
Suppose Suppose that $Y$ and $Z$ are continuous random variables with joint probability density function $f_{y,z}(y,z) = sy^{2}e^{-(\alpha + \beta z)y}$ where $y >0 \;,\; z > 0$ and $\alpha >0 \;,\beta > 0$. Find the value of $s$ that would allow for $f_{y,z}(y,z)$ to be a valid pdf.
My approach
For $f_{y,z}(y,z)$ to be a valid distribution, the following condition must be met.
$$ \iint_{A}f_{y,z}(y,z) \; \mathrm{d}z\mathrm{dy}=1 $$
In context, for the given bounds, I have come up with this result with the given bounds.
\begin{align} \int_{0}^{\infty}\int_{0}^{\infty} sy^{2}e^{-(\alpha + \beta z)y} \; \mathrm{d}z \; \mathrm{d}y \end{align}
Integrating for $\mathrm{d}z$, I have obtained the following:
\begin{align} \int_{0}^{\infty} sy^{2}e^{-(\alpha + \beta z)y} \; \mathrm{d}z &= sy^{2}\int_{0}^{\infty} e^{-(\alpha + \beta z)y} \; \mathrm{d}z \\ &= sy^{2}\int_{0}^{\infty} e^{-y\alpha + \beta zy} \mathrm{d}z \\ &= sy^{2} \cdot -\frac{1}{\beta y} \cdot e^{-y\alpha + \beta zy} \\ \end{align}
Combining and evaluating for the bounds
\begin{align} -\frac{sye^{-y(\alpha + \beta z)}}{\beta} \Biggr|_{0}^{\infty} &= -\frac{sye^{-y(\alpha + \beta (\infty))}}{\beta} - \left(-\frac{{sye^{-y(\alpha + \beta (0))}}}{\beta}\right) \\ &= \frac{sy}{e^{y\alpha + \beta (\infty)}\beta} + \frac{sy}{e^{y(\alpha)}\beta} \end{align}
As, $\lim_{z\rightarrow\infty}$, the first fraction becomes $0$. Thus, leaving only the second result $\frac{sy}{e^{y(\alpha)}\beta}$.
Integrating for $\mathrm{d}y$ over the bounds using integration by parts. \begin{align} \int_{0}^{\infty} \frac{sye^{-y\alpha}}{\beta} \; \mathrm{d}y &= \frac{s}{\beta} \int_{0}^{\infty} ye^{-y\alpha} \; \mathrm{d}y \\ &= \frac{s}{\beta}\cdot\left(\frac{e^{-y(\alpha)}(\alpha y + 1)}{a^2} \Biggr|_{0}^{\infty}\right) \end{align}
Whereby, this yields (I am very much unsure about this component)
\begin{align} \frac{s}{\beta}\cdot\frac{e^{-y(\alpha)}(\alpha y + 1)}{a^2} \Biggr|_{0}^{\infty} &= \frac{s}{\beta}\cdot\left(\frac{e^{-\infty\alpha}(\alpha \infty)}{a^2} - \frac{e^{-\alpha(0)}(\alpha(0) + 1)}{a^2}\right) \end{align}
To satisfy the condition for a valid pdf, this must be equal to 1. \begin{align} \frac{s}{\beta}\cdot - \frac{1}{a^2} &= 1 \\ -\frac{s}{\alpha^2\beta} &= 1 \\ \alpha^2\beta &= -s \\ \therefore s &= -\alpha^2\beta \end{align}
I am very unsure if my approach is correct and my integration is not the greatest aspect. Any input would mean the world to me.
Thanks in advance!
Everything looks correct except the sign is wrong. If $\alpha > 0$ and $\beta > 0$, then your choice $s = -\alpha^2 \beta$ would be negative, which in turn would make the joint density negative.
The marginal density of $Y$ that you obtained is correct: $$f_Y(y) = \frac{s}{\beta} y e^{-\alpha y} \, y > 0.$$ Its integral can be computed via integration by parts, or by recognizing it as a gamma function, or by recognizing it as a gamma density:
$$\begin{align} 1 &= \frac{s}{\beta} \int_{y=0}^\infty ye^{-\alpha y} \, dy \\ &= \frac{s}{\beta} \left( \left[ -\frac{1}{\alpha} y e^{-\alpha y} \right]_{y=0}^\infty + \int_{y=0}^\infty \frac{1}{\alpha} e^{-\alpha y} \, dy \right) \\ &= \frac{s}{\beta} \left( -0 + 0 + \left[-\frac{1}{\alpha^2} e^{-\alpha y} \right]_{y=0}^\infty \right) \\ &= \frac{s}{\beta} \left( -0 + \frac{1}{\alpha^2} \right) \\ &= \frac{s}{\alpha^2 \beta}. \end{align}$$
Equivalently, the substitution $$u = \alpha y, \quad du = \alpha \, dy$$ gives $$\int_{y=0}^\infty ye^{-\alpha y} \, dy = \int_{u=0}^\infty \frac{u}{\alpha} e^{-u} \cdot \frac{1}{\alpha} \, du = \frac{1}{\alpha^2} \int_{u=0}^\infty ue^{-u} \, du = \frac{1}{\alpha^2} \Gamma(2) = \frac{1}{\alpha^2}$$ where we have used the gamma function $$\Gamma(z) = \int_{u=0}^\infty u^{z-1} e^{-u} \, du.$$ Finally, we can also note that $f_Y(y)$ is proportional to a gamma density with shape parameter $a = 2$ and rate parameter $b = \alpha$; i.e., $$f_Y(y) \propto \frac{b^a y^{a-1} e^{-by}}{\Gamma(a)} = \alpha^2 y e^{-\alpha y}.$$ Therefore we require $s/\beta = \alpha^2$, or $s = \alpha^2 \beta$.