Answer
$C=15$
$P\left( {Y \ge {X^{3/2}}} \right) = 0.625$
Work Step by Step
We have
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{Cy}&{{\rm{if}}{\ }0 \le x \le 1{\ }{\rm{and}}{\ }{x^2} \le y \le x}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
Recall the conditions that a joint probability density function must satisfy:
1. First condition: $p\left( {x,y} \right) \ge 0$ for all $x$ and $y$, since probabilities cannot be negative.
Since $y \ge 0$, we require that $C \ge 0$.
2. Second condition: the normalization condition must hold, namely Eq. (5) must be satisfied:
(5) ${\ \ \ \ \ }$ $\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = 1$
From the definition of $p\left( {x,y} \right)$ we obtain the domain ${\cal D}$, where $p$ is nonzero. So, the description of ${\cal D}$ is
${\cal D} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,{x^2} \le y \le x} \right\}$
Then, using Eq. (5) we evaluate:
$\mathop \smallint \limits_{ - \infty }^\infty \mathop \smallint \limits_{ - \infty }^\infty p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = \mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = {x^2}}^x Cy{\rm{d}}y{\rm{d}}x = 1$
$\frac{C}{2}\mathop \smallint \limits_{x = 0}^1 \left( {{y^2}|_{{x^2}}^x} \right){\rm{d}}x = 1$
$\frac{C}{2}\mathop \smallint \limits_{x = 0}^1 \left( {{x^2} - {x^4}} \right){\rm{d}}x = 1$
$\frac{C}{2}\left( {\left( {\frac{1}{3}{x^3} - \frac{1}{5}{x^5}} \right)|_0^1} \right) = 1$
$\frac{C}{2}\left( {\frac{1}{3} - \frac{1}{5}} \right) = 1$
$\frac{C}{{15}} = 1$
So, $C=15$.
Thus, the joint probability density function:
$p\left( {x,y} \right) = \left\{ {\begin{array}{*{20}{c}}
{15y}&{{\rm{if}}{\ }0 \le x \le 1{\ }{\rm{and}}{\ }{x^2} \le y \le x}\\
0&{{\rm{otherwise}}}
\end{array}} \right.$
To calculate the probability that $Y \ge {X^{3/2}}$, we must define the domain ${{\cal D}_1}$ such that $y \ge {x^{3/2}}$ and satisfies the domain of $p\left( {x,y} \right)$ where $p$ is nonzero, that is, $0 \le x \le 1$, ${x^2} \le y \le x$.
We sketch the domain and from the figure attached we obtain the description of ${{\cal D}_1}$:
${{\cal D}_1} = \left\{ {\left( {x,y} \right)|0 \le x \le 1,{x^{3/2}} \le y \le x} \right\}$
Evaluate:
$P\left( {Y \ge {X^{3/2}}} \right) = \mathop \smallint \limits_{} \mathop \smallint \limits_{{{\cal D}_1}}^{} p\left( {x,y} \right){\rm{d}}y{\rm{d}}x = 15\mathop \smallint \limits_{x = 0}^1 \mathop \smallint \limits_{y = {x^{3/2}}}^x y{\rm{d}}y{\rm{d}}x$
$ = \frac{{15}}{2}\mathop \smallint \limits_{x = 0}^1 \left( {{y^2}|_{{x^{3/2}}}^x} \right){\rm{d}}x$
$ = \frac{{15}}{2}\mathop \smallint \limits_{x = 0}^1 \left( {{x^2} - {x^3}} \right){\rm{d}}x$
$ = \frac{{15}}{2}\left( {\left( {\frac{1}{3}{x^3} - \frac{1}{4}{x^4}} \right)|_0^1} \right)$
$ = \frac{{15}}{2}\left( {\frac{1}{3} - \frac{1}{4}} \right) = \frac{5}{8} = 0.625$
So, the probability that $Y \ge {X^{3/2}}$:
$P\left( {Y \ge {X^{3/2}}} \right) = 0.625$
