Question
Let $Y = \sqrt X$, where $X$ has cumulative distribution function
$$F_X(x) = \begin{cases} 0\qquad \mathrm{if}\ x < 0,\\ \frac 1 8 x^3\quad \mathrm{if}\ 0 \leq x \leq 2,\\ 1\qquad \mathrm{if}\ x > 2.\\ \end{cases}$$
Find $\mathbb{P}[Y \leq X]$.
My working
$$\begin{aligned} f_Y(y) & = f_X(x) \frac {\mathrm{d}x} {\mathrm{d}y}\\ & = \frac 3 8 x^2 (2y)\\ & = \frac 3 4 y^5\ \forall\ y \in [0, \sqrt 2] \end{aligned}$$
$$\begin{aligned} \implies \mathbb{P}[Y \leq X] & = \int^{\infty}_{-\infty} \int^x_{-\infty} f_X(x)f_Y(y)\ \mathrm{d}y\mathrm{d}x\\ & = \int^2_0 \int^x_0 \left(\frac 3 8 x^2\right)\left(\frac 3 4 y^5\right)\ \mathrm{d}y\mathrm{d}x\\ & = \int^2_0 \frac 3 {64} x^8\ \mathrm{d}y\mathrm{d}x\\ & = \frac 8 3 \end{aligned}$$
However, this is clearly incorrect as my probability is greater than $1$, but I am not sure where I have gone wrong.
Any intuitive explanations will be greatly appreciated :)
I think the most clear mistake is that $X$ and $Y$ are not indepenedent! So you can't compute the density $f_{X,Y}$ as $f_Xf_Y$.
One way to get a correct answer is note that $Y \leq X$ if and only if $\sqrt{X} \leq X$, i.e. if and only if $\sqrt{X}(\sqrt{X}-1) \geq 0$. (This uses that $X$ and $Y$ are positive.) This inequality holds if $\sqrt{X}=0$ or if $\sqrt{X}\geq 1$, so you just have to compute these two probabilites. The first one gives us $\mathbb{P}[X=0]=0$ as $X$ is continuous, and the second one yields $$ \mathbb{P}[X\geq 1] = 1-\mathbb{P}[X \leq 1] = 1-F_X(1) = 1 -\frac{1}{8}=\frac{7}{8}. $$ (This uses again that $X$ is continuous, so that $\mathbb{P}[X\leq 1]=\mathbb{P}[X=1]$.)