How to prove $Y=X^{2}$ is Beta$\left(0.5,\ 1\right)$ if $X$ is Uniform$(0,\ 1)$ 
I've been reading about uniform distributions but I can't see how $Y=X^{2}$ is Beta$\left(0.5,\ 1\right)$ if $X$ is Uniform$(0,\ 1)$. Is there a way to prove this using the cumulative distribution function method?
My attempt:

 A: Yes, it is incredibly straightforward.  The CDF of $Y$ is $\mbox{Pr} \left[Y \le y \right]$.  Now substitute $Y$ for $X^2$ and square root both sides to obtain $\mbox{Pr} \left[X \le \sqrt{y}\right]$.  The square root operator is a one-to-one transformation for the domain $(0,1)$.  Now the CDF of $X$ is  $\mbox{Pr} \left[X \le x\right] =x$.  Replace $x$ with $\sqrt{y}$ to obtain that the CDF of $Y$ is $\sqrt{y}$.  This is equivalent to the regularized incomplete beta function $I_y(.5,1)$, the CDF of a $Beta(.5,1)$ random variable.
A: A $\text{Beta}(\alpha,\beta)$ distribution has support on $[0,1]$ with density proportional to $x^{\alpha-1}(1-x)^{\beta-1}$.
You have found that $F_Y(y)=y^{0.5}$ when $0<y<1$
so its density is the derivative $f(y)=0.5y^{-0.5} = 0.5y^{0.5-1}(1-y)^{1-1}$
which is clearly a $\text{Beta}(0.5,1)$ density
A: Consider $10,000$ observations x from $\mathsf{Unif}(0,1).$
Then square them. About $1000$ of these $x$-values will
lie in $(0,0.1]$ and their squares will lie in the interval
$(0,0.01].$ In the histograms below both bars are colored
red. [The red bar on the right is tall and thin, so
you may have to look closely to see it is colored red.]
similarly about $1000$ of these $x$-values will lie in
$(0.9, 1]$ and their squares will lie in the interval
$(0.81, 1].$ In the histograms below both bars are colored violet. So the original points and their images
under transformation lie in bars of the same color.
Each bar in each histogram contains about $1000$ observations,
and each bar has area $0.1.$ Heights of the bars on the right differ because the widths of their bases differ.
The histgram on the right is consistent with
a sample from $\mathsf{Beta}(0.5,1),$ as shown
by the density function (black).
set.seed(2022)
x = runif(10^5)
y = x^2


R code for the figure is shown below:
farb = rainbow(11)[1:10]
par(mfrow=c(1,2))
cutp = seq(0,1, by=.1)
 hdr1 = "Density of UNIF(0,1)"
 hist(x, ylim=c(0,10), prob=T, br=cutp, col=farb, main=hdr1)
  curve(dunif(x), add=T, lwd=2)
 hdr2 = "Density of BETA(.5.1)"
 hist(y, prob=T, br=cutp^2, col=farb, main=hdr2)
  curve(dbeta(x,.5,1), add=T, lwd=2)
par(mfrow = c(1,1))

