P(sqrt(X) > Y) Using Joint PDF I'm having difficulty understanding a homework solution from my probability class.  The problem provides the following PDF:

And the problem statement is:

Last but not least, here's the given solution:

I'm struggling to understand 2 things.  First, the logic behind it.  Why would we use the PDF to find P(sqrt(X) > Y)?  From my understanding of the PDF and its uses, it doesn't quite seem to follow.  Secondly, I don't really understand how a constant is derived as the answer.  I use u-sub integration and I still get an answer in terms of x.  Am I just bad at calc 2?  Appreciate any help, sorry I can't embed the images directly.
EDIT: I got 10 points so I've embedded the images.  Thanks whoever gave me the points!
 A: Two random variables, $X$ and $Y$, are jointly continuous if there exists a function $f_{XY}(x,y):\mathbb R^2\rightarrow \mathbb R$ which is non-negative and is called the joint probability density function. It satisfies $$\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}f_{XY}(x,y)dxdy=1$$
and for any set $A$ in $\mathbb R^2$ we have $$P((x,y)\in A)=\int\int_{A}f_{XY}(x,y)dxdy$$
See 1 or 2.

So we have since $x>0$ and $y>0$ here then  $$P(\sqrt{X}>Y)=\int_{0}^{\infty}\int_{0}^{\sqrt{x}}f_{XY}(x,y)dydx$$
$$=\int_{0}^{\infty}\int_{0}^{\sqrt{x}}e^{-x/2}ye^{-y^2}dydx$$
$$=\int_{0}^{\infty}\int_{0}^{\sqrt{x}}e^{-x/2}\frac{d}{dy}\big(-\frac{1}{2}e^{-y^2}\big)dydx$$
$$=\int_{0}^{\infty}e^{-x/2}\big[-\frac{1}{2}e^{-y^2}\big]_{y=0}^{y=\sqrt{x}}dx$$
$$=-\frac{1}{2}\int_{0}^{\infty}e^{-x/2}\big[e^{-x}-1\big]dx$$
$$=-\frac{1}{2}\int_{0}^{\infty}\big[e^{-3x/2}-e^{-x/2}\big]dx$$
$$=-\frac{1}{2}\big[-\frac{2}{3}e^{-3x/2}+2e^{-x/2}\big]_{x=0}^{x\rightarrow\infty}$$
$$=-\frac{1}{2}\big[0-(-\frac{2}{3}+2)\big]=-\frac{1}{2}\cdot-\frac{4}{3}=\frac{2}{3}.$$
