Joint PDF solving

Question

The final answer for the problem says its 7/20. But i dont understand how the solution is real value. I tried to simplify it but got stuck at (3-2rooty-y)/2. Please help

Answer 1

Since $X,Y$ are both almost surely positive, then: $$\begin{align}\mathsf P(X>\surd Y) {}&= \mathsf P(X^2>Y) \\ &= \int_0^1\int_0^{x^2} (x+y)~\mathsf dy~\mathsf d x \end{align}$$

From this you should immediately anticipate a rational result.

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However, it looks like you tried directly: $$\begin{align}\mathsf P(X>\surd Y) {}&= \int_0^1\int_{\surd y}^1 (x+y)~\mathsf dx~\mathsf d y \end{align}$$

This is slightly harder to do by hand, because fractions.   But if you persevere and take care, it does give the same result.