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$X$ and $Y$ are distributed identically, with probability density function:

\begin{equation} f(x)= \begin{cases} 2x, & \text{if}\ 0 < x <1 \\ 0, & \text{otherwise} \end{cases} \end{equation}

And cumulative distribution function:

\begin{equation} F(x)= \begin{cases} 0, & \text{if}\ x \leq 0 \\ x^2, & \text{if}\ x \in ]0,1[ \\ 1, & \text{if}\ x \geq 1 \end{cases} \end{equation}

Upon looking at the answer, it is of the form of a double integral:

$$\int_{0}^{1} \int_{0}^{x^2} (2x)(2y) dydx=\frac{1}{3}$$

Performing the integration is not the issue, it's how to get there, to be honest. I am not sure what property of the independent random variables allows us to solve it in this way. Is it something to do with the fact that:

$$F_{X,Y}(x,y) = F_X(x) \cdot F_Y(y)$$

I saw additionally this post, which shows the same kind of formula. I can't find it anywhere in my probability textbook, so assume it is derived from another one of the properties we have been given, but I'm not sure which.

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The independence of $X$ and $Y$ gives you the joint density: $$f_{X,Y}(x,y)=f(x)f(y).$$

Once you have the joint density, $$ P(X^2>Y)=\iint_{\{(x,y)\in\mathbb{R}^2|x^2>y\}}f_{X,Y}(x,y)\ dxdy\\ =\iint_{\{(x,y)\in\mathbb{R}^2|x^2>y\}}f(x)f(y)\ dxdy. $$

Now, it is an exercise to show (using the definition of $f$) that $$ \iint_{\{(x,y)\in\mathbb{R}^2|x^2>y\}}f(x)f(y)\ dxdy=\int_{0}^{1} \int_{0}^{x^2} (2x)(2y) dydx. $$

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    I think I understand! Regarding the final exercise, is this correct?: $P(X^2 > Y) = P(Y \leq X^2)=P(X^2 < \infty, Y\leq X^2)=\int_{-\infty}^{\infty}\int_{-\infty}^{x^2}f_Y(y) \cdot f_X(x)dydx$ And then upon limiting only to the support we get $\int_{0}^{1}\int_{0}^{x^2}f_Y(y) \cdot f_X(x)dydx$ – 2017-01-04
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    Or one can directly do the double integral on $\{(x,y)|x^2>y\}\cap[0,1]\times[0,1]$ without referring to the event $X^2<\infty$. – 2017-01-04
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    The way that I wrote makes more sense in my head, so if it's correct I think I'll keep to that. Thanks for your help :) – 2017-01-04