Probability that $x^2+y^2 \leq 1$ for uniformly distributed random variables. $X$ and $Y$ are independent random variables uniformly distributed on $(0,1)$

Find $P(X^2+Y^2\leq1|X\leq Y)$

^conditional probability where | represents given

The final answer is $\frac{\pi}{4}$

So I know that I'm going to have to set up some sort of integral to solve this problem, but I'm confused where to start. Do I need to convert to radians first?
Note that $P(X\le Y)=1/2$ and the area of $\{(x,y)\in\mathbb{R}_+^2\ :\ x^2+y^2\le 1 \text{ and }x\le y \}$ is exactly the area of $\{(r\cos\theta,r\sin\theta)\ :\ r\le 1\text{ and }\theta\in[\pi/4,\pi/2]\}$. Therefore,

\begin{align*}P(X^2+Y^2\le1\,|\,X\le Y)&amp;=\frac{1}{P(X\le Y)}P(X^2+Y^2\le1,\ X\le Y)\\
&amp;=2\int_0^1\int_{\pi/4}^{\pi/2}r\,\mathrm{d}r\mathrm{d}\theta\\
&amp;=\frac{\pi}{4}.
\end{align*}
We have $P(X^2 + Y^2 \leq1|X\leq Y)$
Note: $x = rcos\theta, y= r sin\theta$, hence $x^2 +y^2 = r^2$

We have a circle where $x \leq y$, hence a circle $\frac{\pi}{4} \leq \theta \leq \frac{5\pi}{4}$ and radius $0 \leq r \leq 1$

Using this we can construct:
 $$\int_0^1\int_{\frac{\pi}{4}}^{\frac{5\pi}{4}} r^3 \mathrm{d\theta dr}$$
$$=\int_0^1 r^3 * \pi \mathrm{dr}$$
$$=[\frac{r^4 * \pi}{4}]_0^1$$
$$=\frac{\pi}{4}$$

Note: $r^3$ is because $x^2 + y^2 = r^2$ then the Jacobian of transformation is $r$.