Joint density of $(R,X)$ when $(X,Y)$ is uniform on the unit circle and $R^2=X^2+Y^2$ The question is like this: 



A point $(X,Y)$ is picked at random uniformly in the unit circle. 
  Find the joint density of $R$ and $X$, where $R^2 = X^2 + Y^2$.

The TA drew a diagram like this and claim that 
$F_{R,X}(r,t)=r^2$ when $t>r$ and $F_{R,X}(r,t)=0$ when $t<r$

She claimed that it is resulted from the constraints $x^2+y^2 \le r^2 $ and $x\le t$

But
I don't understand why it is like this, isn't it we can always find $y$ whenever there is a $x$, why would $t$ matters? Can anyone please explain, thank you so much

 A: I assume that $F_{R,X}$ means the joint CDF of $R, X$ and $R$ is defined as the non-negative square root: $\sqrt{X^2 + Y^2} \geq 0$.
Then by definition of joint CDF,
$$ F_{R, X}(r, t) = \Pr\{R\leq r, X \leq t\} $$
The event $\{R \leq r\}$ means the random point fall into the blue disc in your figure (the circle centered at origin with radius $r$)
The event $\{X \leq t\}$ means the random point fall into the left half-plane, divided by the vertical green line $X = t$ in your figure.
So when $r < t$, the vertical line is located at the right hand side of the blue disc without any intersection. And thus the disc is a subset of the half-plane, i.e.
$$ \{R \leq r\} \subseteq \{X \leq t\} $$
which implies
$$ \{R \leq r\} \cap \{X \leq t\} = \{R \leq r\}$$
So the joint CDF becomes
$$ F_{R, X}(r, t) = \Pr\{R\leq r, X \leq t\} = \Pr\{R \leq r\} = \frac {\pi r^2} {\pi 1^2} = r^2$$
(The proportion of the area of blue disc to the unit circle, due to the uniform distribution)
Similarly when $t < -r \leq 0$, we have the vertical line being the left of the circle without any intersection. In such case the intersection of the half-plane and the disc is an empty set. i.e.
$$ \{R \leq r\} \cap \{X \leq t\} = \varnothing$$
So the joint CDF $ F_{R, X}(r, t) = 0$ in such case.
(*Maybe your TA should define more clearly and put the negative sign there)
The most complicated one is the remaining case - when $-r \leq t \leq r$. In such case the two regions intersections become a circular segment, and we computed the area by integration,
$$ F_{R, X}(r, t) = \frac {2} {\pi} \int_{-r}^t \sqrt{r^2 - x^2} dx 
= \frac {1} {\pi} \left[r^2 \sin^{-1} \left(\frac {t} {r}\right) + t\sqrt{r^2 - t^2} + \frac {\pi r^2} {2} \right]$$
$2$ is due to the upper and lower circle symmetry, and $\pi$ is the area of the unit circle.
And here we only discussed the non-trivial case of $-1 \leq r \leq 1$, $-1 \leq t \leq 1$. If $r$ or $t$ are outside of this interval, then the event becomes a trivial event and the problem just reduced to a lower dimensional one.
