<s>[INST] Probability and Integrals Suppose $f(x,y) = c$ for $0\lt y\lt x\lt 1$ and $0$ outside. What is $P(X+Y \leq 1)$? What is $P(X^2+Y^2 \leq 1)$?
So 
\begin{equation*}
P(X+Y \leq 1) = \int_{0}^{1} \int_{0}^{1-x} 2 \ dy \ dx? 
\end{equation*}
Likewise, 
\begin{equation*}
$P(X^2+Y^2 \leq 1) = \int_{0}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} 2 \ dy \ dx$? 
\end{equation*}
This is assuming that $c=2$.  
 [/INST] Since the joint density is constant on the support region, you can get your answers by considering areas.  The answer in each case is the area that corresponds to the event in question divided by the total area of the support region.  
Thus, to calculate $P(X + Y \leq 1)$, you want the area of the lower triangular region below (i.e., the region for which $x + y \leq 1$), divided by the total area of the triangle.

This is $$\frac{\frac{1}{4}}{\frac{1}{2}} = \frac{1}{2}.$$
Similarly, to calculate $P(X^2 + Y^2 \leq 1)$, you want the area of the circle sector divided by the total area of the triangle.

This is $$\frac{\frac{\pi}{8}}{\frac{1}{2}} = \frac{\pi}{4}.$$
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