3. Given $x, y \in \mathbf{R}$, and $x^{2}+y^{2} \leqslant 1$. Then the maximum value of $x+y-x y$ is $\qquad$ .
3. 1 .

Notice that, $x+y-x y=x(1-y)+y$.
When $y$ is fixed, the expression can only achieve its maximum value when $x$ is as large as possible; similarly, when $x$ is fixed, $y$ should also be as large as possible. Therefore, we might as well assume that $x$ and $y$ are non-negative, and $x^{2}+y^{2}=1$.
Let $x+y=t(t \in[1, \sqrt{2}])$.
From $t^{2}=x^{2}+y^{2}+2 x y=1+2 x y \Rightarrow x y=\frac{t^{2}-1}{2}$.
Thus, $x+y-x y=t-\frac{t^{2}-1}{2}=1-\frac{(t-1)^{2}}{2} \leqslant 1$, and

when $x=1, y=0$, the equality holds.
Therefore, the maximum value sought is 1.