Vector Fields — Question 2

Question 2 – Solution

Strategy. Intersect the natural domains of the logarithm and square root, keeping track of strict versus non-strict inequalities.

Step 1: Component restrictions The logarithm requires $x-y>0$, while the square root requires $4-x^2-y^2\ge 0$. Therefore $$\boxed{D=\{(x,y):x>y,\ x^2+y^2\le 4\}}.$$

Step 2: Boundary decisions No point of $y=x$ is included because $\ln 0$ is undefined. The circular boundary is included only where $x>y$; there the second component equals $0$ and the first remains defined.

Step 3: Regularity Both component functions are continuous wherever they are defined, so $\mathbf F$ is continuous on all of $D$. Both are differentiable where their defining inequalities are strict: $$\boxed{x>y,\qquad x^2+y^2<4}.$$ At the included circular arc, the square-root component is continuous but its derivatives become singular.

Verification The shaded half-disk excludes its straight diagonal edge but includes the appropriate circular arc, reflecting $>$ versus $\le$.