Functions of Several Variables — Question 1

Question 1 – Solution

Strategy Enforce separately the square-root, logarithm, and nonzero-denominator conditions.

Step 1: Square root Real values require $9-x^2-y^2\ge 0$, hence $x^2+y^2\le 9$.

Step 2: Logarithm The logarithm requires $x-y>0$. Because it is a denominator, $\ln(x-y)\ne 0$, so $x-y\ne 1$.

Step 3: Combine Therefore $$\boxed{D=\{(x,y):x^2+y^2\le 9,\ x>y,\ x-y\ne 1\}}.$$ The circular boundary is included only where the other restrictions hold; $x=y$ and $x-y=1$ are excluded.

Step 4: Test points $(2,0)$ satisfies all conditions. $(0,0)$ fails $x>y$. $(1,0)$ makes the denominator $\ln 1=0$. Thus only $\boxed{(2,0)}$ belongs to $D$.