1. **Problem Statement:** Analyze the functions for domain, range, level curves, boundary, and domain properties. --- ### (i) $f(x,y) = \sqrt{9 - x^2 - y^2}$ 1. **Domain:** The expression under the square root must be non-negative: $$9 - x^2 - y^2 \geq 0$$ which implies $$x^2 + y^2 \leq 9$$ This is the disk of radius 3 centered at the origin. 2. **Range:** Since the square root outputs values from 0 to the maximum when $x^2 + y^2 = 0$: $$0 \leq f(x,y) \leq 3$$ 3. **Level Curves:** For a constant $c$, $$\sqrt{9 - x^2 - y^2} = c \implies 9 - x^2 - y^2 = c^2 \implies x^2 + y^2 = 9 - c^2$$ These are circles centered at the origin with radius $\sqrt{9 - c^2}$ for $0 \leq c \leq 3$. 4. **Boundary of Domain:** The boundary is where the inequality becomes equality: $$x^2 + y^2 = 9$$ 5. **Domain Type:** The domain includes the boundary (since $\leq$), so it is a **closed region**. 6. **Boundedness:** The domain is bounded because it is contained within a finite circle. --- ### (ii) $f(x,y) = \ln(x^2 + y^2 - 4)$ 1. **Domain:** The argument of the logarithm must be positive: $$x^2 + y^2 - 4 > 0 \implies x^2 + y^2 > 4$$ This is the exterior of the circle radius 2, excluding the circle itself. 2. **Range:** Since $x^2 + y^2$ can be arbitrarily large, the argument of $\ln$ can be arbitrarily large, so range is: $$(-\infty, \infty)$$ 3. **Level Curves:** For constant $c$, $$\ln(x^2 + y^2 - 4) = c \implies x^2 + y^2 - 4 = e^c \implies x^2 + y^2 = 4 + e^c$$ These are circles centered at origin with radius $\sqrt{4 + e^c}$. 4. **Boundary of Domain:** The boundary is where the argument equals zero: $$x^2 + y^2 = 4$$ 5. **Domain Type:** The domain excludes the boundary, so it is an **open region**. 6. **Boundedness:** The domain is unbounded since it extends to infinity. --- ### (iii) $f(x,y) = \frac{1}{x^2 + y^2}$ 1. **Domain:** Denominator cannot be zero: $$x^2 + y^2 \neq 0 \implies (x,y) \neq (0,0)$$ Domain is all points except the origin. 2. **Range:** As $(x,y) \to (0,0)$, $f \to \infty$; as $\sqrt{x^2 + y^2} \to \infty$, $f \to 0^+$. Range is: $$(0, \infty)$$ 3. **Level Curves:** For constant $c$, $$\frac{1}{x^2 + y^2} = c \implies x^2 + y^2 = \frac{1}{c}$$ Circles centered at origin with radius $\frac{1}{\sqrt{c}}$. 4. **Boundary of Domain:** The only excluded point is the origin, so boundary is: $$(0,0)$$ 5. **Domain Type:** The domain excludes the boundary point, so it is an **open region**. 6. **Boundedness:** The domain is unbounded. --- ### (iv) $f(x,y) = \sqrt{y - x - 2}$ 1. **Domain:** Expression under root must be non-negative: $$y - x - 2 \geq 0 \implies y \geq x + 2$$ This is the half-plane above the line $y = x + 2$. 2. **Range:** Since $y - x - 2 \geq 0$, $f(x,y) \geq 0$ and can be arbitrarily large. Range: $$[0, \infty)$$ 3. **Level Curves:** For constant $c$, $$\sqrt{y - x - 2} = c \implies y - x - 2 = c^2 \implies y = x + 2 + c^2$$ These are lines parallel to $y = x + 2$ shifted upward by $c^2$. 4. **Boundary of Domain:** The boundary is the line where the root is zero: $$y = x + 2$$ 5. **Domain Type:** The domain includes the boundary, so it is a **closed region**. 6. **Boundedness:** The domain is unbounded. --- **Summary:** - (i) Domain: disk $x^2 + y^2 \leq 9$, closed, bounded. - (ii) Domain: exterior $x^2 + y^2 > 4$, open, unbounded. - (iii) Domain: all except origin, open, unbounded. - (iv) Domain: half-plane $y \geq x + 2$, closed, unbounded.