Worked examples · Sheet 1

Domains of functions of two variables

Theory used: Natural domain Open set and closed set Bounded set and compact set

Go to: Exercise 1 Exercise 2 Exercise 3 Exercise 4

Exercise 1

Determine and describe the domain of

f (x, y) = 4 - x^{2} - y^{2} .

4 - x^{2} - y^{2} \geq 0

x^{2} + y^{2} \leq 4

D = {(x, y) \in R^{2} : x^{2} + y^{2} \leq 4}

The square root requires an argument greater than or equal to zero, so the condition is $4 - x^{2} - y^{2} \geq 0$ , that is, $x^{2} + y^{2} \leq 4$ . The solution set is the closed disk with center the origin and radius $2$ , boundary circle included. Since the inequality is non-strict, the domain is closed; it is also bounded, because it is contained in the disk of radius $2$ , so it is compact.

Exercise 2

Determine and describe the domain of

f (x, y) = ln (y - x^{2}) .

y - x^{2} > 0

y > x^{2}

D = {(x, y) \in R^{2} : y > x^{2}}

y = x^{2}

The domain

D

: the points above the parabola

y = x^{2}

, parabola excluded (dashed)

The logarithm requires a strictly positive argument, so the condition is $y > x^{2}$ . The domain is the region of points lying strictly above the parabola $y = x^{2}$ , parabola excluded. The inequality is strict between continuous functions, therefore $D$ is open; it is not bounded, because it contains points with arbitrarily large ordinate, and it is connected, since any two of its points can be joined by a curve above the parabola.

Exercise 3

Determine and describe the domain of

f (x, y) = x^{2} + y^{2} - 1 + ln (4 - x^{2} - y^{2}) .

{x^{2} + y^{2} - 1 \geq 0 4 - x^{2} - y^{2} > 0

{x^{2} + y^{2} \geq 1 x^{2} + y^{2} < 4

D = {(x, y) \in R^{2} : 1 \leq x^{2} + y^{2} < 4}

The root and the logarithm impose two conditions that must be satisfied simultaneously, so they are written as a system: the argument of the root greater than or equal to zero and the argument of the logarithm strictly positive. The system is equivalent to $1 \leq x^{2} + y^{2} < 4$ , therefore the domain is the annulus centered at the origin between the radii $1$ and $2$ , with the inner circle included and the outer one excluded. The annulus is not open, since the points of the inner circle are not interior, and is not closed, since the outer circle consists of boundary points that do not belong to $D$ ; it is bounded and connected.

Exercise 4

Determine and describe the domain of

f (x, y) = \frac{x y}{x ^{2} + y ^{2} - 9} .

{x y \geq 0 x^{2} + y^{2} \neq = 9

D = {(x, y) \in R^{2} : x y \geq 0, x^{2} + y^{2} \neq = 9}

The domain

D

: first and third quadrants with the axes, without the points of the circle

x^{2} + y^{2} = 9

(dashed arcs and open points on the axes)

The root in the numerator requires $x y \geq 0$ and the denominator must be different from zero, so the two conditions form a system. The inequality $x y \geq 0$ identifies the first and third quadrants, axes included, since the product of two numbers is greater than or equal to zero when the two numbers have the same sign or at least one of them is zero. The condition $x^{2} + y^{2} \neq = 9$ removes from this region the points of the circle with center the origin and radius $3$ . The domain is not bounded, because it contains the points $(x, 0)$ with $∣ x ∣ > 3$ , and is neither open nor closed: the points of the axes are not interior, while the removed points of the circle are boundary points of $D$ without belonging to it. Finally, $D$ is not connected: the removed circle separates the part with $x^{2} + y^{2} < 3^{2}$ , which contains the origin, from the two outer parts in the first and third quadrants, and these three parts cannot be joined by curves contained in $D$ .