Worked examples · Sheet 1

Domains of functions of two variables

Theory used: Natural domain Interior and boundary points Open, closed and bounded sets

Exercise 1

Find the natural domain of

f(x, y) = \ln(2x - y + 4)

and state whether it is an open, closed, bounded set.

2x - y + 4 > 0 \iff y < 2x + 4

(0, 0) : \quad 2 \cdot 0 - 0 + 4 = 4 > 0

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

y = 2x + 4

The domain of

ln (2 x - y + 4)

is the half-plane below the line

y = 2 x + 4

, line excluded (dashed)

The argument of the logarithm must be strictly positive, so the only condition is $2x - y + 4 > 0$ , which is equivalent to $y < 2x + 4$ . The line $y = 2x + 4$ divides the plane into two half-planes. The origin is not on the line and satisfies the condition, because $4 > 0$ , therefore the domain is the half-plane containing the origin, that is, the one below the line.

The boundary of the domain is the line, which does not belong to the domain because the inequality is strict. The domain contains none of its boundary points, so it is open; it is not closed, because the points of the line are boundary points and do not belong to it. The half-plane contains the points $(x, 0)$ with $x > -2$ , which move as far from the origin as desired, so it is not bounded.

Exercise 2

Find the natural domain of

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

and state whether it is an open, closed, bounded set.

\begin{cases} x^2 + y^2 - 1 > 0 \\[4pt] 4 - x^2 - y^2 \geq 0 \end{cases} \iff \begin{cases} x^2 + y^2 > 1 \\[4pt] x^2 + y^2 \leq 4 \end{cases}

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

x^2 + y^2 = 4

x^2 + y^2 = 1

The domain is the annulus

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

; the outer circle belongs to the domain (solid line), the inner one does not (dashed)

The logarithm requires a strictly positive argument and the root an argument greater than or equal to zero; the two conditions must hold together, so they are written as a system. The quantity $x^2 + y^2$ is the square of the distance of $(x, y)$ from the origin. The first condition therefore says that the distance of the point from the origin is greater than $1$ , that is, the point lies outside the circular region of radius $1$ ; the second says that the distance is at most $2$ , that is, the point lies in the circular region of radius $2$ , circle included. The domain is the circular annulus between the two circles.

The boundary consists of the two circles. The domain contains the points of the outer circle, which are boundary points, so it is not open; it does not contain the points of the inner circle, which are also boundary points, so it is not closed. The domain is therefore neither open nor closed. Each of its points has distance at most $2$ from the origin, so it is bounded.

Exercise 3

Find the natural domain of

f(x, y) = \frac{\sqrt{y - x^2}}{\ln(4 - y)}

and state whether it is an open, closed, bounded set.

\begin{cases} y - x^2 \geq 0 \\[4pt] 4 - y > 0 \\[4pt] \ln(4 - y) \neq 0 \end{cases} \iff \begin{cases} y \geq x^2 \\[4pt] y < 4 \\[4pt] y \neq 3 \end{cases}

D = \{ (x, y) \in \mathbb{R}^2 : x^2 \leq y < 4, \ y \neq 3 \}

y = x^2

The domain is the region between the parabola

y = x^{2}

(included) and the line

y = 4

(excluded), with the segment on the line

y = 3

removed

The conditions come from the three elements of the formula. The root in the numerator requires $y - x^2 \geq 0$ , that is, the point must lie above the parabola $y = x^2$ or on it. The logarithm requires $4 - y > 0$ , that is, $y < 4$ . The logarithm is in the denominator, so it must also be nonzero. Since $\ln u = 0$ only for $u = 1$ , the condition $\ln(4 - y) \neq 0$ is equivalent to $4 - y \neq 1$ , that is, $y \neq 3$ .

The domain is the region between the parabola and the line $y = 4$ , from which the segment of the line $y = 3$ interior to the region, with endpoints $\left( \pm\sqrt{3}, 3 \right)$ , is removed. The domain contains the points of the parabola, which are boundary points, so it is not open. It does not contain the point $(0, 4)$ , which is a boundary point because each of its neighborhoods contains points of the domain below the line and points with $y > 4$ , so it is not closed. The domain is therefore neither open nor closed. In the domain $x^2 \leq y < 4$ , so $|x| < 2$ and $0 \leq y < 4$ , and the domain is bounded, because it is contained in the rectangle $[-2, 2] \times [0, 4]$ .