Question

At what points of \(\mathbb{R}^{2}\) are the following functions continuous? $$f(x, y)=\sqrt{4-x^{2}-y^{2}}$$

Short answer

If so, at which points is it continuous? Answer: Yes, the function is continuous at all points within the closed disc centered at the origin with a radius of 2, denoted by the inequality \(x^2 + y^2 \leq 4\).

Step-by-step solution

Determine the domain of the function

To be defined, the radicand inside the square root must be non-negative. Therefore, we should find the values of x and y for which: $$4 - x^2 - y^2 \geq 0$$ Rearranging the inequality, we get: $$x^2 + y^2 \leq 4$$ This represents the equation of a closed circle with a center at the origin (0, 0) and a radius of 2. Thus, the function is defined on the closed disc centered at the origin, with a radius of 2.

Check the function's continuity within the domain

Since the function is defined as a square root over a continuous function (polynomial), it is continuous wherever it is defined. Given that the polynomial is continuous at every point within the closed disc centered at the origin with a radius of 2, the square root function is also continuous within this domain. Therefore, the function \(f(x, y) = \sqrt{4 - x^2 - y^2}\) is continuous at all the points of \(\mathbb{R}^{2}\) within the closed disc centered at the origin with a radius of 2, denoted by the inequality \(x^2 + y^2 \leq 4\).