Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.1 Exercises - Page 876: 25

Answer

The domain is given by $$(x,y)\in\{(x,y)|x^2+y^2\leq 4\}.$$ The range is given by $$f\in[0,2].$$

Work Step by Step

1) Domain The given function is $f(x,y) = \sqrt{4-x^2-y^2}.$ We have the restriction that any expression under a square root must not be negative. This means that $$4-x^2-y^2\geq 0 \Rightarrow x^2+y^2 \leq 4.$$ The expression $x^2+y^2 = 4 = 2^2$ is the equation of an infinite cylinder which axis of symmetry is $z$ axis and its radius is $2$. Since we demand that $x^2+y^2 \leq 4$, the ordered pair $(x,y)$ can take values from the surface and from the inside of this cylinder. We will denote the set of these ordered pairs by $s=\{(x,y)|x^2+y^2\leq 4\}$ so the domain is given by $(x,y)\in s$. 2) Range We can rewrite the function as $$f(x,y) = \sqrt{4-x^2-+y^2} =\sqrt{4-(x^2+y^2)}.$$ The sum $x^2+y^2$ is always nonnegative. When that sum is maximal then the value of $f$ is minimal and when it is minimal the value of $f$ is maximal (because it is subtracted). The minimal value of $x^2+y^2$ is $0$ and it occurs when $(x,y) = (0,0)$, so the maximal value of $f$ is $\sqrt{4-0} = 2$. The maximal value of $x^2+y^2$ is $4$ as we have shown when finding the domain so the minimal value for $f$ is $\sqrt{4-4} = 0.$ $f$ can take any values in between as well so the range is given by $$f\in[0,2].$$
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