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Chapter 13: Problem 19

Describe the region \(R\) in the \(x y\) -plane that corresponds to the domain of the function. $$ z=\sqrt{4-x^{2}-y^{2}} $$

Short Answer

Expert verified
The region \(R\) corresponding to the domain of the function \(z=\sqrt{4-x^{2}-y^{2}}\) is the interior of a circle with radius \(2\) and center at the origin, including points lying on the circle itself.

Step by step solution

01

Define the Domain of Function

The domain of the given function \(z=\sqrt{4-x^{2}-y^{2}}\) is determined by the condition that the expression under the square root must be greater than or equal to \(0\). Hence the domain is determined by the inequality \(4-x^{2}-y^{2} \geq 0\).
02

Rearrange the Inequality

Rearranging the inequality \(4-x^{2}-y^{2} \geq 0\), gives \(x^{2} + y^{2} \leq 4\). This inequality is the equation of a circle with radius \(2\) and the center at the origin \((0, 0)\). The less than or equal to sign (\(\leq\)) indicates that the area of the circle is included in the region.
03

Describe the Region R

The region \(R\) in the \(x y\) -plane that corresponds to the domain of the function \(z=\sqrt{4-x^{2}-y^{2}}\) represents the set of all \((x, y)\) points that lie in the circle of radius 2 centered at the origin, including the points on the circle.

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