Calculus 8th Edition

${(x,y)∈R^{2}|x^{2}+y^{2}\leq 4}$ and $-1\leq x\leq 1$
Given: $f(x,y)=\sqrt {4-x^{2}-y^{2}}+\sqrt {1-x^{2}}$ Since we can only take the square root of non-negatives quantities, we want $4-x^{2}-y^{2}\geq 0$ This implies $x^{2}+y^{2}\leq 4$ and $1-x^{2}\geq 0$ This implies $x^{2}\leq 1$ or $-1\leq x\leq 1$ Therefore, the domain for given function is ${(x,y)∈R^{2}|x^{2}+y^{2}\leq 4}$ and $-1\leq x\leq 1$ Sketch the graph for the domain as shown below: