a)Determine the domain of the multivariable function.

\(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{\sqrt{{{16}-{x}^{{2}}-{y}^{{2}}}}}\)

\(\displaystyle{16}-{x}^{{2}}-{y}^{{2}}\ge{0}\)

\(\displaystyle-{x}^{{2}}-{y}^{{2}}\ge-{16}\)

\(\displaystyle-{1}\times{\left(-{x}^{{2}}-{y}^{{2}}\right)}\le-{1}\times-{16}\)

\(\displaystyle{x}^{{2}}+{y}^{{2}}\le{16}\)

Also, \(\displaystyle\sqrt{{{16}-{x}^{{2}}-{y}^{{2}}}}{>}{0}\)

\(\displaystyle{16}-{x}^{{2}}-{y}^{{2}}{>}{0}\)

\(\displaystyle{x}^{{2}}+{y}^{{2}}{<}{16}\)</span>

From, \(\displaystyle{x}^{{2}}+{y}^{{2}}\le{16}{\quad\text{and}\quad}{x}^{{2}}+{y}^{{2}}{<}{16}\)</span>

The domain for f(x,y) is (x,y) \(\displaystyle\in\mathbb{R}^{{2}}:{x}^{{2}}+{y}^{{2}}{<}{16}.\)</span>

b) Evaluate the range for the given function.

Range for \(\displaystyle{16}−{x}^{{2}}−{y}^{{2}}{i}{s}{\left({0},{16}\right]}.\)

Therefore, range for f(x,y) is

\(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{\sqrt{{{16}-{x}^{{2}}-{y}^{{2}}}}}\)

As \(\displaystyle{16}−{x}^{{2}}−{y}^{{2}}\to{0},{f{{\left({x},{y}\right)}}}\to\infty\)

As \(\displaystyle{16}−{x}^{{2}}−{y}^{{2}}\to{16},{f{{\left({x},{y}\right)}}}\to{14}\)

Therefore, range for \(\displaystyle{f{{\left({x},{y}\right)}}}\in{\left[\frac{{1}}{{4}},\infty\right)}.\)

c)Determine the boundary of the domain using the domain for f(x, y).

The domain for f(x,y) is (x,y) \(\displaystyle\in\mathbb{R}^{{2}}:{x}^{{2}}+{y}^{{2}}{<}{16}.\)</span>

The equation \(\displaystyle{x}^{{2}}+{y}^{{2}}{<}{16}\)</span> is a circle.

The boundary for the domain is points lying in the circle \(\displaystyle{x}^{{2}}+{y}^{{2}}={16}.\)

d) The domain for the function f(x, y) do not contain the boundary points. Therefore, the domain for the function is an open region.

e) The domain for the function is enclosed in the circle.

Hence, the domain is bounded in the circle of equation \(\displaystyle{x}^{{2}}+{y}^{{2}}{<}{16}.\)</span>

\(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{\sqrt{{{16}-{x}^{{2}}-{y}^{{2}}}}}\)

\(\displaystyle{16}-{x}^{{2}}-{y}^{{2}}\ge{0}\)

\(\displaystyle-{x}^{{2}}-{y}^{{2}}\ge-{16}\)

\(\displaystyle-{1}\times{\left(-{x}^{{2}}-{y}^{{2}}\right)}\le-{1}\times-{16}\)

\(\displaystyle{x}^{{2}}+{y}^{{2}}\le{16}\)

Also, \(\displaystyle\sqrt{{{16}-{x}^{{2}}-{y}^{{2}}}}{>}{0}\)

\(\displaystyle{16}-{x}^{{2}}-{y}^{{2}}{>}{0}\)

\(\displaystyle{x}^{{2}}+{y}^{{2}}{<}{16}\)</span>

From, \(\displaystyle{x}^{{2}}+{y}^{{2}}\le{16}{\quad\text{and}\quad}{x}^{{2}}+{y}^{{2}}{<}{16}\)</span>

The domain for f(x,y) is (x,y) \(\displaystyle\in\mathbb{R}^{{2}}:{x}^{{2}}+{y}^{{2}}{<}{16}.\)</span>

b) Evaluate the range for the given function.

Range for \(\displaystyle{16}−{x}^{{2}}−{y}^{{2}}{i}{s}{\left({0},{16}\right]}.\)

Therefore, range for f(x,y) is

\(\displaystyle{f{{\left({x},{y}\right)}}}=\frac{{1}}{\sqrt{{{16}-{x}^{{2}}-{y}^{{2}}}}}\)

As \(\displaystyle{16}−{x}^{{2}}−{y}^{{2}}\to{0},{f{{\left({x},{y}\right)}}}\to\infty\)

As \(\displaystyle{16}−{x}^{{2}}−{y}^{{2}}\to{16},{f{{\left({x},{y}\right)}}}\to{14}\)

Therefore, range for \(\displaystyle{f{{\left({x},{y}\right)}}}\in{\left[\frac{{1}}{{4}},\infty\right)}.\)

c)Determine the boundary of the domain using the domain for f(x, y).

The domain for f(x,y) is (x,y) \(\displaystyle\in\mathbb{R}^{{2}}:{x}^{{2}}+{y}^{{2}}{<}{16}.\)</span>

The equation \(\displaystyle{x}^{{2}}+{y}^{{2}}{<}{16}\)</span> is a circle.

The boundary for the domain is points lying in the circle \(\displaystyle{x}^{{2}}+{y}^{{2}}={16}.\)

d) The domain for the function f(x, y) do not contain the boundary points. Therefore, the domain for the function is an open region.

e) The domain for the function is enclosed in the circle.

Hence, the domain is bounded in the circle of equation \(\displaystyle{x}^{{2}}+{y}^{{2}}{<}{16}.\)</span>