\(\displaystyle{w}{\left({x},{y}\right)}=\sqrt{{{y}-{4}{x}^{{2}}}}\)

\(\displaystyle{y}−{4}{x}^{{2}}\ge{0}\)

\(\displaystyle−{4}{x}^{{2}}\ge−{y}\)

\(\displaystyle−{1}\times{\left(−{4}{x}^{{2}}\right)}\le−{1}×{\left(−{y}\right)}\)

\(\displaystyle{4}{x}^{{2}}\le{y}\)

\(\displaystyle{x}^{{2}}{>}{0}\)

\(\displaystyle\Rightarrow{4}{x}^{{2}}{>}{0}\)

\(\displaystyle\Rightarrow{y}\ge{0}\)

\(\displaystyle{D}=\mathbb{R}^{{2}}\) such that \(\displaystyle{y}\ge{0}{\quad\text{and}\quad}{4}{x}^{{2}}\le{y}.\)

\(\displaystyle\sqrt{{{y}-{4}{x}^{{2}}}}\ge{0}\)

So, \(\displaystyle{w}\ge{0}\)

\(\displaystyle{R}\in\mathbb{R}\) such \(\displaystyle{w}{\left({x},{y}\right)}\ge{0}\)

\(\displaystyle{y}−{4}{x}^{{2}}\ge{0}\)

\(\displaystyle−{4}{x}^{{2}}\ge−{y}\)

\(\displaystyle−{1}\times{\left(−{4}{x}^{{2}}\right)}\le−{1}×{\left(−{y}\right)}\)

\(\displaystyle{4}{x}^{{2}}\le{y}\)

\(\displaystyle{x}^{{2}}{>}{0}\)

\(\displaystyle\Rightarrow{4}{x}^{{2}}{>}{0}\)

\(\displaystyle\Rightarrow{y}\ge{0}\)

\(\displaystyle{D}=\mathbb{R}^{{2}}\) such that \(\displaystyle{y}\ge{0}{\quad\text{and}\quad}{4}{x}^{{2}}\le{y}.\)

\(\displaystyle\sqrt{{{y}-{4}{x}^{{2}}}}\ge{0}\)

So, \(\displaystyle{w}\ge{0}\)

\(\displaystyle{R}\in\mathbb{R}\) such \(\displaystyle{w}{\left({x},{y}\right)}\ge{0}\)