Step 1

Given that

\(\displaystyle{F}{\left({x},{y}\right)}={\left({y}^{{{2}}}-{2}{x}\right)}{i}+{2}{x}{y}{j}\)

we need to

determine whether or not F is conservative vector field

Recall that if F(x,y)=P(x,y)i+Q(x,y)j, than in order to check the conservative vector field we need to chack whether

\(\displaystyle{\frac{{\partial{P}}}{{\partial{y}}}}={\frac{{\partial{Q}}}{{\partial{x}}}}\)

in the domian of F.Here \(\displaystyle{P}{\left({x},{y}\right)}={y}^{{{2}}}-{2}{x}\text{and}{Q}{\left({x},{y}\right)}={2}{x}{y}\).

Step 2

Observe that

\(\displaystyle{\frac{{\partial{P}}}{{\partial{y}}}}={2}{y}\)

\(\displaystyle{\frac{{\partial{Q}}}{{\partial{x}}}}={2}{y}\).

Since on the domain of F, which is \(\displaystyle{R}^{{{2}}},{f}{a}{r}{c}{\left\lbrace\partial{P}\right\rbrace}{\left\lbrace\partial{y}\right\rbrace}={\frac{{\partial{Q}}}{{\partial{x}}}}\). Note that \(\displaystyle{R}^{{{2}}}\) is simply connected and open and hence using Theorem 6, F is a concervative vector field and hence there exists a function f such that

\(\displaystyle{F}=\triangle{d}{o}{w}{n}{f}\).

Our next goal is to determine the f. From the above relation, we have

\(\displaystyle{F}={t}{r}{i}{\quad\text{and}\quad}\le{d}{o}{w}{n}{f}\Rightarrow{\left({y}^{{{2}}}-{2}{x}\right)}{i}+{2}{x}{y}{j}={{f}_{{{x}}}{\left({x},{y}\right)}}{i}+{{f}_{{{y}}}{\left({x},{y}\right)}}{j}\)

\(\displaystyle\Rightarrow{{f}_{{{x}}}{\left({x},{y}\right)}}={y}^{{{2}}}-{2}{x},\text{and}{{f}_{{{y}}}{\left({x},{y}\right)}}={2}{x}{y}\).

Given that

\(\displaystyle{F}{\left({x},{y}\right)}={\left({y}^{{{2}}}-{2}{x}\right)}{i}+{2}{x}{y}{j}\)

we need to

determine whether or not F is conservative vector field

Recall that if F(x,y)=P(x,y)i+Q(x,y)j, than in order to check the conservative vector field we need to chack whether

\(\displaystyle{\frac{{\partial{P}}}{{\partial{y}}}}={\frac{{\partial{Q}}}{{\partial{x}}}}\)

in the domian of F.Here \(\displaystyle{P}{\left({x},{y}\right)}={y}^{{{2}}}-{2}{x}\text{and}{Q}{\left({x},{y}\right)}={2}{x}{y}\).

Step 2

Observe that

\(\displaystyle{\frac{{\partial{P}}}{{\partial{y}}}}={2}{y}\)

\(\displaystyle{\frac{{\partial{Q}}}{{\partial{x}}}}={2}{y}\).

Since on the domain of F, which is \(\displaystyle{R}^{{{2}}},{f}{a}{r}{c}{\left\lbrace\partial{P}\right\rbrace}{\left\lbrace\partial{y}\right\rbrace}={\frac{{\partial{Q}}}{{\partial{x}}}}\). Note that \(\displaystyle{R}^{{{2}}}\) is simply connected and open and hence using Theorem 6, F is a concervative vector field and hence there exists a function f such that

\(\displaystyle{F}=\triangle{d}{o}{w}{n}{f}\).

Our next goal is to determine the f. From the above relation, we have

\(\displaystyle{F}={t}{r}{i}{\quad\text{and}\quad}\le{d}{o}{w}{n}{f}\Rightarrow{\left({y}^{{{2}}}-{2}{x}\right)}{i}+{2}{x}{y}{j}={{f}_{{{x}}}{\left({x},{y}\right)}}{i}+{{f}_{{{y}}}{\left({x},{y}\right)}}{j}\)

\(\displaystyle\Rightarrow{{f}_{{{x}}}{\left({x},{y}\right)}}={y}^{{{2}}}-{2}{x},\text{and}{{f}_{{{y}}}{\left({x},{y}\right)}}={2}{x}{y}\).