\(\begin{aligned}\rm F(x,y) &= \nabla {\rm{f}}\\{{\rm{f}}_{\rm{x}}}{\rm{i + }}{{\rm{f}}_{\rm{y}}}\rm j &= \left( {{\rm{4}}{{\rm{x}}^{\rm{3}}}{{\rm{y}}^{\rm{2}}}{\rm{ - 2x}}{{\rm{y}}^{\rm{3}}}} \right){\rm{i + }}\left( {{\rm{2}}{{\rm{x}}^{\rm{4}}}{\rm{y - 3}}{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + 4}}{{\rm{y}}^{\rm{3}}}} \right){\rm{j}}\\{{\rm{f}}_{\rm{x}}}\rm &= 4{{\rm{x}}^{\rm{3}}}{{\rm{y}}^{\rm{2}}}{\rm{ - 2x}}{{\rm{y}}^{\rm{3}}}\\{{\rm{f}}_{\rm{y}}}\rm &= 2{{\rm{x}}^{\rm{4}}}{\rm{y - 3}}{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + 4}}{{\rm{y}}^{\rm{3}}}\end{aligned}\)
Integrate in terms of \({\rm{x}}\) while keeping \({\rm{y}}\)constant.
\({\rm{f(x,y) = }}\int {\rm{4}} {{\rm{x}}^{\rm{3}}}{{\rm{y}}^{\rm{2}}}{\rm{ - 2x}}{{\rm{y}}^{\rm{3}}}{\rm{dx}}\)
\({\rm{f(x,y) = }}{{\rm{x}}^{\rm{4}}}{{\rm{y}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{3}}}{\rm{ + p(y)}} \to {\rm{(1)}}\)
Note that \({{\rm{P}}_{\rm{y}}}\) is the Integration constant. To get \({\rm{y}}\), partial differentiation \({\rm{Eqn(1)}}\).
\({{\rm{f}}_{\rm{y}}}{\rm{ = 2}}{{\rm{x}}^{\rm{4}}}{\rm{y - 3}}{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{p}}^{\rm{'}}}{\rm{(y)}}\)
Known that \({{\rm{f}}_{\rm{y}}}{\rm{ = 2}}{{\rm{x}}^{\rm{4}}}{\rm{y - 3}}{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{2}}}{\rm{ + 4}}{{\rm{y}}^3}\)
\({{\rm{p}}^{\rm{'}}}{\rm{(y) = 4}}{{\rm{y}}^{\rm{3}}}\)
Integrate both sides. To get
\({\rm{p(y) = }}{{\rm{y}}^{\rm{4}}}{\rm{ + K}}\)
Replace 'expression' with a substitute.\({\rm{p(y)}}\)in \({\rm{Eqn(1)}}\)
To get \({\rm{f(x,y) = }}{{\rm{x}}^{\rm{4}}}{{\rm{y}}^{\rm{2}}}{\rm{ - }}{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{3}}}{\rm{ + }}{{\rm{y}}^{\rm{4}}}{\rm{ + K}}\)
Line Integrals are defined by the fundamental theorem of line integrals.
\(\int_{\rm{C}} {\rm{F}} \cdot d{\rm{r}} = \int_C \nabla f \cdot {\rm{dr = f}}\left( {{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{y}}_{\rm{2}}}} \right){\rm{ - f}}\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{y}}_{\rm{1}}}} \right)\)
Where \(C\) denotes a curve extending from \(\left( {{{\rm{x}}_{\rm{1}}}{\rm{,}}{{\rm{y}}_{\rm{1}}}} \right){\rm{to}}\left( {{{\rm{x}}_{\rm{2}}}{\rm{,}}{{\rm{y}}_{\rm{2}}}} \right)\) Given this,
\({\rm{r(t) = (t + sin\pi t)i + (2t + cos\pi t)j,}}\;\;\;0 \le t \le 1\)
Hence \(C\) denotes a curve extending from \(\left( {{\rm{0,1}}} \right){\rm{to}}\left( {{\rm{1,1}}} \right)\)
Therefore,
\(\begin{aligned}\int_{\rm{C}} {\rm{F}} \cdot \rm dr &= f(1,1) - f(0,1)\\\rm &= (1 - 1 + 1 + K) - (0 - 0 + 1 + K)\\\rm &= 0\end{aligned}\)
