Question

Let

\({\rm{F}}\left( {{\rm{x,y}}} \right){\rm{ = }}\frac{{\left( {{\rm{2}}{{\rm{x}}^{\rm{3}}}{\rm{ + 2x}}{{\rm{y}}^{\rm{2}}}{\rm{ - 2y}}} \right){\rm{i + }}\left( {{\rm{2}}{{\rm{y}}^{\rm{3}}}{\rm{ + 2}}{{\rm{x}}^{\rm{2}}}{\rm{y + 2x}}} \right){\rm{j}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}\)

Evaluate \(\oint_{\rm{C}} {{\rm{F}} \cdot {\rm{dr}}} \), where \(C\)is shown in the figure.

Short answer

The \(\oint_C F \cdot {\rm{dr = 4\pi }}\).

Step-by-step solution

Expand the equation.

\(\begin{aligned}\int_{\rm{C}} {\rm{F}} \cdot {\rm{dr = }}\int_{\rm{C}} {\frac{{{\rm{2}}{{\rm{x}}^{\rm{3}}}{\rm{ + 2x}}{{\rm{y}}^{\rm{2}}}{\rm{ - 2y}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}} {\rm{dx + }}\frac{{{\rm{2}}{{\rm{y}}^{\rm{3}}}{\rm{ + 2}}{{\rm{x}}^{\rm{2}}}{\rm{y + 2x}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}{\rm{dy}}\\\int_{\rm{C}} {\left( {{\rm{2x - }}\frac{{{\rm{2y}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}} \right)} {\rm{dx + }}\left( {{\rm{2y + }}\frac{{{\rm{2x}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}} \right){\rm{dy}}\end{aligned}\)

Because the partial derivatives\(\frac{{\partial P}}{{\partial y}}\)and\(\frac{{\partial Q}}{{\partial x}}\)do not exist at the origin, Green's theorem cannot be applied to a closed-loop enclosing the origin.

Re-write this

\(\int_{\rm{C}} {\rm{2}} {\rm{xdx + 2ydy + }}\int_{\rm{C}} {\frac{{{\rm{ - 2ydx + 2xdy}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}} {\rm{ = }}{{\rm{I}}_{\rm{1}}}{\rm{ + }}{{\rm{I}}_{\rm{2}}}\)

\({I_{\rm{1}}}{\rm{ = }}\int_{\rm{C}} {{\rm{2xdx + 2ydy}}} \)

When measured in polar coordinates,

\(\begin{aligned}\rm x &= rcos\theta \\ \Rightarrow \rm dx &= \frac{{\partial {\rm{x}}}}{{\partial {\rm{r}}}}{\rm{dr + }}\frac{{\partial {\rm{x}}}}{{\partial {\rm{\theta }}}}{\rm{d\theta }}\\\rm &= cos\theta dr - rsin\theta d\theta \end{aligned}\)

And

\(\begin{aligned}\rm y &= rsin\theta \\ \Rightarrow \rm dy &= \frac{{\partial y}}{{\partial r}}dr{\rm{ + }}\frac{{\partial y}}{{\partial \theta }}d\theta \\\rm &= sin\theta dr + rcos\theta d\theta \end{aligned}\)

\({{\rm{I}}_{\rm{1}}}{\rm{ = }}\int_{\rm{C}} {\rm{2}} {\rm{rcos\theta (cos\theta dr - rsin\theta d\theta ) + 2rsin\theta (sin\theta dr + rcos\theta d\theta )}}\)

Rearrange the items as follows:

\(\begin{aligned}{{\rm{I}}_{\rm{1}}}\rm &= \int_{\rm{C}} {\left( {{\rm{2rco}}{{\rm{s}}^{\rm{2}}}{\rm{\theta + 2rsi}}{{\rm{n}}^{\rm{2}}}{\rm{\theta }}} \right)} {\rm{dr + }}\left( {{\rm{ - 2}}{{\rm{r}}^{\rm{2}}}{\rm{cos\theta sin\theta + 2}}{{\rm{r}}^{\rm{2}}}{\rm{sin\theta cos\theta }}} \right){\rm{d\theta }}\\{{\rm{I}}_{\rm{1}}}\rm &= \int_{\rm{C}} {{\rm{(2r)}}} {\rm{dr + (0)d\theta }}\end{aligned}\)

\({{\rm{I}}_{\rm{1}}}{\rm{ = 2}}\int_{\rm{C}} {\rm{r}} {\rm{dr}}\)

Evaluate the equation.

Because \(c\) is a closed loop, its value is zero.

Starting at a position where \(r\) equals \({\rm{k}}\) and moving once around the loop, integrate from \({\rm{k}}\) to \({\rm{k}}\).

\(\begin{aligned}{{\rm{I}}_{\rm{1}}}\rm &= 2\int_{\rm{k}}^{\rm{k}} {\rm{r}} {\rm{dr}}\\\rm &= 0\\{{\rm{I}}_{\rm{2}}}\rm &= \int_{\rm{C}} {\frac{{{\rm{ - 2ydx + 2xdy}}}}{{{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}}}} \end{aligned}\)

In the polar coordinate system

\({{\rm{I}}_{\rm{2}}}{\rm{ = }}\int_{\rm{C}} {\frac{{{\rm{ - 2rsin\theta (cos\theta dr - rsin\theta d\theta ) + 2rcos\theta (sin\theta dr + rcos\theta d\theta )}}}}{{{{\rm{r}}^{\rm{2}}}}}} \)

Rearrange the items in the order indicated below.

\(\begin{aligned}{{\rm{I}}_{\rm{2}}}\rm &= \int_{\rm{C}} {\frac{{{\rm{( - 2rsin\theta cos\theta + 2rsin\theta cos\theta )dr + 2}}{{\rm{r}}^{\rm{2}}}\left( {{\rm{si}}{{\rm{n}}^{\rm{2}}}{\rm{\theta + co}}{{\rm{s}}^{\rm{2}}}{\rm{\theta }}} \right){\rm{d\theta }}}}{{{{\rm{r}}^{\rm{2}}}}}} \\{{\rm{I}}_{\rm{2}}}\rm &= \int_{\rm{C}} {\frac{{{\rm{(0)dr + 2}}{{\rm{r}}^{\rm{2}}}{\rm{(1)d\theta }}}}{{{{\rm{r}}^{\rm{2}}}}}} {{\rm{I}}_{\rm{2}}}\\\rm &= 2\int_{\rm{C}} {\rm{d}} {\rm{\theta }}\end{aligned}\)

The loop grows \({\rm{\theta }}\) from \({\rm{0}}\) to\({\rm{2\pi }}\) in a counter clockwise orientation.

Integrate from\({\rm{0}}\) to as a result \({\rm{2\pi }}\).

\(\begin{aligned}{{\rm{I}}_{\rm{2}}}\rm &= 2\int_{\rm{0}}^{{\rm{2\pi }}} {\rm{d}} {\rm{\theta }}\\\rm &= 4\pi \\\int_{\rm{C}} {\rm{F}} \cdot \rm dr &= {{\rm{I}}_{\rm{1}}}{\rm{ + }}{{\rm{I}}_{\rm{2}}}\\\rm &= 0 + 4\pi \\\rm &= 4\pi \end{aligned}\)

Therefore, the \(\oint_C F \cdot {\rm{dr = 4\pi }}\).