Question

Use Green’s Theorem to evaluate the line integral along the given positively oriented curve.

\(\int_C {\cos } ydx + {x^2}\sin ydy\),

\(C\)is the rectangle with vertices \((0,0),(5,0),(5,2)\), and \((0,2)\)

Short answer

The value of line integral \(\oint_C {\cos } ydx + {x^2}\sin ydy\) by Green's Theorem is \(30(1 - \cos (2))\).

Step-by-step solution

Given information from question

The line integral is \(\oint_C {\cos } ydx + {x^2}\sin ydy\).

The curve \(C\) is a rectangle with vertices \((0,0),(5,0),(5,2)\), and \((0,2)\).

Green’s Theorem

Green’s Theorem said that,

"Let\(C\)be a positively oriented, piecewise-smooth, simple closed curve in the plane and let\(D\)be the region bounded by\(C\). If\(P\)and\(Q\)have continuous partial derivatives on an open region that contains\(D\), then ……. (1)

Compare the two expressions\(\int_C P dx + Qdy\)and\(\oint_C {\cos } ydx + {x^2}\sin ydy\)

The curve \(C\) is positively oriented, piecewise-smooth, and simply closed curve with domain \(D = \left\{ {\begin{aligned}{*{20}{l}}{0 \le x \le 5}\\{0 \le y \le 2}\end{aligned}} \right.\) and hence Green's theorem is applicable.

Compare the two expressions,

\(\int_C P dx + Qdy\)and \(\oint_C {\cos } ydx + {x^2}\sin ydy\).

\(\begin{aligned}P & = \cos y\\Q & = {x^2}\sin y\end{aligned}\)

The value of \(\frac{{\partial P}}{{\partial y}}\).

\(\begin{aligned}\frac{{\partial P}}{{\partial y}} & = \frac{\partial }{{\partial y}}(\cos y)\\ & = - \sin y\end{aligned}\)

The value of \(\frac{{\partial Q}}{{dx}}\).

\(\begin{aligned}\frac{{\partial Q}}{{dx}} & = \frac{\partial }{{dx}}\left( {{x^2}\sin y} \right)\\ & = \sin y\frac{\partial }{{\partial x}}\left( {{x^2}} \right)\\ & = 2x\sin y\quad \end{aligned}\)

Calculate the value of integral

Equation (1) can be rewritten as,

\(\int_C P dx + Qdy = \int_{{x_1}}^{{x_2}} {\int_{{y_1}}^{{y_2}} {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)} } dydx\) ……. (2)

Substitute the values in equation (2),

\(\begin{aligned}\oint_C {\cos } ydx + {x^2}\sin ydy & = \int\limits_0^5 {\int\limits_0^2 {(2x\sin y + \sin y)dydx} } \\ & = \int\limits_0^5 {\left( { - 2x\cos y - \cos y} \right)} dx\\ & = \int\limits_0^5 {\left( {( - 2xcos(2) - \cos (2)) - ( - 2x\cos (0) - \cos (0))} \right)dx} \end{aligned}\)

Solve the equation further.

\(\begin{aligned}\oint_C {\cos } ydx + {x^2}\sin ydy & = \int\limits_0^5 {\left( {\cos (2)( - 2x - 1) - ( - 2x - 1)} \right)dx} \\ & = (\cos (2) - 1)\int\limits_0^5 {\left( {( - 2x - 1)} \right)dx} \\ & = (cos(2) - 1){\left( {( - {x^2} - x} \right)^5}_0\end{aligned}\)

The above result can be simplified as follows.

\(\begin{aligned}\oint_C {\cos } ydx + {x^2}\sin ydy & = (\cos (2) - 1)(( - 25 - 5))\\ & = - 30(\cos (2) - 1)\\ & = 30 - 30\cos (2)\\ & = 30(1 - \cos (2))\end{aligned}\)

Thus, the value of line integral \(\oint_C {\cos } ydx + {x^2}\sin ydy\) by Green's Theorem is \(30(1 - \cos (2))\).