Question

Evaluate the line integral by two methods: (a) directly and (b) using Green's Theorem.

\(\oint_C x ydx + {x^2}dy\),

\(C\)is the rectangle with vertices \((0,0),(3,0),(3,1)\), and \((0,1)\)

Short answer

a) The value of line integral\(\oint_C x ydx + {x^2}dy\)is\(4.5\).

b) The value of line integral \(\oint_C x ydx + {x^2}dy\) using Green's Theorem is \(4.5\).

Step-by-step solution

Given data

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

The curve\(C\)is rectangle with vertices\((0,0),(3,0),(3,1)\), and\((0,1)\).

Green’s Theorem

Green's Theorem:

"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 ."

Find the value of the line integral by direct method

a)

Draw the rectangular curve with vertices \((0,0),(3,0),(3,1)\), and \((0,1)\) as shown below in Figure \(1\)

From Figure 1, write the expressions for curve\({C_1},0 \le t \le 3\).

\(\begin{aligned}{l}x = t\,\,\,\,\,\,\,\,\,\,\,(1)\\y = 0\,\,\,\,\,\,\,\,\,\,(2)\end{aligned}\)

Differentiate the equation (1) with respect to\(t\).

\(\begin{align} & x=t\,\,\,\,\,\,\,\,\,\,\,(1) \\ & y=0\,\,\,\,\,\,\,\,\,\,(2) \\ \end{align}\)

Differentiate the equation\((2)\)with respect to\(t\).

\(\begin{align} & \frac{dx}{dt}=\frac{d}{dt}(t) \\ & \frac{dx}{dt}=1\left\{ \because \frac{d}{dt}(t)=1 \right\} \\ & dx=dt \end{align}\)

From Figure 1, write the expressions for curve\({C_2},0 \le t \le 1\).

\(\begin{aligned}{l}x = 3\,\,\,\,\,\,\,\,\,\,\,(3)\\y = t\,\,\,\,\,\,\,\,\,\,\,\,\,(4)\end{aligned}\)

Differentiate the equation\((3)\)with respect to\(t\).

\(\begin{aligned}{l}\frac{{dx}}{{dt}} & = \frac{d}{{dt}}(3)\\\frac{{dx}}{{dt}} & = 0\\dx & = 0\end{aligned}\)

Differentiate the equation\((4)\)with respect to\(t\).

\(\begin{align} & \frac{dy}{dt}=\frac{d}{dt}(t) \\ & \frac{dy}{dt}=1\left\{ \because \frac{d}{dt}(t)=1 \right\} \\ & dy=dt \\ \end{align}\)

From Figure 1, write the expressions for curve\({C_3},0 \le t \le 3\).

\(\begin{aligned}{l}x & = 3 - t\,\,\,\,\,\,\,\,\,\,(5)\\y & = 1\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(6)\end{aligned}\)

Differentiate the equation\(\left( 5 \right)\)with respect to\(t\).

\(\begin{aligned}{l}\frac{{dx}}{{dt}} & = \frac{d}{{dt}}(3 - t)\\\frac{{dx}}{{dt}} & = 0 - 1\\dx & = - dt\end{aligned}\)

Differentiate the equation\(\left( 6 \right)\)with respect to\(t\).

\(\begin{aligned}{l}\frac{{dy}}{{dt}} & = \frac{d}{{dt}}(1)\\\frac{{dy}}{{dt}} & = 0\\dy & = 0\end{aligned}\)

From Figure 1, write the expressions for curve\({C_4},0 \le t \le 1\).

\(\begin{aligned}{l}x & = 0\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(7)\\y & = 1 - t\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(8)\end{aligned}\)

Differentiate the equation\((7)\)with respect to\(t\).

\(\begin{align} & \frac{dx}{dt}=\frac{d}{dt}(0) \\ & \frac{dx}{dt}=0\left\{ \because \frac{d}{dt}(k)=0 \right\} \\ & dx=0 \\ \end{align}\)

Differentiate the equation\(\left( 8 \right)\)with respect to\(t\).

\(\begin{aligned}{l}\frac{{dy}}{{dt}} & = \frac{d}{{dt}}(1 - t)\\\frac{{dy}}{{dt}} & = 0 - 1\\dy & = - dt\end{aligned}\)

Find the value of line integral.

Find the value of line integral\(\oint_C x ydx + {x^2}dy\).

\(\begin{aligned}\oint_C x ydx + {x^2}dy & = \oint_{{C_1} + {C_2} + {C_3} + {C_4}} x ydx + {x^2}dy\\ & = \oint_{{C_1}} x ydx + {x^2}dy + \oint_{{C_2}} x ydx + {x^2}dy + \oint_{{C_3}} x ydx + {x^2}dy + \oint_{{C_4}} x ydx + {x^2}dy\\ & = \left( {\begin{aligned}{*{20}{l}}{\int_0^3 t (0)(dt) + {t^2}(0)dt + \int_0^1 3 (t)(0)dt + {3^2}(dt) + }\\{\int_0^3 {(3 - t)} (1)( - dt) + {{(3 - t)}^2}(0)dt + \int_0^1 {(1 - t)} (0)(0)dt + {0^2}( - dt)}\end{aligned}} \right)\\ & = \left( {0 + 9(t)_0^1 - \left( {3t - \frac{{{t^2}}}{2}} \right)_0^3 + 0} \right)\end{aligned}\)

Simplify further as,

\(\begin{aligned}\oint_C x ydx + {x^2}dy & = \left( {9 - 9 + \frac{9}{2}} \right)\\ & = \frac{9}{2}\\ & = 4.5\end{aligned}\)

Thus, the value of line integral \(\oint_C x ydx + {x^2}dy\) in direct method is 4.5.

Find the value of the line integral by Green's Theorem

b)

From Figure 1, the curve is positively oriented, piecewise-smooth, and simply closed curve with domain\(D = \left\{ {\begin{array}{*{20}{l}}{0 \le x \le 3}\\{0 \le y \le 1}\end{array}} \right.\)and hence Green's theorem is applicable.

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

\(\begin{array}{l}P = xy\\Q = {x^2}\end{array}\)

Find the value of\(\frac{{\partial P}}{{\partial y}}\)

\(\begin{array}{c}\frac{{\partial P}}{{\partial y}} = \frac{\partial }{{\partial y}}(xy)\\ = x\end{array}\)

Find the value of\(\frac{{\partial Q}}{{\partial x}}\).

\(\begin{align} & \frac{\partial Q}{\partial x}=\frac{\partial }{\partial x}\left( {{x}^{2}} \right) \\ & =\frac{\partial }{\partial x}\left( {{x}^{2}} \right) \\ & =2x\quad \left\{ \because \frac{\partial }{\partial t}\left( {{t}^{2}} \right)=2t \right\} \end{align}\)

Substitute\(xy\)for\(P,{x^2}\)for\(Q,x\)for\(\frac{{\partial P}}{{\partial y}},2x\)for\(\frac{{\partial Q}}{{\partial x}},0\)for\({x_1},3\)for\({x_2},0\)for\({y_1}\), and 1 for\({y_2}\)

\(\begin{align} & \oint_{C}{x}ydx+{{x}^{2}}dy=\int_{0}^{3}{\int_{0}^{1}{(2x-x)}}dydx \\ & =\int _{0}^{3}[xy]dx\quad \left\{ \because \int tdt=\frac{{{t}^{2}}}{2} \right\} \\ & =\int _{0}^{3}[x]dx \\ & =\left[ \frac{{{x}^{2}}}{2} \right]_{0}^{3} \end{align}\)

Simplify the equation as,

\(\begin{array}{c}\oint_C x ydx + {x^2}dy = \left( {\frac{{{x^2}}}{2}} \right)_0^3\\ = \frac{{{3^2}}}{2}\\ = \frac{9}{2}\\ = 4.5\end{array}\)

Thus, the value of line integral using Green's Theorem is\(4.5\)\(\).