Question

Use Green's Theorem to evaluate \(\int_C {\bf{F}} \cdot d{\bf{r}}\). (Check the orientation of the curve before applying the theorem.)

\({\bf{F}}(x,y) = \left\langle {\sqrt {{x^2} + 1} ,{{\tan }^{ - 1}}x} \right\rangle ,\)C is the triangle from \((0,0)\) to \((1,1)\)to\((0,1)\)to\((0,0)\)

Short answer

The line integral of vector field \({\bf{F}}(x,y) = \langle y\cos x - xy\sin x,xy + x\cos x\rangle \) using Green's Theorem is \(\frac{\pi }{4} - \frac{1}{2}\ln 2\).

Step-by-step solution

Given information from question 

The line integral of vector field \({\bf{F}}(x,y) = \left\langle {\sqrt {{x^2} + 1} ,{{\tan }^{ - 1}}x} \right\rangle ,\)and curve C is a triangle from \((0,0)\) to \((1,1)\)to\((0,1)\)to\((0,0)\).

Green’s Theorem

Green’s Theorem states that

Consider a positively oriented curve\(C\)which is piece-wise smooth, simple closed curve in plane with domain\(D\). Then the line integration of vector field\({\bf{F}}(x,y) = \langle P(x,y),Q(x,y)\rangle \)over curve\(C\)is,

……. (1)

Here,

\(\frac{{\partial P}}{{\partial y}}\)is continuous first-order partial derivative of\(P\),

\(\frac{{\partial Q}}{{\partial x}}\)is continuous first-order partial derivative of\(Q\), and

\(P\)and\(Q\)have continuous partial derivatives.

Compare the two vector fields \({\bf{F}}(x,y) = \langle P(x,y),Q(x,y)\rangle \) and \({\bf{F}}(x,y) = \left\langle {\sqrt {{x^2} + 1} ,{{\tan }^{ - 1}}x} \right\rangle .\)

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

Compare the two vector fields

\({\bf{F}}(x,y) = \langle P(x,y),Q(x,y)\rangle \)and\({\bf{F}}(x,y) = \left\langle {\sqrt {{x^2} + 1} ,{{\tan }^{ - 1}}x} \right\rangle .\)

\(\begin{array}{l}P = \sqrt {{x^2} + 1} \\Q = {\tan ^{ - 1}}x\end{array}\)

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

\(\begin{align} \frac{\partial P}{\partial y} & =\frac{\partial }{\partial y}\left( \sqrt{{{x}^{2}}+1} \right) \\ & =0\left\{ \because \frac{\partial }{\partial t}(k)=0 \right\} \end{align}\)

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

\(\begin{align} \frac{\partial Q}{\partial x} & =\frac{\partial }{\partial x}\left( {{\tan }^{-1}}x \right) \\ & =\frac{1}{1+{{x}^{2}}}\left\{ \because \frac{\partial }{\partial t}\left( {{\tan }^{-1}}t \right)=\frac{1}{1+{{t}^{2}}} \right\} \end{align}\)

Calculate the line integral of vector field 

Modify the equation (1)

\(\int_C {\bf{F}} \cdot d{\bf{r}} = \int_{{x_1}}^{{x_2}} {\int_{{y_1}}^{{y_2}} {\left( {\frac{{\partial Q}}{{\partial x}} - \frac{{\partial P}}{{\partial y}}} \right)} } dydx\)

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

\(\begin{aligned}\int_C {\bf{F}} \cdot d{\bf{r}} & = \int_0^1 {\int_x^1 {\left( {\frac{1}{{1 + {x^2}}} - 0} \right)} } dydx\\ & = \int_0^1 {\int_x^1 {\left( {\frac{1}{{1 + {x^2}}}} \right)} } dydx\\ & = \int_0^1 {\left( {\frac{1}{{1 + {x^2}}}} \right)} (y)_x^1dx\\ & = \int_0^1 {\left( {\frac{1}{{1 + {x^2}}}} \right)} (1 - x)dx\end{aligned}\)

Expand the equation

\(\begin{align} \int_{C}{\mathbf{F}}\cdot d\mathbf{r} & =\int_{0}^{1}{\left( \frac{1}{1+{{x}^{2}}}-\frac{x}{1+{{x}^{2}}} \right)}dx \\ & =\left[ {{\tan }^{-1}}x-\frac{1}{2}\ln \left( 1+{{x}^{2}} \right) \right]_{0}^{1}\left\{ \because \int{\frac{\frac{1}{1+{{t}^{2}}}dt={{\tan }^{-1}}t,}{\int{\frac{t}{1+{{t}^{2}}}}dt=\frac{1}{2}\ln \left( 1+{{t}^{2}} \right)}} \right\} \\ & =\left( {{\tan }^{-1}}(1)-\frac{1}{2}\ln \left( 1+{{1}^{2}} \right) \right)-\left( {{\tan }^{-1}}(0)-\frac{1}{2}\ln \left( 1+{{0}^{2}} \right) \right) \end{align}\)

Solve the equation further

\(\begin{array}{l}\int_C {\bf{F}} \cdot d{\bf{r}} = \frac{\pi }{4} - \frac{1}{2}\ln 2 - 0 + 0\\\int_C {\bf{F}} \cdot d{\bf{r}} = \frac{\pi }{4} - \frac{1}{2}\ln 2\end{array}\)

Thus, the line integral of vector field \({\bf{F}}(x,y) = \langle y\cos x - xy\sin x,xy + x\cos x\rangle \) using Green's Theorem is \(\frac{\pi }{4} - \frac{1}{2}\ln 2\).