Finding a Line Integral Using Green’s Theorem

Use Green’s Theorem to find the line integral \[ \begin{equation*} \oint_{C}[ ( e^{-x^{2}}+y^{2}) \,dx+( \ln y-x^{2}) \,dy] \end{equation*} \]

where \(C\) is the square illustrated in Figure 34.

Solution  Recall that \(\int e^{-x^{2}}dx\) cannot be expressed in terms of elementary functions, so finding the integral directly is impossible. Since both the curve \(C\) and the region \(R\) enclosed by \(C\) satisfy the conditions of Green’s Theorem, we use it to find the line integral. If \( P(x,y)=e^{-x^{2}}+y^{2}\) and \( Q(x,y)=\ln y-x^{2}\) then \(\dfrac{\partial P}{ \partial y}=2y\) and \(\dfrac{\partial Q}{\partial x}=-2x\). Since \(P\) and \(Q\) have continuous first-order derivatives in \(R\), by Green’s Theorem, We have \[ \begin{eqnarray*} &&\oint_{C}[(e^{-x^{2}}+y^{2})\,dx+(\ln y-x^{2})]\,dy =\iint\limits_{R}(-2x-2y)\,dx\,dy \\ &&\hspace{10.4pc}=-2\int_{1}^{2}\int_{0}^{1}(x+y) \,dx\,dy=-2\int_{1}^{2}\left[ \dfrac{x^{2}}{2}+xy\right] _{0}^{1}\,dy \\ &&\hspace{9.1pc}\underset{\color{#0066A7}{R\hbox{ is } y\hbox{-simple}}}{\color{#0066A7}{\uparrow}}\\ &&\hspace{10.4pc}=-2\int_{1}^{2}\left( \dfrac{1}{2}+y\right) dy=-2\left[ \dfrac{1}{2}y+ \dfrac{y^{2}}{2}\right] _{1}^{2}=-4 \end{eqnarray*} \]