Problem statement:

Calculate $\int_{\gamma }(3e^{(y-3x)^{2}}-y)\mathrm{dx}+(-e^{(y-3x)^{2}}+2x)\mathrm{dy}$ where $\gamma$ is the curve $y=x^2$ from $(0,0)$ to $(3,9)$.


First idea was to parametrize the curve: $f(t,t^2), t\in [0,3]$ but using this approach I ended up with an integral whose primitive could not be expressed in elementary functions.

Secondly I noticed that "$\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 3$" so Green's theorem could be suitable. So I enclosed a region using a vertical and a horizontal line segment. However, these couldn't be solved in elementary functions.

What are some other ways?


Let $\Omega$ be the domain which is enclosed by $y=3x$ and $y=x^2$, and $\Gamma$ be the line $y=3x$, by the Green's Formula, we have \begin{eqnarray*} \int_{\partial\Omega}Pdx+Qdy&=&\int_{\Omega}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)\ dxdy=3|\Omega|\\ &=&3\int_0^3 (3x-x^2)\ dx=\frac{27}{2}. \end{eqnarray*}

On the other hand, we have \begin{eqnarray*} \int_{\partial\Omega}Pdx+Qdy&=&\int_{\gamma}Pdx+Qdy-\left(\int_\Gamma Pdx+Qdy\right). \end{eqnarray*}

Notice that \begin{eqnarray*} \int_\Gamma Pdx+Qdy&=&\int_0^3\left[(3-3x)+(-1+2x)\right]\ dx=\int_0^3[2-x]\ dx=\frac{3}{2}. \end{eqnarray*}

Hence we get \begin{eqnarray*} \int_{\gamma}Pdx+Qdy=15. \end{eqnarray*}

  • $\begingroup$ Looks ok to me..but not correct answer according to key. $\endgroup$ – EricAm Feb 26 '14 at 18:44
  • $\begingroup$ Last step is wrong but otherwise it is solid. Thanks. $\endgroup$ – EricAm Feb 26 '14 at 19:48
  • $\begingroup$ Yes, it should be 15. Thanks. $\endgroup$ – m-agag2016 Feb 27 '14 at 15:04

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