So i'm supposed to calculate the line integral

$$\int_C\mathbf{F}\cdot d\mathbf{l}$$

where $\mathbf{F}=(xy^2+2y)\vec{\mathbf{x}}+(x^2y+2x)\vec{\mathbf{y}}$

  1. Through curve $C_1$ which contains two straight lines that crosses the points $(0,0),(a,0)$ and $(a,b)$ i found out this to be $(\int_{c_1}+\int_{c_1})\mathbf{F}\cdot d\mathbf{l}=\frac{a^2b^2}{2}+2ab$

  2. Through the curve $C_2$ which contains a single straight line that connects the points $(0,0)$ and $(a,b)$ and i found out i'll get the same answer going as through $C_1$ $\int\mathbf{F}\cdot d \mathbf{l}=\frac{a^2b^2}{2}+2ab$

  3. Verify using Stoke's Theorem and explain why you gotthe same answer in 1. and 2. using Stoke's theorem (Aka use stokes theorem to explain why $C_1$ and $C_2$ yielded the same answer).

I don't know how to use Stoke's theorem in a 2d vector field. What would the upper and limits be? How should I approach this problem using Stoke's? Since we're on the $xy$ plane then $\mathbf{n}=\mathbf{k}$ right?

  • $\begingroup$ $z$ would be $0$? if i'm not mistaken? $\endgroup$ – Sami Shafi Sep 18 '18 at 21:43
  • $\begingroup$ also i know that the curl in 2d is given simply $\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}=4$ $\endgroup$ – Sami Shafi Sep 18 '18 at 22:07
  • $\begingroup$ What version of Stokes’ theorem are you meant to use? I might be more concerned about the fact that neither curve is closed than by what the equivalent of curl is in 2-D. $\endgroup$ – amd Sep 18 '18 at 23:29
  • 1
    $\begingroup$ Clue: The fact that the two integrals are the same is an indication that you have a conservative vector field. This means that there is a potential function $V$ such that $\bf{F}=\frac{\partial V}{\partial x}\bf{x}+\frac{\partial V}{\partial y}\bf{y}$. In such a case the value of the integral is going to be $V(end)-V(start)$ and so on. $\endgroup$ – Jap88 Sep 19 '18 at 2:03

Note that $$ Q'_x=P'_y = 2xy + 2, $$ and $F$ is well defined on $\mathbb{R}^2$, hence $$ \oint_{\partial D}Fdr=\int\int_D(Q'_x-P'_y)dxdy=0, $$ hence $F$ is conservative field. As such, $$ V(x,y) = \int_x P(x,y)dx = \frac{x^2 y^ 2}{ 2 } + 2yx + g(y), $$ thus $$ V'_y=x^2y+2x+g'_y(y)= x^ 2y+2x, $$ hence $g(y) = 0$, thus $$ V(x,y) = \frac{x^2 y^ 2}{ 2 } + 2yx + C. $$ Therefore, path integral from $(0,0)$ to $(a,b)$ is $$ V(a,b) - V(0,0)= \frac{a^2b^2}{2} + 2ab, $$ for any path.


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