1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $z$ 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
0
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.