So i'm supposed to calculate the line integral
$$\int_C\mathbf{F}\cdot d\mathbf{l}$$
where $\mathbf{F}=(xy^2+2y)\vec{\mathbf{x}}+(x^2y+2x)\vec{\mathbf{y}}$
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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 $(\int_{c_1}+\int_{c_1})\mathbf{F}\cdot d\mathbf{l}=\frac{a^2b^2}{2}+2ab$
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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 $C_1$ $\int\mathbf{F}\cdot d \mathbf{l}=\frac{a^2b^2}{2}+2ab$
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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?
Best Answer
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.