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$\newcommand{\F}{\mathbf{F}}$$\newcommand{\R}{\mathbb{R}}$Consider the vector field $\F(x,y)=\big(1-x^2y, xy^2+\exp(y^2)\cdot\cos(y)\big)$ for $(x,y)\in\R^2$ and the curves $$C_1=\{(x,y)\in\R^2\colon x\in[-1,1], y=0\}\text{ and }C_2=\{(x,y)\in\R^2\colon x^2+y^2=1, y=0\}.$$ Let $R$ be the region in $\R^2$ that is enclosed by $C_1$ and $C_2$. Let $C$ be the union of the curves $C_1$ and $C_2$ with a counter-clockwise orientation.

How can I determine the line integral $\oint_{\partial R}\F\bullet\mathrm d\mathbf r$, where $\partial R$ is the boundary of $R$ (which consists of $C_1\cup C_2$). I feel like Green's theorem is the easiest way, since the orientation is positive. I have tried a direct computation with a parametrisation, but this makes the line integral quite difficult, due to the $\exp(y^2)\cos(y)$ in $\F$.

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As I understand it, you are trying to do a line integral over the curve formed by the top half of the unit circle and [-1,1] on the x-axis? By Green's Theorem we get $$\int_R (1-x^2y)\mathrm{d}x + (xy^2+\exp(y^2)⋅\cos(y))\mathrm{d}y] = \int\int_D (x^2 + y^2)\mathrm{d}x\mathrm{d}y$$ by taking the partials of $(1-x^2y)$ with respect to $y$ and $(xy^2+\exp(y^2)⋅\cos(y))$ with respect to $x$ to get one expression $\mathrm{d}y\mathrm{d}x$ and one $\mathrm{d}x\mathrm{d}y$, respectively. $\mathrm{d}y\mathrm{d}x = -\mathrm{d}x\mathrm{d}y$ so we get $(x^2 + y^2)\mathrm{d}x\mathrm{d}y$. The rest of it should be simple using polar coordinates. I hope that was helpful!

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Thank you for your help. I've just found out I made a mistake: there was an calculation error in one of my partial derivative. – gsc May 23 '12 at 21:32

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