# How to compute this line integral in $\mathbb R^2$?

$\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\}.$$Let R be the region in \R^2 that is enclosed by C_1 and C_2. Let C be the union of the curves C_1 and C_2 with a counter-clockwise orientation. How can I determine the line integral \oint_{\partial R}\F\bullet\mathrm d\mathbf r, where \partial R is the boundary of R (which consists of C_1\cup C_2). I feel like Green's theorem is the easiest way, since the orientation is positive. I have tried a direct computation with a parametrisation, but this makes the line integral quite difficult, due to the \exp(y^2)\cos(y) in \F. - ## 1 Answer As I understand it, you are trying to do a line integral over the curve formed by the top half of the unit circle and [-1,1] on the x-axis? By Green's Theorem we get$$\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