So...I have found a problem where I have to solve the integral of the vector field:

$F(x,y)=(\sin(x)ln(x)+y^2) a_x + (\cos(y)e^y-x^2) a_y$

Along the borders of the region bounded by the circumferences:

C1 with R=1 centered on $(1,0)$

C2 with R=2.centered on $(2,0)$

Which look like this (a half moon):


The vector field is not conservative, but it shouldn't be such a big deal since I could either plan the integral or just apply Green's theorem. The issue comes from having two circumferences there since so far I have only made line integrals over a single curve, so I suspect I should apply Green's theorem, though I am not sure how to show (if) that the region is enclosed by a smooth, closed, positively oriented curve. Or maybe there are other means to solve it?

Any help is welcome


In fact it is still a simple curve. We can start from the origin and follow C1 back to the origin. Then from the origin start following C2 to complete the closed curve. Hence use the direct integration or Green's theorem.

For direct integration use 2 integrals, add up the one for C1 and one for C2.

C1 : $x - 1 = cos t$ and $y = sin t$

C2 : $x - 2 = 2cos t$ and $y = 2sin t$ $0 \le t \le 2\pi$

However the integrals are very difficult to solve. Hence it is easier to use Green's theorem

$$I = \int_{C1+C2} (sin(x)ln(x) + y^2)dx + (e^{y}cos(y) - x^2)dy$$

$$P = sin(x)ln(x) + y^2 \implies \frac{\partial P}{\partial y} = 2y$$ $$Q = e^y cos(y) - x^2 \implies \frac{\partial Q}{\partial x} = -2x$$

$$I = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)dxdy$$

$$= \int_{2}^{4} \int_{-\sqrt{4 - (x - 2)^2}}^{\sqrt{4 - (x - 2)^2}}(-2x - 2y)dydx + \int_{0}^{2} \int_{\sqrt{1 - (x - 1)^2}}^{\sqrt{4 - (x - 2)^2}} (-2x - 2y)dydx + $$ $$\int_{0}^{2} \int_{-\sqrt{4 - (x - 2)^2}}^{-\sqrt{1 - (x - 1)^2}} (-2x -2y) dydx$$

$$= 8\left(\pi - \frac{4}{3}\right)$$

Note that the sum of the last two terms is 0.


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