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I am having trouble with the following problem. I would like to see how to set up the problem and if there is any other tips I should use to solve similar problems. Thank you.

Let $F(x,y) = (2x + 3y, 3x + 4y)$ and let C be the upper part of the circle of radius $\sqrt{5}$ about the point $(2,0)$ which runs from the point $(1,2)$ to $(3,2)$. Determine the line integral $\int_{\gamma} F \gamma dt $.

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Note that Green's theorem doesn't apply here because the path isn't closed.

First find an appropriate parameterization for the path $\gamma$. Recall that a circle centered at the point $(a,b)$ with radius $r$ can be described by the set of parametric equations, $$\begin{cases}x=r\cos t+a\\[1ex]y=r\sin t+b\end{cases}$$ where $0\le t\le2\pi$ for one complete revolution in the counterclockwise. (You can check this is valid by substituting into the equation $(x-a)^2+(y-b)^2=r^2$.) This means you can parameterize $\gamma$ by $$\begin{cases}x=\sqrt5 \cos t+2\\[1ex]y=\sqrt5\sin t\end{cases}$$ Because the path starts at $(1,2)$ and ends at $(3,2)$, you are traveling clockwise, so you need to modify the parameterization a bit. This is just a matter of reflecting the $x$ coordinate across the circle's axis of symmetry - this amounts to changing $\cos t$ to $-\cos t$, and hence $$\gamma:\begin{cases}x=-\sqrt5 \cos t+2\\[1ex]y=\sqrt5\sin t\end{cases}\implies\begin{cases}\mathrm{d}x=\sqrt5\sin t\,\mathrm{d}t\\[1ex]\mathrm{d}y=\sqrt5\cos t\,\mathrm{d}t\end{cases}$$ or equivalently by the vector-valued function $$\vec{r}(t)=\langle -\sqrt5\cos t+2,\sqrt5\sin t\rangle\implies\mathrm{d}\vec{r}=\langle \sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm{d}t$$

You're also not interested in a complete revolution, so you'll need to figure out an appropriate range for $t$, which amounts to solving the system of equations below: $$\begin{cases}-\sqrt5\cos t_1+2=1\\[1ex] \sqrt5\sin t_1=2\\[1ex] -\sqrt5\cos t_2+2=3\\[1ex] \sqrt5\sin t_2=2\end{cases}\implies\begin{cases}t_1=2\arctan\dfrac{2\sqrt5}{5+\sqrt5}\\[1ex]t_2=2\arctan\dfrac{2\sqrt5}{5-\sqrt5}\end{cases}$$ The line integral $\displaystyle\int_\gamma F(x,y)\bullet\mathrm{d}\vec{r}$ is then given by $$\int_{t_1}^{t_2}\langle-2\sqrt5\cos t+3\sqrt5\sin t+4,-3\sqrt5\cos t+4\sqrt5\sin t+6\rangle\bullet\langle \sqrt5\sin t,\sqrt5\cos t\rangle\,\mathrm{d}t$$ $$\int_{t_1}^{t_2}\bigg(6\sqrt5\cos t-15\cos2t+4\sqrt5\sin t+5\sin2t\bigg)\,\mathrm{d}t$$

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