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Evaluate the integral $\int_c(3x-y)ds$,Where $c$ is the line segment from $(1,2)$ to $(3,3)$ followed by the portion of circle $x^2+y^2=18$ from $(3,3)$ to $(3,-3)$ in clockwise sense.

My try:For line segment $x(t)=1+2t$,$y(t)=2+t$ ant $0\leq t\leq 1$ and finally got $\int_0^1(5t+1)\sqrt5dt$ and for second portion $x(t)=3\sqrt2\cos t$,$y(t)=3\sqrt2\sin t$ then?

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For part two I would use a parametrisation where $t$ starts at $0$ and goes to $\pi/2$, because we're travelling along a quarter of the edge of the circle. Then the parametrisation is given by $x(t) = 3\sin(t+\pi/4), y(t) = 3\cos(t+\pi/4)$. Hence $x'(t) = 3\cos(t+\pi/4)$ and $y'(t) = -3\sin(t+\pi/4)$. This segment of the curve can now be calculated by: \begin{align*} &\int_0^{\frac{\pi}{2}}\big(3x(t)-y(t)\big)\sqrt{x'(t)^2+y'(t)^2}\ dt\\ =& \int_0^{\frac{\pi}{2}}\big(9\sin(t+\pi/4)-3\cos(t+\pi/4)\big)\sqrt{(3\cos(t+\pi/4))^2+(-3\sin(t+\pi/4))^2}\ dt\\ =& \int_0^{\frac{\pi}{2}}\big(9\sin(t+\pi/4)-3\cos(t+\pi/4)\big)\sqrt{9\cos^2(t+\pi/4)+9\sin^2(t+\pi/4)}\ dt\\ =& \int_0^{\frac{\pi}{2}}\big(9\sin(t+\pi/4)-3\cos(t+\pi/4)\big)3\ dt\\ =&\ 27\int_0^{\frac{\pi}{2}}\sin(t+\pi/4)\ dt-9\int_0^{\frac{\pi}{2}}\cos(t+\pi/4)\ dt\\ =&\ 27\big[-\cos(t+\pi/4)\big]_{0}^{\pi/2}-9\big[\sin(t+\pi/4)\big]_{0}^{\pi/2}\\ =&\ 29\sqrt{2} \end{align*} You can do the rest, combining it with the other integral and so on. Good luck!

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