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Evaluate the integral $$ \int_c (x+xy+y) \, ds $$ where $C$ is the path of the arc along the circle given by $x^2+y^2=4$ starting at the point $(2,0)$ going counterclockwise making an inscribed angle of $\frac76\pi$

I started this problem by letting $x=2\cos t$ and $y=2\sin t$. I know that because it is counterclockwise it is positive. Then I started by taking the integral $$ \int_0^{\frac76\pi} (2\cos t +4\cos t\sin t + 2\sin t)\,dt $$

I solve the integral and got $-2.5+\sqrt3$ but this answer is incorrect. I think the values I have for $x$ and $y$ is correct but am I supposed to take the intergral from $0$ to $2$? Can someone help me?

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The general way of evaluating a path integral along a curve $C$ with parametrization $r: \mathbb [a,b] \to \mathbb R^2$ for a scalar field $f: \mathbb R^2 \to \mathbb R$ is: $$\int_C f ds = \int_a^b f(r(t)) \|r'(t)\| dt$$ and it seems you have forgotten the $\|r'(t)\|$ part. You have: $$r(t) = (2 \cos t, 2 \sin t)$$ so $$r'(t) = (-2 \sin t, 2 \cos t)$$ so $$\|r'(t)\| = \sqrt{4 sin^2 t + 4 \cos^2 t} = 2.$$ Your integral becomes $$\int_0^{\frac76\pi} 2(2\cos t +4\cos t\sin t + 2\sin t)\,dt $$ so your answer should only be of by a factor of 2. We can check: $$\begin{align} 2 \int_0^{\frac76\pi}& 2(2\cos t +4\cos t\sin t + 2\sin t)\,dt = 2\left[ 2\sin t - 2 \cos^2 t - 2 \cos t \right]_0^{\frac76\pi}\\ &=2 \left( -2\frac{1}{2} - 2\frac{3}{4} + 2\frac{\sqrt{3}}{2} - (0 - 2 - 2) \right ) = 2(-1 - 3/2 + \sqrt{3} + 4) = \\ &= 2(3/2 + \sqrt{3}) = 3 + 2\sqrt{3} \end{align}$$ so it seems you like you have some sign error in your calculations as well.

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