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# Evaluate the line integral, where C is the given curve. integral C x^2dx+y^2dy, C consists of the arc of the circle x^2+y^2=4 from (2, 0) to (0, 2) followed by the line segment from (0, 2) to (4, 3)

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We want to find the line integral

$\int_C x^2dx +y^2dy,$

Note that $C = C_1+C_2$, where $C_1$ is the arc of the circle joining $(2,0)$ to $(0,2)$ and $C_2$ is the line segment joining $(0,2)$ to $(4,3)$. Thus,

$\int_C x^2~dx+y^2~dy = \int_{C_1+C_2}x^2dx+y^2dy.$

• $\text{\textcolor{#4257b2}{On $C_1:$}}$

\begin{align*} & x = 2\cos t \implies dx = -2\sin t~dt\\ & y = 2\sin t \implies dy = 2\cos t~dt\\ & 0\le t\le \dfrac{\pi}{2}. \end{align*}

• $\text{\textcolor{#4257b2}{On $C_2:$}}$ The equation of the line can be parametrized as

\begin{align*} C_2 & : \left\{(0,2)(1-t) + (4,3)t:0\le t\le 1\right\} \\ & = \left\{(4t, 2-2t+3t): 0\le t\le 1 \right\}\\ & = \left\{(4t,2+t): 0\le t\le 1 \right\}. \end{align*}

Thus,


\begin{align*} & x = 4t \implies dx = 4dt\\ & y = 2+t \implies dy = dt\\ \end{align*}

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