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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

Cx2dx+y2dy,\int_C x^2dx +y^2dy,

Note that C=C1+C2C = C_1+C_2, where C1C_1 is the arc of the circle joining (2,0)(2,0) to (0,2)(0,2) and C2C_2 is the line segment joining (0,2)(0,2) to (4,3)(4,3). Thus,

Cx2 dx+y2 dy=C1+C2x2dx+y2dy.\int_C x^2~dx+y^2~dy = \int_{C_1+C_2}x^2dx+y^2dy.

  • $\text{\textcolor{#4257b2}{On C1:C_1:}}$

x=2cost    dx=2sint dty=2sint    dy=2cost dt0tπ2.\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*}

  • $\text{\textcolor{#4257b2}{On C2:C_2:}}$ The equation of the line can be parametrized as

C2:{(0,2)(1t)+(4,3)t:0t1}={(4t,22t+3t):0t1}={(4t,2+t):0t1}.\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,

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

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