You are watching: Evaluate the line integral, where c is the given curve.
How do you solve this by using the following parametrics? I split them up but got a negative answer of -1/3. What"s wrong?
For $C_1$ got, $\langle t, t/2\rangle$, $0 \leq t \leq 2$.
For $C_2$ got, $\langle t, 3-y\rangle$, $2 \leq t \leq 3$.
Hints:
$$\;\;\;(0,0)\to(2,1)\,:\;\;\; 0\le x\le 2\;,\;\;y=\frac x2\implies$$
$$\int\limits_{(0,0)}^{(2,1)}(x+2y)dx+x^2dy=\int\limits_0^2 (x+x)dx+x^2\left(\frac12\,dx\right)=\int\limits_0^2\left(\frac12x^2+2x\right)dx=$$
$$=\frac16\cdot8+4=\frac{16}3$$
and something similar with the other one...
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Added:
$$(2,1)\to(3,0):\;\;\;2\le x\le 3\;,\;\;y=-x+3\implies$$
$$\int\limits_{(2,1)}^{(3,0}(x+2y)dx+x^2dy=\int\limits_2^3 (x+2(-x+3))dx+x^2\left((-1)\,dx\right)=\int\limits_2^3\left(-x^2-x+6\right)dx=$$
$$\left.-\frac13x^3\right|_2^3-\left.\left.\frac12x^2\right|_2^3+6x\right|_2^3=-\frac{19}3-\frac52+6=-\frac{17}6$$
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edited Apr 30 "13 at 21:05
answered Apr 30 "13 at 20:56
DonAntonioDonAntonio
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Try writing these integrals as$$\int_C (x+2y,x^2)\cdot (dx,dy)$$where $\cdot$ is the usual inner product.Now, for $C_1$, make $(x,y)=(x(t),y(t))=(t,\frac{t}{2})$ (if your parametrization is correct), from which we have $dx=dt$ and $dy=\frac{dt}{2}$. You can now easily integrate with respect to $t$, within the range where $t$ varies.
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answered Apr 30 "13 at 20:50
MarraMarra
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