You are watching: Evaluate the line integral, where c is the given curve.
How do you fix this by using the following parametrics? I break-up them up yet got a an adverse answer of -1/3. What"s wrong?
For $C_1$ got, $langle t, t/2 angle$, $0 leq t leq 2$.
For $C_2$ got, $langle t, 3-y angle$, $2 leq t leq 3$.
$$;;;(0,0) o(2,1),:;;; 0le xle 2;,;;y=frac x2implies$$
$$intlimits_(0,0)^(2,1)(x+2y)dx+x^2dy=intlimits_0^2 (x+x)dx+x^2left(frac12,dx ight)=intlimits_0^2left(frac12x^2+2x ight)dx=$$
and something similar with the other one...
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$$(2,1) o(3,0):;;;2le xle 3;,;;y=-x+3implies$$
$$intlimits_(2,1)^(3,0(x+2y)dx+x^2dy=intlimits_2^3 (x+2(-x+3))dx+x^2left((-1),dx ight)=intlimits_2^3left(-x^2-x+6 ight)dx=$$
$$left.-frac13x^3 ight|_2^3-left.left.frac12x^2 ight|_2^3+6x ight|_2^3=-frac193-frac52+6=-frac176$$
edited Apr 30 "13 in ~ 21:05
answered Apr 30 "13 at 20:56
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Try writing these integrals as$$int_C (x+2y,x^2)cdot (dx,dy)$$where $cdot$ is the normal inner product.Now, for $C_1$, do $(x,y)=(x(t),y(t))=(t,fract2)$ (if her parametrization is correct), from which we have actually $dx=dt$ and also $dy=fracdt2$. You can now easily integrate with respect to $t$, in ~ the variety where $t$ varies.
answered Apr 30 "13 in ~ 20:50
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