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Let $F(x,y)=(x,y)$. Let $C$ be the portion of the ellipse $(x^2/a^2)+(y^2/b^2)=1$ in quadrants 1 and 2. Show how to evaluate $\int_C F \cdot dr.$

this is the question i am given. i am not sure if what i am doing is right but this is what i have.

$$x=a \cos t \ \ \ \implies \ \ \ dx/dt=-a \sin t $$

$$y=b \sin t \ \ \ \implies \ \ \ dy/dt = b \cos t$$

$$0<t<\pi$$

ds=srt(a^2sin^2t+b^2cos^2t)dt

(integral along the curve)xdx+(integral along the curve)ydy

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    $\begingroup$ F(x,y) is a conservative force. $\endgroup$ – Doug M Dec 8 '16 at 17:53
  • $\begingroup$ what do you mean by srt() ? $\endgroup$ – Jean Marie Dec 8 '16 at 18:25
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$x = a\cos t\\y= b\sin t\\ dr = (\frac {dx}{dt},\frac {dy}{dt})dt$

That part is fine.

Now you substitute into $F(x,y)\to F(t) = (a\cos t, b\sin t)$

$\int_c F(t)\cdot dr = \int (a\cos t, b\sin t)\cdot (-a\sin t, b\cos t) dt = \int (b^2-a^2) \cos t\sin t\;dt$

Now $F(x,y) = \nabla (\frac 12 x^2 + \frac 12 y^2)$ Which means that F(x,y) is a conservative force and $\int_c F(x,y)\cdot dr$ depends only on the endpoints and not on the path taken. And $\oint F(x,y)\cdot dr = 0$ for any closed contour

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