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1. Evaluate $\displaystyle\oint_C\vec{F}.d\vec{r}$, where $\vec{F}=(x^2-3y^2)\hat{i}+(y^2-2x^2)\hat{j}$ and the closed curve $C$ is given by $x=3\cos{t}, y=2\sin{t}$, where $0 \leq t<2 \pi$ in the $xy$ plane.

Applying the usual procedure, i.e. changing $\vec{F}.d\vec{r}$ to $(9\cos^2{t}-12\sin^2{t})(-3\sin{t})dt+(4\sin^2{t}-18\cos^2{t})(2\sin{t})dt$ and putting the limit $0$ to $2\pi$, I get zero. Although the answer to the problem is given to be $\frac{5}{3}.$ Is there anything wrong in my process?

[*Typo $2\cos {t}$ instead of $2\sin{t }$]

  1. Evaluate $\displaystyle\oint_C\vec{F}.d\vec{r}$, where $\vec{F}=(2x-y+4z)\hat{i}+(x+y-z^2)\hat{j}+(3x-2y+4z^3)\hat{k} $, and $C$ is the curve given be $x^2+y^2=9$, $z=0$.

To parameterise the curve, we put $x=3\cos{t}, y=3\sin{t}, z=0$. Proceeding just as above, and setting the lower and the upper limit $0$ and $2\pi$ respectively (counter-clockwise), my answer turns out to be $18 \pi$, whereas the answer is $-18 \pi$ (Maybe they assumed clockwise rotation?!). Have I committed any mistake?

It would be of great help if someone checks out my procedure/ post their own answer. Thank you.

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    $\begingroup$ They should have specified the orientation of the curve. In your calculation, you assumed counterclockwise. Clearly, they assumed the opposite. $\endgroup$ – GReyes Feb 21 at 8:47
  • $\begingroup$ Regarding 1., the integral evaluates to zero. Both your version and the one corrected by @Fred. $\endgroup$ – PierreCarre Feb 21 at 9:07
  • $\begingroup$ That was a typo. Thanks for the help :) $\endgroup$ – Subhasis Biswas Feb 21 at 9:20
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You have

$\vec{F}.d\vec{r}=(9\cos^2{t}-12\sin^2{t})(-3\sin{t})dt+(4\sin^2{t}-18\cos^2{t})(2\sin{t})dt$,

but this is wrong. Correct is

$\vec{F}.d\vec{r}=(9\cos^2{t}-12\sin^2{t})(-3\sin{t})dt+(4\sin^2{t}-18\cos^2{t})(2\cos{t})dt$,

since $ \frac{d}{dt}y(t)=2 \cos t .$

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  • $\begingroup$ Ohhhoo.. That's a typo. Edited question now. $\endgroup$ – Subhasis Biswas Feb 21 at 9:17

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