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 $2\sin{t }$]
- 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.
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 .$