# Evaluation of a few line integrals

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.

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

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