Let $\vec{F}=ay\hat{i}+z\hat{j}+x\hat{k}$ and $C$ be the positively oriented closed curve given by $x^2+y^2=1,z=0$. If $\int{\vec{F}\vec{dr}}=\pi$, then the value of $a$ is : A) $-1$ B) $0$ C) $\frac12$ D) $1$
Firstly, I have checked that the force is not conservative by finding the partial derivatives and seeing they are not equal. So the integral is not independent of the path and I cannot parameterize it. Then, I proceeded to use Stoke's Theorem.
$Curl \vec{F}=-\hat{i}-\hat{j}-a\hat{k}$
I picked the base region as given by $x^2+y^2=1$ and so my perpendicular is $\hat{n}=\hat{k}$.
Then, $\int{\vec{F}\vec{dr}}= \int \int Curl \vec{F} \cdot \hat{n} dA= -a\pi$
By given conditions then, $-a\pi=\pi$, which gives me $a=-1$. However, the answer is given as $\frac12$. What am I doing wrong? Please help me out!
I would have preferred to do the line integral directly instead of using the Stokes' Theorem. So, the curve C is given by the circle $x^2+y^2=1$. I change it to polar coordinates$(r,\theta)$ such that $x=r\cos \theta, y=r \sin \theta.$ So we also have that $dx=-r\sin \theta d\theta$ and $dy=r\cos \theta d\theta$ since $r=1$ is constant for a circle and $dz=0$ as $z=0$ is the surface on which the circle lies.
Hence the integral becomes $$\oint \vec F\cdot d\vec r$$ $$=\oint (ay\hat{i}+z\hat{j}+x\hat{k})\cdot(dx\hat{i}+dy\hat{j}+dz\hat{k})$$ $$=\oint (aydx+zdy+xdz)$$ $$=\oint (ar \sin \theta)\cdot (-r\sin \theta d\theta)+0\cdot dy+x\cdot 0)$$ $$=-ar^2\oint (\sin^2 \theta d\theta)$$ $$=-a(1)^2\frac{1}{2}\int^{2\pi}_{0} (1-\cos 2\theta) d\theta$$ $$=-a\cdot\frac{1}{2}[2\pi-\frac{1}{2}\cdot (0-0)] =-\pi a$$
So your answer is right. The value of $a$ is indeed $-1$. I think the answer to this question is given wrong in the solution. Because my procedure is logical and both our answers match.