Let S be the surface of the cone $z =\sqrt{x^2 +y^2}$ bounded by the planes $z =0$ and $z =3$ and Let C be the closed curve forming the boundary of of Surface S.
A vector field $\vec{F}$ such that
$\nabla \times F = -x\hat{i} -y\hat{j}$ Find the absolute value of the line integral
$\displaystyle \int \vec{F}.dr$
Now, I have a problem with this question Firstly Using the Vector Identity we should have
$\nabla. (\nabla \times F) = 0$, but here $\nabla. (\nabla \times F) = -2$
Suppose I ignore this for a moment and try to solve the problem
Then by Stokes theorem $\displaystyle \int \vec{F}.dr = \displaystyle \int \nabla \times F \hat{n} ds$
and Since in Stoke's Theorem I can take any curve that forms outer boundary for given surface, Suppose I take the boundary curve $x^2 + y^2 =9$ then here,
Normal vector = $\hat{k}$, so clearly $ (\nabla \times F). \hat{n} = 0 $ and hence
the Line integral should be zero.
However, the answer is given to me as $18\pi$.
Can anyone please resolve my doubts and tell me the correct way to solve this question ?
Thank you .
Best Answer
Your doubts are more than justified. They've given you a problem with a contrived hypothesis that cannot exist. But you do seem to have a misunderstanding. You choose a surface whose boundary curve is $C$. One such example would be the disk $z=3$, $x^2+y^2\le 9$. The normal $\vec n$ to that surface is indeed $\vec k$, so the flux of the curl will be $0$.
But what if you choose the original cone $S$ (and don't worry about the sharp point at $z=0$)? Then you can compute and get the $18\pi$.
How can you get two different answers? Easily, because they gave you a curl field that cannot possible be a curl. It's $\text{div}(\text{curl}\vec F) = 0$ that says this flux will not depend on the surface you pick with boundary curve $C$.