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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 .

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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 $C$.

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