# How to calculate Line Integral for given Closed Curve.

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 .

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 $$C$$.