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
$\displaystyle\int F.dr$
So, this question was on my test, I searched the site and noticed that this question is almost identical of this How to calculate Line Integral for given Closed Curve.
Only difference being the vector fields, here $F =x\hat{i} – y\hat{j}$ there
$F = -x\hat{i} – y\hat{j}$
Now, I want to solve this via stoke's theorem
The circle $z = 3$ is an outward boundary to the surface so, normal Vector
$\hat{n} = \hat{k}$
Now, Line Integral =
$\nabla \times F.\hat{n} ds$
$= 0$
But, answer given to me is $18\pi$.
Can anyone please check my solution and tell me whether is it correct or Not ?
Thank you.
Best Answer
Yes it's correct, to verify this we can use try to use a different surface such as the cone $z=\sqrt{x^2+y^2}$ , by defining $$g\left(x,y,z\right)=z-\sqrt{x^2+y^2}=0\:$$ our surface integral becomes $$\int _s\:\:(F.\nabla g)\:dA$$ over the surface of the cone, which by using cylinderical coordinates becomes $$\int _0^{2\pi }\int _0^3\:r^2\left(sin^2\left(\theta \right)-cos^2\left(\theta \right)\right)\:dr\:d\theta $$ which evaluates to zero.