Verifying Stokes’ Thm. $\vec F = \left< -y,x,-2\right>$ where $S$ is the region defined by $z^2 = x^2 + y^2, \space z \in [0,4]$

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I want to verify Stokes’ Theorem by evaluating both the left and right side of the following eqn.

$$\int_C \vec F \cdot d\vec r = \int \int_S curl \vec F \cdot d\vec S$$

where $\vec F = \left< -y,x,-2\right>$ and $S$ is the region defined by $z^2 = x^2 + y^2, \space z \in [0,4]$.

When I use the parameterization $\vec r = \left< \cos \theta, \sin \theta, z\right>$

I get a normal vector $\vec n = \left< \cos\theta, \sin \theta,0\right>$ and since $curl \vec F = \left<0,0,2 \right>$

$$curlf \vec F \cdot \vec n = 0$$

so the right hand side equals zero.

Here is what I am not sure since it has been a while since I have evaluated line integrals.

I believe that $C$ is the projection of $S$ onto the $xy$-plane, so it will be a circle with radius $16$.

So, the parameterization I choose is $\vec r = \left< 4\cos \theta, 4\sin \theta, 0\right>$

and because $\vec F = \left<-4\sin\theta, 4\cos\theta,-2 \right>$

$$\vec F \cdot \vec r = 0$$

hence the left side is equal to zero.

Can anyone confirm that I did this correctly?

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No, both RHS and LHS have mistakes that have led to you getting integrals as zero.

A) For RHS, your parametrization is not correct. The correct parametrization is

$\vec r(\rho,\theta) = (\rho \cos\theta, \rho \sin\theta, \rho)$, as $z = \rho$ from the equation of the cone.

$r'_{\rho} \times r'_{\theta} = (\rho \cos\theta, \rho \sin \theta, - \rho)$

$0 \leq \rho \leq 4, 0 \leq \theta \leq 2\pi$.

Can you now complete the integral?

B) For LHS which is line integral, you should have the $z$ component in your parametrization though it would not matter here.

$\vec r(\theta) = (4\cos\theta, 4\sin\theta, 4)$

Now we are supposed to do $\vec F \cdot \vec {r} \ '$ and not $\vec F \cdot \vec r$. That was the mistake in your working of LHS.

$r'_{\theta} = (- 4 \sin\theta, 4 \cos\theta, 0)$

$0 \leq \theta \leq 2\pi$.

Can you complete the working now?

Last point - the outward normal vector for the given cone surface is pointing downward and that is why we have taken $z$ component negative in the first working. Now align the orientation of the boundary curve to the orientation of the surface in your line integral.