Evaluate the flux of $\operatorname{curl}\mathbf F$ through the given surface.

Question

Consider the vector field $\textbf{F} = \langle yz, -xz, z^3 \rangle$ and the surface S given by the part of the cone $z = \sqrt{x^2 + y^2}$ that lies between the two planes $z = 1$ and $z = 3$, oriented with an upward facing normal vector.

Use a method of your choice to evaluate the flux of $\text{curl}(\textbf{F})$ through the given surface

My two issues are as follows: First, with finding the correct parameterization of the boundary curve in order to apply Stokes Thereom, and second I am not sure how the part of the problem saying "oriented with an upward pointing normal vector" factors in.

Answer 1

Since the divergence of curl is zero, the total flux of the curl through the closed frustum surface formed by the top cap, bottom cap and conical side will be zero $\Phi_T+\Phi_B+\Phi_C=0$, the flux through the sides of the cone will be offset by that thru the top and bottom of the cone ... so flux through sides will be negative that thru top plus bottom combined. The $\hat{z}$ component of the curl (the only one needed here) is $\frac{\partial(-xz)}{\partial x}-\frac{\partial(yz)}{\partial y}=-2z$.
Answer is -$52\pi$ by computing flux through top and bottom caps (where normal vectors are merely $+\hat{z}$ and $−\hat{z}$ respectively) and by using area of top face is $\pi3^2$ and that of bottom is $\pi1^2$, and the magnitude of dot product of close surface normal vector and curl is -6 for top surface and 2 for bottom surface, also taking into account that the normal vector as stated in the problem points opposite that of outward normal vector for the entire closed capped surface.

Answer 2

$S$ has two boundaries, the circles $x^2+y^2=1$ and $x^2+y^2=9$, in the planes $z=1$ and $z=3$, respectively. So if you choose to use Stokes' theorem, you need to compute two integrals.

We want the normal vector to $S$ to point upward, which is to say the normal is pointing toward the interior of the cone. Then the "outer" boundary (the circle in the plane $z=9$) should be parameterized with standard, counter-clockwise orientation, while the "inner" boundary (the circle in the plane $z=1$) needs to oriented in the opposite direction.

Call these circles $C_1$ and $C_2$ respectively, and parameterize them by

$$\vec r_1(t)=\langle3\cos t,3\sin t,3\rangle$$

$$\vec r_2(t)=\langle\cos t,-\sin t,1\rangle$$

both for $t\in[0,2\pi]$.

Then by Stokes' theorem,

$$\begin{align*} \iint_S\operatorname{curl}\vec F\cdot\mathrm d\vec S&=\sum_{i=1}^2\int_{C_i}\vec F\cdot\mathrm d\vec r_i\\[1ex] &=\int_0^{2\pi}\langle9\sin t,-9\cos t,27\rangle\cdot\langle-3\sin t,3\cos t,0\rangle\,\mathrm dt\\ &\quad\quad{}+\int_0^{2\pi}\langle-\sin t,-\cos t,1\rangle\cdot\langle-\sin t,-\cos t,0\rangle\,\mathrm dt\\[1ex] &=-26\int_0^{2\pi}\mathrm dt=\boxed{-52\pi} \end{align*}$$