Multivariable divergence theorem work

Question

5. Find the flux of the vector field $\vec{G}=\operatorname{curl} \vec{F}$, where $$ \vec{F}(x, y, z)=y^{3} \vec{i}+x^{3} \vec{j}+z^{3} \vec{k} $$ through the upward oriented part of paraboloid $$ z=1-x^{2}-y^{2} $$ lying over the plane $z=0$. What i evaluated is

Is that the right way to evaluate

Answer 1

Gauss's Law aka the Divergence Theorem states that the volume integral of the divergence of a vector field equals the total flux of that vector field through the entire closed boundary surface of the volume.

You have identified a volume between the plane $z=0$ and the function $z = 1-x^2 - y^2$. That volume has two pieces to the boundary: the flat circle within the plane $z=0$, and the piece of the paraboloid above it. So the outward (downward) flux through the circle plus the flux outward (upward) through the paraboloid equals the volume integral of the divergence, which you have already found to be $0$.

Your goal is the upward flux through the paraboloid, so find the downward flux through the circle and reverse the sign to get your answer.

For the flat circle, the unit normal vector is $-\hat z$, so $\overrightarrow F \cdot \hat n = <y^3,x^3, z^3>\cdot<0,0,-1> = -z^3$. Since you are in the plane $z=0$, $\overrightarrow F \cdot \hat n = 0$ and the flux through the circle is $0$, meaning that the flux through the paraboloid is also $0$.

Answer 2

Two approaches:

a) You can simply use Stokes' theorem to convert double integral of the curl of the vector field to the line integral of the vector field over the boundary curve of the surface, which is $x^2+y^2 = 1, z = 0$. We can parametrize the curve as $C: r(t) = (\cos t, \sin t, 0), 0 \leq t \leq 2\pi$.

$\vec{F}(x, y, z) = (y^{3}, x^{3}, z^{3})$
$r'(t) = (-\sin t, \cos t, 0)$
$\vec F(r(t)) = (sin^3t, \cos^3 t, 0)$

Applying Stokes' theorem,

$\displaystyle \iint_S (\nabla \times \vec F) \cdot \hat n \ dS = \int_C \vec F \ dr = \int_0^{2\pi} (\cos^4t - \sin^4 t) \ dt$

which is clearly zero as over $2\pi$, integral of $\cos^4 t$ and $\sin^4 t$ will be same.

b) You can find surface integral of the curl of $\vec F$ over any surface with the same boundary curve which is what you arrived at using divergence theorem. Continuing with your approach,

To find surface integral of the curl of the above vector field over paraboloid surface $S1: z = 1 - x^2 - y^2, 0 \leq z \leq 1$.

Find curl: $\nabla \times \vec F = (0, 0, 3x^2-3y^2)$

To apply divergence theorem, we must have a closed surface. So we first need to close the surface with a disk $S2: x^2+y^2 \leq 1$ at $z = 0$.

Of course the divergence of this vector field (as it is curl of another vector field) is zero. So the total flux through the closed surface is zero.

Now to find flux through paraboloid surface, we must subtract the flux through the disk at $z = 0$.

Now note that the outward unit normal vector to the disk is same as the unit normal vector to the plane it resides in (plane $z = 0$). That is $\hat n = (0, 0, -1)$. So integral to find flux through disk is,

$\displaystyle \iint_{S2} 3(y^2-x^2) \ dS \ $. This integral over a disk centered at origin is clearly zero as integral of $y^2$ and $ - x^2$ will cancel each other out.