Is unity normal vector verifying Divergence Theorem for Cone $\frac{(r \cos \theta, r \sin \theta , -r)}{r \sqrt{2}}$?

Question

Let $z=\sqrt{x^2+y^2}$, $z\leq 3$ and we wish to verify the divergence theorem for

a) $F={2x, 2y,1]$, b) $F={2x, 2y,2z]$ c) $F={2x, 2y,4z]$.

Volumintegral of these are

a) $36 \pi$ b) $54 \pi$ c) $72 \pi$.

For example: $\int_0^{2\pi} \int_0^3 \int_r^3 4 r dz dr d\theta = \int_0^{2\pi} \int_0^3 4r (3-r) dr d\theta = 2 \pi (6r^2-4/3 r^3)_0^3= 36 \pi$.

The otherside of divergence theorem for the top of the cone gives:

a) $9 \pi$ b) $54 \pi$ c) $108 \pi$

and the surface integral for the cone sides are:

a) $27 \pi$ b) $0$ c) $-36 \pi$
The unity normal vector used for the top is [0,0,1] and for side is $\frac{[rcos \theta, r\sin \theta, -r]}{r\sqrt{2}}$. Am I calculating correct?

Answer 1

The unit normal vector used for the top is $(0,0,1)$ and for side is $(−r\cos\theta,−r\sin\theta,r)$

No, this is not right. The cone has its apex at the origin and its base at $z=3$ (it is "standing on its tip") so the outward pointing normal points out radially in the $r$ direction and slightly down the $z$ axis. (You have the right normal for the base of the cone).

Also I don't know how you go that formula but the norm of the vector is $r\sqrt{2}\neq 1$. The correct outward-pointing cone normal is $$\frac{1}{\sqrt{2}} \left(\cos\theta, \sin\theta, -1\right).$$

Answer 2

As you know one should use Jacobi and the length of unity vector is same as the Jacobi and $r\sqrt{2}$ is deleted: $\frac{(r\cos \theta. r\sin \theta, -r)}{r\sqrt{2}} \cdot r\sqrt{2}=(r\cos \theta. r\sin \theta, -r)$

Answer 3

The side of the cone can be parametrized as $\mathbf{r}(z,\theta)=(z\cos(\theta),z\sin(\theta),z)$ for $0 \leq z \leq 3$ and $0 \leq \theta \leq 2\pi$. Then $\mathbf{r}_z \times \mathbf{r}_\theta=(-z\cos(\theta),-z\sin(\theta),z)$. This is an inward normal, as you can see easily by just thinking about the x and y components. (It takes a little bit more thinking to see why the $z$ component of the inward normal has the sign that it does.) So you need to flip it over to make an outward normal for comparison with the divergence theorem. This normal isn't unit, but for purposes of computing the surface integral with a parametrization that doesn't actually hurt anything (because you wind up multiplying $\mathbf{F} \cdot \mathbf{n}$ by the magnitude of this vector anyway to compute the scalar surface integral), so the surface integral becomes $\int_0^3 \int_0^{2\pi} \mathbf{F} \cdot (z\cos(\theta),z\sin(\theta),-z) d \theta dz.$ So the first case gives an integrand of $2z^2 - z$, the second case gives an integrand of $0$, and the third case gives an integrand of $-2z^2$. So the results for the fluxes through the side should be $27\pi,0$ and $-36\pi$ respectively.

The flux through the top is easily calculated as it is just the constant value of the $z$ component times the area of the circle which is $9\pi$. So these are $9\pi,54\pi$ and $108\pi$ respectively.

Your volume integral for the first case seems to be incorrect as well, since the divergence of the field is just $4$ so you get $4$ times the volume of the cone which is $36\pi$, consistent with what I obtained for the surface integrals.

In short, the normal vector doesn't get changed; you just had an error in your volume integral in part (a) that made you think that your incorrect surface integral computation was correct.