Surface integral over a cone above the xy plane

Question

From Schaum's vector analysis

My approach:

First, parametrize the equation $x^2 +y^2 = z^2 $

$ x = \rho cos \phi$ , $y= \rho sin\phi$ , $z= \rho$

Then,

$\vec A = 4 \rho^2 cos \phi \hat i + \rho^4 cos \phi sin \phi \hat j + 3 \rho \hat k$

$\vec n = \nabla S = 2x \hat i + 2y \hat j - 2z \hat k$

Then the unit normal $ \hat n = \frac {1}{\sqrt 2} ( cos \phi \hat i + sin \phi \hat j - \hat k)$

$ \vec A . \hat n = \frac {1}{\sqrt 2} ( 4 \rho^2 cos^2 \phi + \rho^4 cos \phi sin^2 \phi - 3 \rho)$

Surface area projection: $dS = \frac {dxdy}{| \hat n . \hat k|} = \sqrt 2 dxdy = \sqrt 2 \rho d \rho d \phi$

$ \iint_S \vec A . \hat n dS = \iint_0^4 (4 \rho^3 cos^2 \phi + \rho^5 cos \phi sin^2 \phi - 3 \rho^2) d \phi = \int_0^{2 \pi} [ \rho^4 cos^2 \phi + \frac { \rho^6}{6} cos \phi sin^2 \phi - \rho^3 ] d \phi$

then,

$ \iint_S \vec A . \hat n dS = \rho^4 [ \int_0^{2 \pi} cos^2 \phi d \phi + \frac { \rho^6}{6} \int_0^{2 \pi} cos \phi sin^2 \phi d \phi - \rho^3 \int_0^{2 \pi} d \phi$

= $ \frac { \rho^4}{2} [ \phi + \frac {sin 2 \phi}{2} ] - \rho^3 [\phi]$ , $ \phi \in [0, 2\pi]$

Finally,

$ \iint_S \vec A . \hat n dS = 2 \pi [ \frac {256}{2} - 64 ] = 128 \pi $

However the answer is $320 \pi$ , why? Where did I go wrong? Thanks.

Answer 1

The problem says "the entire surface". You need to calculate the surface integral over the plane $z = 4$, $\rho \leq 4$ and $0\leq \phi \leq 2\pi$. It is just a matter of repeating the same you did already with

$$ \hat{n} = \hat{z} $$

So that

$$ \int {\bf A}\cdot {\rm d}^2{\bf S} = \int 12r{\rm d}\phi {\rm d}\rho = 192\pi $$

So the total integral is

$$ 192\pi + 128\pi = 320\pi $$

Answer 2

As you probably know, you can use the divergence theorem.

The surface integral over the cone and the plane transforms to the volume integral over the interior of the cone. We have $\operatorname{div} \mathbf{A} = 4z+xz^2+3$ so $$\int\limits_{\text{surface of cone}} \mathbf{A}\cdot d\mathbf{S} = \int\limits_{\text{volume of cone}}\operatorname{div} \mathbf{A} \,dV = \int\limits_{\text{volume of cone}}(4z+xz^2+3) \,dV$$ The integral of $xz^2$ vanishes because of symmetry, and the integral of $3$ is just three times the volume of the cone, or $64\pi$.

We have $$\int\limits_{\text{volume of cone}}4z\,dV = 4\int_{\phi = 0}^{2\pi}\int_{z = 0}^4\int_{\rho = 0}^z z\, \rho\,d\rho\,dz\,d\phi = 8\pi \int_{z=0}^4 \frac{z^3}2\,dz = 4^4\pi = 256\pi$$

Hence the entire integral is $320\pi$.