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.
Best Answer
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
$$ 192\pi + 128\pi = 320\pi $$