I need help to find the solution to the following problem:
$$ I=\iint_S \vec A \cdot\ d \vec s$$
over the entire surface of the region above the $xy$-plane bounded by the cone $x^2 + y^2 = z^2$ and the plane $z = 4$ where $\vec A = 4xz \hat i + xyz^2 \hat j + 3z \hat k$.
The answer is given to be $320 \pi$ but mine comes out to be different.
On
Since we are talking about bounded region, we should use Gauss Divergence Theorem:
$\int_{S} \vec{V}.\vec{n}$ $dS= \int_{\tau} \nabla. \vec{V}$ $d\tau= \iiint \nabla. \vec{V}$ $dxdydz$.
$\nabla. \vec{V}= 4z+xz^2+3$.
$\iint_{R} \int_{z=\sqrt(x^2+y^2)}^{4} \nabla. \vec{V} $ $dzdR=$
$\iint_{R} \int_{z=\sqrt(x^2+y^2)}^{4} 4z+xz^2+3 $ $dzdR=$
$\iint_{R} 2z^2 + \frac{x}{3} z^3 + 3z \Big|_{\sqrt(x^2+y^2)}^4$ $dR=$
$\iint_{R} [44+\frac{64x}{3}-2(x^2+y^2)-\frac{x}{3} (x^2+y^2)^{\frac{3}{2}} - 3\sqrt(x^2+y^2)]$ $dR= I$.
Now, we are dealing with a double integral, and we should use polar coordinates as we have a disc in the $xy-$plane:
We should set: $x= rcos\theta$, $y=rsin\theta$, $z= \sqrt(x^2+y^2) = \sqrt(r^2cos^2\theta + r^2sin^2\theta)= r$ and $dxdy= rdrd\theta$.
After substituting these in $I$ and evaluating the resulting integral, you will get $320\pi$.
Calculate the divergence $\text{div} A=\nabla\cdot A=4z+xz^2+3$ and apply Gauss-Ostrogradsky theorem \begin{align} I&=\iiint_{\text{the region}}\text{div} A\,dxdydz=\int_0^4\iint_{x^2+y^2\le z^2}(4z+xz^2+3)\,dxdy\,dz=\\ &=\int_0^4\Bigl((4z+3)\underbrace{\iint_{x^2+y^2\le z^2}1\,dxdy}_{=\pi z^2}+z^2\underbrace{\iint_{x^2+y^2\le z^2}x\,dxdy}_{=0}\Bigr)\,dz. \end{align} Remark: the first integral is the area of the disc of radius $z$ and the second one has the symmetric integrand $x$ wrt the integration region.
Finally, you get the simple one-dimensional integral wrt $z$ that is easily calculated to be $320\pi$.