Evaluate $\iint_s $$\vec{F}\cdot $$\vec{dS}$ where $\vec{F}=4xi-2y^2j+z^2k$ and $S$ is the surface bounding the region $x^2+y^2=4, z=0$ and $z=3$
When I solve it by Gauss Divergence Theorem it's answer was 84$\pi$ but again when I tried to verified it by direct integral it doesn't gives me the same answer.
Please Don't use Cylindrical Coordinate system :)
By the divergence theorm.
$\iiint 4+2y+2z\ dV$
Integrating with respect to $z$
$\iint (\int_0^3 4+2y+2z \ dz) \ dA$
$\iint 12 + 3y+ 9 \ dA$
The region is symmetric around the y axis, that term will drop on integration.
$21(4\pi) = 84\pi$ we agree on that!
Evaluating the surfaces independently. We have two disks and the wall of the cyinder.
$z = 0$ The normal is $(0,0,-1).$ Negative, because we want the normals pointing out.
$F\cdot n = -2z$ But z is zero.
$0$ on that disk.
$z = 3, n = (0,0,1)$
$\iint 6 \ dA = 24\pi$
You don't want to convert to cylindrical, we can paramertize by $x,z$
$y = \pm\sqrt{4-x^2}$
We can take the positive root and double the integral to cover both the positive and negative cases.
$n = (-\frac {\partial y}{\partial x}, 1, -\frac {\partial y}{\partial z}) = (\frac{x}{\sqrt {4-x^2}}, 1 , 0)$
$F\cdot n = \frac {4x^2}{\sqrt {4-x^2}} + 2\sqrt {4-x^2}$
$2\int_0^3 \int_{-2}^2 \frac {4x^2}{\sqrt{4-x^2}} + 2\sqrt {4-x^2} \ dx \ dz\\ 2\int_0^3 \int_{-2}^2 \frac {4x^2 + 8 - 2x^2}{\sqrt {4-x^2}}\ dx \ dz\\ 2\int_0^3 \int_{-2}^2 \frac {2x^2}{\sqrt{4-x^2}} + \frac{8}{\sqrt {4-x^2}}\ dx \ dz$
$x = 2\sin\theta, dx = 2\cos \theta.$ While this is a Calc 1 substitution, it is effectively the same thing as converting to cylindrical...
$2\int_0^3 \int_{-\frac{\pi}{2}}^{\frac \pi2} 4\sin^2\theta + 8 \ dx \ dz\\ 2\int_0^3 10\pi \ dz\\ 60\pi$
$60\pi + 24\pi = 84\pi$