Im struggling to find an answer to this problem. I know I have to use Guass's Divergence Theorem and I get stuck at calculating the volume.
Let F: $\mathbb{R}^3 \to \mathbb{R}^3$ be a vector field. Where F$(x,y,z)= ( 2x, y, 3 )$ and $S$ is the surface of a sphere $x^2+y^2$+$(z-2)^2=8$ that lies above the $xy$ plane and the normal is pointing outwards.
Calculate on S $$\iint \textbf{F} dS$$
On
From Gauss-Divergence theorem we know that $$ \iint \limits_S \mathbf{F} ds=\iiint \limits \nabla \cdot \mathbf{F} dV $$ \begin{align*} \iint \limits_S \mathbf{F} ds & =\iiint \limits \nabla \cdot \mathbf{F}\\ & = \iiint 3 dV=3 (\text{Volume of the sphere})\\ & = 3\times \frac{4}{3}\pi (2\sqrt{2})^3=64 \sqrt{2}\ \pi \end{align*}
Parameterization of hemisphere: $r(\theta,\phi) = \langle \sqrt{8} \sin \phi \cos \theta, \sqrt{8} \sin \phi \sin \theta, \sqrt{8} \cos \phi \rangle$ for $0 \le \theta \le 2\pi$ and $0 \le \phi \le \pi/2$. This is from spheerical coordinates.
Then $r_\theta \times r_\phi = \langle - \sqrt{8} \sin \phi \sin \theta, \sqrt{8} \sin \phi \cos \theta, 0 \rangle \times \langle \sqrt{8} \cos \phi \cos \theta, \sqrt{8} \cos \phi \sin \theta, -\sqrt{8} \sin \phi \rangle = \langle -8 \cos \theta \sin^2 \phi, -8 \sin \theta \sin^2 \phi, -16 \sin \phi \cos \phi \rangle.$
Looking at $\phi=0$ and the $z$-component, we see that this corresponds to inwards normal, so negate the above. Then,
The surface integral equals \begin{align*} \int_0^{2\pi} \int_0^{\pi/2} F(r(\theta,\phi)) \cdot (r_\theta \times r_\phi) \; d \phi d \theta &= \int_0^{2\pi} \int_0^{\pi/2} \langle 2 \sqrt{8} \sin \phi \cos \theta, \sqrt{8} \sin \phi \sin \theta, 3 \rangle \cdot (r_\theta \times r_\phi) \; d \phi \, d \theta \\ &= \int_0^{2\pi} \int_0^{\pi/2} 16 \sqrt{8} \sin^3 \phi \cos^2 \theta + 8 \sqrt{8} \sin^3\phi \sin^2 \theta + 48 \sin \phi \cos \phi \; d \phi \; d \theta \end{align*} From here I think you just have to grind it out. I have to go now, let me know if you have questions/ think i made an error.