Evaluate$\mathbf{}\mathbf{\iint }\mathit{V}\mathbf{\cdot }\mathit{n}\mathit{d}\mathit{\sigma }$over the curved surface of the hemisphere${\mathit{x}}^{\mathbf{2}}\mathbf{+}{\mathit{y}}^{\mathbf{2}}\mathbf{+}{\mathit{z}}^{\mathbf{2}}\mathbf{=}\mathbf{9}\mathbf{,}\mathit{z}\mathbf{\ge }\mathbf{0}$, if$\mathbf{V}\mathbf{=}\mathbf{yi}\mathbf{+}\mathbf{xzj}\mathbf{+}\mathbf{(}\mathbf{2}\mathbf{z}\mathbf{}\mathbf{-}\mathbf{}\mathbf{1}\mathbf{)}\mathbf{k}$.Careful: See Problem 9.

The given integrals is $\iint V\cdot nd\sigma$.

The divergence theorem, often known as Gauss' theorem or Ostrogradsky's theorem, is a theorem that connects the flow of a vector field across a closed surface to the field's divergence in the volume enclosed. The surface integral of a vector field over a closed surface, also known as the flux through the surface, equals the volume integral of the divergence over the region inside the surface, according to this theorem.

See the XY plane.

If$n=-\stackrel{⏜}{k}$then the condition is mentioned below.

If$z=0$then the condition is mentioned below.

$V·n=-1$${\iint }_{uppersurface}V\cdot nd\sigma =\oint V\cdot nd\sigma -{\iint }_{XYcontribution}V\cdot nd\sigma$

Apply Gauss' theorem and use the fact that$\nabla ·V=2$

$$\begin{gathered} \int \int \int 2dt={\int }_0^{\mathrm{\pi }}{\int }_0^{\frac{\mathrm{\pi }}{2}}{\int }_0^{3}2r^2\sin \theta drd\theta d\varphi \\ =\left(2\right)\left(\pi \right)\left(1\right)\left(9\right) \\ =\left(2\pi \right)\left(9\right) \\ =\mathit{18}\pi \end{gathered}$$

So, the solution is$\oint V\cdot nd\sigma =18\pi$

It has been known.

$$\begin{gathered} \int {\int }_{XY\mathrm{contribution}}V.nd\sigma =-1{\int }_{circleonXY}d\sigma \\ =-1\mathit{\left(}\mathit{9}\pi \mathit{\right)}=-\mathit{9}\pi \end{gathered}$$

Then, the new solution becomes as mentioned below.

$$\begin{gathered} \int {\int }_{uppersurface}V.nd\sigma =1\mathit{8}\pi \mathit{+}\mathit{9}\pi \\ =27\pi \end{gathered}$$

Hence, the solution of the integrals is found to be as mentioned below.

$\int {\int }_{\mathrm{upper}\mathrm{surface}}V.nd\sigma =\mathit{27}\pi$.