If$\mathbf{F}\mathbf{}\mathbf{=}\mathbf{}\mathbf{xi}\mathbf{+}\mathbf{yj}$,calculate $\mathbf{\iint }\mathbf{F}\mathbf{\cdot }\mathbf{nd\sigma }$over the part of the surface $\mathbf{z}\mathbf{=}\mathbf{4}\mathbf{-}{\mathbf{x}}^{\mathbf{2}}\mathbf{-}{\mathbf{y}}^{\mathbf{2}}$that is above the (x, y) plane, by applying the divergence theorem to the volume bounded by the surface and the piece that it cuts out of the $\left(x,y\right)$plane. Hint: What is $\mathbf{F}\mathbf{·}\mathbf{n}$on the $\left(x,y\right)$plane?

The given integrals is $z=4-x^2-y^2$over the part of the surface $z=4-x^2-y^2$.

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.

$$\begin{gathered} n=-\stackrel{⏜}{k} \\ F.n=0. \end{gathered}$$

Now write the integral

${\iint }_{uppersurface}F\cdot nd\sigma +{\iint }_{XYcontribution}F\cdot nd\sigma =\oint F\cdot nd\sigma$

Then the value becomes as mentioned below.

${\iint }_{uppersurface}F\cdot nd\sigma =\oint F\cdot nd\sigma$

Apply Gauss' theorem and use the fact that$\nabla .F=2$

$$\begin{gathered} \int \int \int 2dt={\int }_0^4{\int }_0^2{\int }_0^{\sqrt{4-z}}2\rho d\rho d\theta dz \\ =\left(2\mathrm{\pi }\right){\int }_0^{4}4-zdz=\left(2\mathrm{\pi }\right)\left(8\right) \\ =16\mathrm{\pi } \end{gathered}$$

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

${\iint }_{upperserface}F\cdot nd\sigma =16\pi$