91Ó°ÊÓ

Consider the given question,

$\mathbf{F}(x,y,z)=\mathbf{i}+\mathbf{j}+\mathbf{k}$

If a surface S is the graph of $z=z\left(x,y\right)$, then the Flux of $\mathit{F}\left(x,y,z\right)$ through S is given below,

$\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={âˆ\neg }_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})∥z_x×z_y∥dA\\ ={âˆ\neg }_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})\sqrt{{\left(\frac{\mathrm{∂}z}{\mathrm{∂}x}\right)}^2+{\left(\frac{\mathrm{∂}z}{\mathrm{∂}y}\right)}^2+1}dA…...\left(i\right)\end{array}$

Here, the surface S is the lower half of the unit sphere, so its equation will be,

$z=−\sqrt{1−x^2−y^2}$

Now first find $\frac{\mathrm{∂}z}{\mathrm{∂}x}$. The first partial derivates of z are given below,

$$\begin{gathered} \frac{\mathrm{∂}z}{\mathrm{∂}x}=\frac{\mathrm{∂}}{\mathrm{∂}x}\left(-\sqrt{1-x^2-y^2}\right) \\ =\frac{x}{\sqrt{1-x^2-y^2}} \end{gathered}$$

On finding the value of $\frac{\mathrm{∂}z}{\mathrm{∂}\mathrm{y}}$,

role="math" localid="1650342662824" $$\begin{gathered} \frac{\mathrm{∂}z}{\mathrm{∂}\mathrm{y}}=\frac{\mathrm{∂}}{\mathrm{∂}\mathrm{y}}\left(-\sqrt{1-x^2-y^2}\right) \\ =\frac{y}{\sqrt{1-x^2-y^2}} \end{gathered}$$

Then,

$\begin{array}{r}\sqrt{{\left(\frac{\mathrm{∂}z}{\mathrm{∂}x}\right)}^2+{\left(\frac{\mathrm{∂}z}{\mathrm{∂}y}\right)}^2+1}=\sqrt{{\left(\frac{x}{\sqrt{1−x^2−y^2}}\right)}^2+{\left(\frac{y}{\sqrt{1−x^2−y^2}}\right)}^2+1}\\ =\sqrt{\frac{x^2}{1−x^2−y^2}+\frac{y^2}{1−x^2−y^2}+1}\\ =\frac{\sqrt{1−x^2−y^2}}{\sqrt{1}}\end{array}$

The choice of n should have pointing upwards. Then the following vector, is normal to the surface.

Now, for $z=−\sqrt{1−x^2−y^2}$, gives the following vector perpendicular to this surface,

Then the desired normal vector will be,

The value of $\mathbf{F}(x,y,z)â‹\dots \mathbf{n}$will be,

Substituting these values in equation (i),

$\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={âˆ\neg }_D \left(\mathbf{F}\right(x,y,z)â‹\dots \mathbf{n})\left(\sqrt{{\left(\frac{\mathrm{∂}z}{\mathrm{∂}x}\right)}^2+{\left(\frac{\mathrm{∂}z}{\mathrm{∂}y}\right)}^2+1}\right)dA\\ ={âˆ\neg }_D \left(\sqrt{1−x^2−y^2}\left(1−\frac{x+y}{\sqrt{1−x^2−y^2}}\right)\right)\left(\frac{1}{\sqrt{1−x^2−y^2}}\right)dA\\ ={âˆ\neg }_D \left(1−\frac{x+y}{\sqrt{1−x^2−y^2}}\right)dA…….\left(ii\right)\end{array}$

The surface S is the lower half of the unit sphere, so the region of integration D is the unit disk in the xy-plane and centered at the origin.

Hence, in polar coordinates, the region of integration will be,

$D=\left\{\right(r,\mathrm{θ})âˆ\pounds 0≤r≤1,0≤\mathrm{θ}≤2\mathrm{Ï€}\}$

In this case,

localid="1650343222895" $\begin{array}{r}x=r\cos \mathrm{θ},\\ y=r\sin \mathrm{θ}\end{array}$

Using equation (ii), the flux of the vector field through the surface S,

$\begin{array}{r}{∫}_S \mathbf{F}(x,y,z)â‹\dots \mathbf{n}dS={âˆ\neg }_D \left(1−\frac{x+y}{\sqrt{1−x^2−y^2}}\right)dA\\ ={∫}_0^{2\mathrm{Ï€}} {∫}_0^1 \left(1−\frac{r\cos \mathrm{θ}+r\sin \mathrm{θ}}{\sqrt{1−r^2}}\right)rdrd\mathrm{θ}\\ ={∫}_0^{2\mathrm{Ï€}} \left[{∫}_0^1 1dr−(\cos \mathrm{θ}+\sin \mathrm{θ}){∫}_0^1 \frac{r^2}{\sqrt{1−r^2}}dr\right]d\mathrm{θ}\\ ={∫}_0^{2\mathrm{Ï€}} \left[[1−0]−(\cos \mathrm{θ}+\sin \mathrm{θ})\left[\frac{1}{2}\left(−1\sqrt{1−1^2}+{\sin }^{−1}1\right)−\frac{1}{2}\left(−0\sqrt{1−0^2}+{\sin }^{−1}0\right)\right]\right]d\mathrm{θ}\\ \begin{array}{r}\begin{array}{r}={∫}_0^{2\mathrm{Ï€}} \left[1−\frac{\mathrm{Ï€}}{4}(\cos \mathrm{θ}+\sin \mathrm{θ})\right]d\mathrm{θ}\\ ={\left[\mathrm{θ}−\frac{\mathrm{Ï€}}{4}(\sin \mathrm{θ}−\cos \mathrm{θ})\right]}_0^{2\mathrm{Ï€}}\\ =\left(2\mathrm{Ï€}−\frac{\mathrm{Ï€}}{4}(\sin 2\mathrm{Ï€}−\cos 2\mathrm{Ï€})−\left(0−\frac{\mathrm{Ï€}}{4}(\sin 0−\cos 0)\right)\right.\end{array}\\ =2\mathrm{Ï€}\end{array}\end{array}$