91Ó°ÊÓ

The integral of the derivative of a function over an open surface area is equal to the volume integral of the function.

$\mathbf{âˆ\circledR }\mathbf{(}\mathbf{∇}\mathbf{·}\mathit{v}\mathbf{)}\mathbf{·}\mathit{d}\mathit{Ï„}\mathbf{=}\mathbf{âˆ\circledR }\mathit{v}\mathbf{}\mathit{d}\mathit{a}\mathbf{.}$

Consider the vector point function $F(x,y,z)=F_1(x,y,z)\mathrm{i}+F_2(x,y,z)\mathrm{j}+F_3(x,y,z)\mathrm{k},$where $F_1,F_2,F_3$are components of $F(x,y,z)$
.

The divergence of function$F(x,y,z)$is computed as follows:

$∇F(x,y,z)=\frac{∂F_1}{∂x}+\frac{∂F_2}{∂y}+\frac{∂F_3}{∂z}$

Here,$\frac{∂F_1}{∂x},\frac{∂F_2}{∂y},\frac{∂F_3}{∂z}$ are the partial derivatives of function$F(x,y,z)$ with respect to$x,y,z$ .

The divergence of vector function in spherical coordinates is

.$∇F(r,\mathrm{θ},\mathrm{ϕ})=\frac{1}{r^2}\frac{∂\left(r^2{F}_1\right)}{∂r}+\frac{1}{r\sin \mathrm{θ}}\frac{∂(\sin \mathrm{θ}{F}_2)}{∂\mathrm{θ}}+\frac{1}{r\sin \mathrm{θ}}\frac{∂F_3}{∂\mathrm{ϕ}}$

Here, $r,\mathrm{θ},\mathrm{ϕ}$are the spherical coordinates.

(a)

The given function is $v=s(2+{\sin }^2\mathrm{θ})s+\left(s\sin \mathrm{θ}\cos \mathrm{θ}\right)\mathrm{ϕ}+3z\hat{z}$ , and the del [S1] operator is defined as . The divergence of vector v is computed as follows:

$$\begin{gathered} ∇·v=∇F(r,\mathrm{θ},\mathrm{ϕ})=\frac{1}{s}\frac{∂\left(s{v}_s\right)}{∂s}+\frac{1}{s}\frac{∂\left(v_{\mathrm{ϕ}}\right)}{∂\mathrm{θ}}+\frac{1}{r\sin \mathrm{θ}}\frac{∂\left(v_z\right)}{∂z} \\ =\frac{1}{s}\frac{∂\left(s\right(s\left(2\right)+{\sin }^2\mathrm{θ}\left)\right))}{∂s}+\frac{1}{s}\frac{∂\left(s\sin \mathrm{ϕ}\cos \mathrm{ϕ}\right)}{∂\mathrm{ϕ}}+\frac{∂\left(3z\right)}{∂z} \\ =\frac{1}{s}\frac{∂\left(s^2\right(2)+{\sin }^2\mathrm{θ}))}{∂s}+\frac{1}{s}\frac{∂\left(s\sin \mathrm{ϕ}\cos \mathrm{ϕ}\right)}{∂\mathrm{ϕ}}+\frac{∂\left(3z\right)}{∂z} \\ =4+2{\sin }^2\mathrm{ϕ}+{\cos }^2\mathrm{ϕ}-{\sin }^2+3 \end{gathered}$$.

Solving further

$$\begin{gathered} ∇.v=4+({\sin }^2\mathrm{ϕ}+{\cos }^2\mathrm{ϕ})-3 \\ =4+\left(1\right)+3 \\ =8 \end{gathered}$$

Thus, the divergence of the function is$∇.v=8$

(b)

The cylinder mentioned has a height of 5 units and a radius of 2 units, as shown below:

The surface integral of vector , along the path (i) is computed as follows:

$$\begin{gathered} ∫(∇v)·d\mathrm{τ}={∫}_0^5{∫}_0^{n/2}{∫}_0^{2}8sdsd\mathrm{ϕ}dz \\ =8{∫}_0^{2}sds{∫}_0^{n/2}d\mathrm{ϕ}{∫}_0^{5}dz \\ =8\left(2\right)\left(\frac{\mathrm{π}}{5}\right)5 \\ =40\mathrm{π} \end{gathered}$$

……. (1)

The right side of the gauss divergence theorem is $âˆ\circledR v·da$. The surface area has the top and bottom areas. Thus, the right side can be written as follows:

$$\begin{gathered} âˆ\circledR v·da={âˆ\circledR }_{\left(i\right)}v·da+{âˆ\circledR }_{\left(ii\right)}v·da++{âˆ\circledR }_{\left(iii\right)}v·\mathrm{da}.\mathrm{for}\mathrm{surface}\left(i\right){âˆ\circledR }_{\left(i\right)}v·=âˆ\circledR v_r·da,a=and \\ d{a}_1=sd\mathrm{Ï•}dz\hat{z} \end{gathered}$$

Thus, the areal vector for surface (i) is computed as follows:

$$\begin{gathered} {âˆ\circledR }_{\left(i\right)}v·da={∫}_{z-0}^5{∫}_{\mathrm{Ï•}-0}^{\frac{n}{2}}{s}^2(2+{\sin }^2\mathrm{Ï•})d\mathrm{φ}dz \\ ={∫}_{z-0}^5{{\left(2\mathrm{Ï•}+\left(\frac{\mathrm{Ï•}}{2}-\frac{\sin 2\mathrm{Ï•}}{4}\right)\right)}_0}^{\frac{\mathrm{Ï€}}{2}}dz \\ =4\left(\frac{5\mathrm{Ï€}}{2}\right){\left(z\right)}_0^5=25\mathrm{Ï€} \end{gathered}$$

For surface (ii) ${âˆ\circledR }_{\left(i\right)}v·da=âˆ\circledR v_z·da$, and $d{a}_2=sd\mathrm{Ï•}dz\hat{z},$where z = 5 .

Thus, the areal vector for surface (ii) is computed as follows:

$$\begin{gathered} {âˆ\circledR }_{\left(ii\right)}v·da={∫}_{z-0}^2{∫}_{\mathrm{Ï•}-0}^{\frac{n}{2}}\left(3z\right)sd\mathrm{Ï•}dz \\ =15\mathrm{Ï€} \end{gathered}$$

For surface (iii) ${âˆ\circledR }_{\left(iii\right)}v·da=âˆ\circledR v·da,$and $d{a}_3=sdsd\mathrm{Ï•}\hat{z},$where z = 0 .

Thus, the areal vector for surface (ii) is computed as follows:

$$\begin{gathered} {âˆ\circledR }_{\left(iii\right)}v·da={∫}_{z-0}^2{∫}_{\mathrm{Ï•}-0}^{\frac{n}{2}}(-3z)sd\mathrm{Ï•}dz \\ =0 \end{gathered}$$

Thus, the total areal vector for the surface area is computed as follows:

$$\begin{gathered} {âˆ\circledR }_{\left(i\right)}v·da={âˆ\circledR }_{\left(i\right)}v·da+{âˆ\circledR }_{\left(ii\right)}v·da+{âˆ\circledR }_{\left(iii\right)}v·da \\ =25\mathrm{Ï€}+15\mathrm{Ï€}+0 \\ =40\mathrm{Ï€} \end{gathered}$$

…… (2)

From equations (1) and (2), the left and right side of the gauss divergence theorem is satisfied.

(c)

The curl of vector is calculated as follows:

$$\begin{gathered} ∇×v=\left\{\frac{1}{s}\left(\frac{∂v_z}{∂\mathrm{ϕ}}\right)-\frac{∂v_{\mathrm{ϕ}}}{∂\mathrm{ϕ}}\right\}\hat{s}\left(\frac{∂v_z}{∂z}-\frac{∂v_z}{∂s}\right)\widehat{\mathrm{ϕ}}+\frac{1}{s}\left\{\frac{1}{∂s}\left(s{v}_{\mathrm{ϕ}}\right)-\frac{∂v_s}{∂\mathrm{ϕ}}\right\}\hat{z} \\ =\left\{\frac{1}{s}\left(\frac{∂v_z}{∂\mathrm{ϕ}}\right)-\frac{∂(s\sin \mathrm{ϕ}\cos \mathrm{ϕ})}{∂\mathrm{ϕ}}\right\}\hat{s}+\left(\frac{∂\left(s\right(2+{\sin }^2\mathrm{ϕ})}{∂z}-\frac{∂\left(\right(3z\left)\right)}{∂s}\right)\widehat{\mathrm{ϕ}} \\ =+\frac{1}{s}\left\{\frac{1}{∂s}\left(s\right(s\sin \mathrm{ϕ}\cos \mathrm{ϕ}\left)\right)-\frac{∂\left(s\right(2+{\sin }^2\mathrm{ϕ}\left)\right)}{∂\mathrm{ϕ}}\right\}\hat{z} \\ =\left\{\frac{1}{s}×0-0\right\}\hat{s}+\left\{0+0\right\}\widehat{\mathrm{ϕ}}\frac{1}{s}\left\{2s\sin \mathrm{ϕ}\cos \mathrm{ϕ}-2s\sin \mathrm{ϕ}\cos \mathrm{ϕ}\right\}\hat{z} \\ =0 \end{gathered}$$

Therefore, the curl of v is$∇×v=0$