Chapter MA, Problem 2.6MA

To determine

Obtain the divergence of A and verify the divergence theorem on S.

Answer to Problem 2.6MA

The divergence of A is $\underset{\_}{2y+3z+5}$ and divergence theorem is verified.

Explanation of Solution

Calculation:

Consider the expression to find divergence of the vector.

$\begin{array}{c}\text{Divergence}=\nabla \cdot A\\ =\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot A\end{array}$

Substitute $2xy{a}_x+3zy{a}_y+5z{a}_z$ for A.

$\begin{array}{c}\text{Divergence}=\left(\frac{\partial }{\partial x}{a}_x+\frac{\partial }{\partial y}{a}_y+\frac{\partial }{\partial z}{a}_z\right)\cdot \left(2xy{a}_x+3zy{a}_y+5z{a}_z\right)\\ =\left(\frac{\partial }{\partial x}\left(2xy\right)+\frac{\partial }{\partial y}\left(3zy\right)+\frac{\partial }{\partial z}\left(5z\right)\right)\\ =2y+3z+5\end{array}$

Consider the mathematical form of divergence theorem.

${\displaystyle {\oint }_{S}A\cdot dS}={\displaystyle {\int }_v\nabla \cdot Adv}$

Find ${\displaystyle {\oint }_{S}A\cdot dS}$:

$\begin{array}{c}{\displaystyle {\oint }_{S}A\cdot dS}=\left[\begin{array}{l}{\displaystyle {\int }_4^6{\displaystyle {\int }_{-3}^4{\left(A\right)}_{x=-2}\cdot \left(-dydz{a}_x\right)}}+{\displaystyle {\int }_4^6{\displaystyle {\int }_{-3}^4{\left(A\right)}_{x=3}\cdot \left(dydz{a}_x\right)}}+\\ {\displaystyle {\int }_{-3}^4{\displaystyle {\int }_{-2}^3{\left(A\right)}_{y=4}\cdot \left(-dxdz{a}_y\right)}}+{\displaystyle {\int }_{-3}^4{\displaystyle {\int }_{-2}^3{\left(A\right)}_{y=6}\cdot \left(dxdz{a}_y\right)}}+\\ {\displaystyle {\int }_{-2}^3{\displaystyle {\int }_4^6{\left(A\right)}_{z=-3}\cdot \left(-dxdy{a}_z\right)}}+{\displaystyle {\int }_{-2}^3{\displaystyle {\int }_4^6{\left(A\right)}_{z=4}\cdot \left(-dxdy{a}_z\right)}}\end{array}\right]\\ =\left[\begin{array}{l}{\displaystyle {\int }_4^6{\displaystyle {\int }_{-3}^4{\left(2xy{a}_x+3zy{a}_y+5z{a}_z\right)}_{x=-2}\cdot \left(-dydz{a}_x\right)}}+\\ {\displaystyle {\int }_4^6{\displaystyle {\int }_{-3}^4{\left(2xy{a}_x+3zy{a}_y+5z{a}_z\right)}_{x=3}\cdot \left(dydz{a}_x\right)}}+\\ {\displaystyle {\int }_{-3}^4{\displaystyle {\int }_{-2}^3{\left(2xy{a}_x+3zy{a}_y+5z{a}_z\right)}_{y=4}\cdot \left(-dxdz{a}_y\right)}}+\\ {\displaystyle {\int }_{-3}^4{\displaystyle {\int }_{-2}^3{\left(2xy{a}_x+3zy{a}_y+5z{a}_z\right)}_{y=6}\cdot \left(dxdz{a}_y\right)}}+\\ {\displaystyle {\int }_{-2}^3{\displaystyle {\int }_4^6{\left(2xy{a}_x+3zy{a}_y+5z{a}_z\right)}_{z=-3}\cdot \left(-dxdy{a}_z\right)}}+\\ {\displaystyle {\int }_{-2}^3{\displaystyle {\int }_4^6{\left(2xy{a}_x+3zy{a}_y+5z{a}_z\right)}_{z=4}\cdot \left(dxdy{a}_z\right)}}\end{array}\right]\\ =\left[\begin{array}{l}{\displaystyle {\int }_4^6{\displaystyle {\int }_{-3}^4\left(4y\right)dydz}}+{\displaystyle {\int }_4^6{\displaystyle {\int }_{-3}^{4}6ydydz}}+\\ {\displaystyle {\int }_{-3}^4{\displaystyle {\int }_{-2}^3-12zdxdz}}+{\displaystyle {\int }_{-3}^4{\displaystyle {\int }_{-2}^{3}18zdxdz}}+\\ {\displaystyle {\int }_{-2}^3{\displaystyle {\int }_4^{6}15dxdy}}+{\displaystyle {\int }_{-2}^3{\displaystyle {\int }_4^{6}20dxdy}}\end{array}\right]\\ =\left[{\displaystyle {\int }_4^6{\displaystyle {\int }_{-3}^{4}10ydydz}}+{\displaystyle {\int }_{-3}^4{\displaystyle {\int }_{-2}^{3}6zdxdz}}+{\displaystyle {\int }_{-2}^3{\displaystyle {\int }_4^{6}35dxdy}}\right]\end{array}$

Simplify further.

$\begin{array}{c}{\displaystyle {\oint }_{S}A\cdot dS}=\left[{\displaystyle {\int }_{-3}^{4}5{\left[y^2\right]}_4^{6}dz}+{\displaystyle {\int }_{-2}^{3}3{\left[z^2\right]}_{-3}^{4}dx}+{\displaystyle {\int }_4^{6}35{\left[x\right]}_{-2}^{3}dy}\right]\\ =\left[100{\displaystyle {\int }_{-3}^{4}dz}+21{\displaystyle {\int }_{-2}^{3}dx}+175{\displaystyle {\int }_4^{6}dy}\right]\\ =100{\left[z\right]}_{-3}^4+21{\left[x\right]}_{-2}^3+175{\left[y\right]}_4^6\\ =700+105+350\end{array}$

${\displaystyle {\oint }_{S}A\cdot dS}=1,155$

Find ${\displaystyle {\int }_v\nabla \cdot Adv}$:

$\begin{array}{c}{\displaystyle {\int }_v\nabla \cdot Adv}={\displaystyle {\int }_{x=-2}^{x=3}{\displaystyle {\int }_{y=4}^{y=6}{\displaystyle {\int }_{z=-3}^{z=4}\left(2y+3z+5\right)dxdydz}}}\\ ={\displaystyle {\int }_{x=-2}^{x=3}{\displaystyle {\int }_{y=4}^{y=6}{\left[2yz+\frac{3}{2}{z}^2+5z\right]}_{-3}^{4}dxdy}}\\ ={\displaystyle {\int }_{x=-2}^{x=3}{\displaystyle {\int }_{y=4}^{y=6}\left[\left(8y+24+20\right)-\left(-6y+\frac{27}{2}-15\right)\right]dxdy}}\\ ={\displaystyle {\int }_{x=-2}^{x=3}{\displaystyle {\int }_{y=4}^{y=6}\left(14y+\frac{91}{2}\right)dxdy}}\end{array}$

$\begin{array}{c}{\displaystyle {\int }_v\nabla \cdot Adv}={\displaystyle {\int }_{x=-2}^{x=3}{\left[7y^2+\frac{91}{2}y\right]}_4^{6}dx}\\ ={\displaystyle {\int }_{x=-2}^{x=3}\left(140+91\right)dx}\\ =231{\displaystyle {\int }_{x=-2}^{x=3}dx}\\ =231{\left[x\right]}_{-2}^3\end{array}$

${\displaystyle {\int }_v\nabla \cdot Adv}=1,155$

Both the values ${\displaystyle {\oint }_{S}A\cdot dS}$ and ${\displaystyle {\int }_v\nabla \cdot Adv}$ are equal. Hence, the divergence theorem is verified.

Conclusion:

Thus, the divergence of A is $\underset{\_}{2y+3z+5}$ and divergence theorem is verified.