Chapter 9, Problem 16P

Let vector $\stackrel{\to }{c}$ be given $\stackrel{\to }{G}=4xz\stackrel{\to }{i}-y^2\stackrel{\to }{i}+yz\stackrel{\to }{k}$and let V be the volume of a cube of unit length with its corner at the origin, bounded by $x=0$to 1, $y=0$to 1, $z=0$to 1 (Fig. 9-16) Ares A is the surface area of the cube. Perform both integrals of the divergence theorem and verify that they are equal. Show all your work.

To determine

If both integrals of the divergence theorem are equal.

Answer to Problem 16P

Both integrals of the divergence theorem are equal.

Explanation of Solution

Given information:

The vector is $\overrightarrow{G}=4xz\overrightarrow{i}-y^2\overrightarrow{j}+yz\overrightarrow{k}$, the volume of the cube is $V$, and surface area of the cube is $A$.

The following figure shows the faces of the cube.

  

  Figure-(1)

Consider flow is steady and incompressible.

Write the expression for the divergence theorem.

  ${\displaystyle {\int }_V\overrightarrow{\nabla }\cdot \overrightarrow{G}dV}={\displaystyle {\int }_A\overrightarrow{G}\cdot \overrightarrow{n}dA}$   ....... (I)

Here, unit vector is $\overrightarrow{n}$, volume integral $dV$, and area integral is $dA$.

Write the reference equation for $\overrightarrow{G}$.

  $\overrightarrow{G}=G_x\overrightarrow{i}+G_y\overrightarrow{j}+G_z\overrightarrow{k}$   ....... (II)

Here, unit vector in x-direction is $\overrightarrow{i}$, unit vector in y-direction is $\overrightarrow{j}$, and unit vector in z-direction is $\overrightarrow{k}$.

Compare the Equation (II ) with given vector.

  $G_x=4xz$   ....... (III)

  $G_y=-y^2$   ...... (IV)

  $G_z=yz$   ...... (V)

Partially differentiate Equation (III) with respect to the $x$.

  $\frac{\partial G_x}{\partial x}=4z$

Partially differentiate Equation (IV) with respect to the $y$.

  $\frac{\partial G_y}{\partial y}=-2y$

Partially differentiate Equation (V) with respect to the $z$.

  $\frac{\partial G_z}{\partial z}=y$

Write the expression for the volume integral.

  ${\displaystyle {\int }_V\overrightarrow{\nabla }\cdot \overrightarrow{G}dV}={\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}\left(\frac{\partial G_x}{\partial x}+\frac{\partial G_y}{\partial y}+\frac{\partial G_z}{\partial z}\right)dzdydx}}}$   ....... (VI)

Write the expression for $\overrightarrow{n}$.

  $\overrightarrow{n}=x\overrightarrow{i}+y\overrightarrow{j}+z\overrightarrow{k}$   ...... (V)

Write the expression for $\overrightarrow{G}\cdot \overrightarrow{n}$.

  $\overrightarrow{G}\cdot \overrightarrow{n}=\left(4xz\overrightarrow{i}-y^2\overrightarrow{j}+yz\overrightarrow{k}\right)\overrightarrow{n}$   ...... (VII)

Write the expression for the surface integral.

  ${\displaystyle {\int }_A\overrightarrow{G}\cdot \overrightarrow{n}dA}=\left[\begin{array}{l}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}{\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_{1}dxdz}}+{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_{2}dzdy}}+{\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_{3}dydx}}\\ +{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_{4}dzdy}}+{\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_{5}dydx}}+{\displaystyle \underset{z=0}{\overset{z=1}{\int }}{\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_{6}dxdz}}\end{array}\right]$   ...... (VIII)

Calculation:

Substitute $4z$ for $\frac{\partial G_x}{\partial x}$, $-2y$

for $\frac{\partial G_y}{\partial y}$,and $y$ for $\frac{\partial G_z}{\partial z}$ in the Equation (IV).

  $\begin{array}{c}{\displaystyle {\int }_V\overrightarrow{\nabla }\cdot \overrightarrow{G}dV}={\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}\left(4z-2y+y\right)dzdydx}}}\\ ={\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}\left(4z-y\right)dzdydx}}}\\ ={\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\left[\frac{4z^2}{2}-yz\right]}_{z=0}^{z=1}dydx}}\\ ={\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left(2-y\right)dydx}}\end{array}$

  $\begin{array}{c}{\displaystyle {\int }_V\overrightarrow{\nabla }\cdot \overrightarrow{G}dV}={\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\left[2y-\frac{y^2}{2}\right]}_{y=0}^{y=1}}dx\\ ={\displaystyle \underset{x=0}{\overset{x=1}{\int }}\frac{3}{2}}dx\\ =\frac{3}{2}{\left[x\right]}_{x=0}^{x=1}\\ =\frac{3}{2}\left[1-0\right]\end{array}$

  ${\displaystyle {\int }_V\overrightarrow{\nabla }\cdot \overrightarrow{G}dV}=\frac{3}{2}$   ....... (IX)

The unit vector $\overrightarrow{n}$ for face 1 is $\left(0,-1,0\right)$

Substitute $0$ for $x$, $-1$ for $y$, and $0$ for $z$ in the Equation (V).

  $\begin{array}{c}\overrightarrow{n}=0\overrightarrow{i}-1\overrightarrow{j}+0\overrightarrow{k}\\ =-\overrightarrow{j}\end{array}$

The unit vector $\overrightarrow{n}$ for face 2 is $\left(1,0,0\right)$.

Substitute $1$ for $x$, $0$ for $y$, and $0$ for $z$ in the Equation (V).

  $\begin{array}{c}\overrightarrow{n}=1\overrightarrow{i}+0\overrightarrow{j}+0\overrightarrow{k}\\ =\overrightarrow{i}\end{array}$

The unit vector $\overrightarrow{n}$ for face 3 is $\left(0,0,-1\right)$.

Substitute $0$ for $x$, $0$ for $y$, and $-1$ for $z$ in the Equation (V).

  $\begin{array}{c}\overrightarrow{n}=0\overrightarrow{i}+0\overrightarrow{j}-1\overrightarrow{k}\\ =-\overrightarrow{k}\end{array}$

The unit vector $\overrightarrow{n}$ for face 4 is $\left(-1,0,0\right)$

Substitute $-1$ for $x$, $0$ for $y$, and $0$ for $z$ in the Equation (V).

  $\begin{array}{c}\overrightarrow{n}=-1\overrightarrow{i}+0\overrightarrow{j}+0\overrightarrow{k}\\ =-\overrightarrow{i}\end{array}$

The unit vector $\overrightarrow{n}$ for face 5 is $\left(0,0,1\right)$.

Substitute $0$ for $x$, $0$ for $y$, and $1$ for $z$ in the Equation (V).

  $\begin{array}{c}\overrightarrow{n}=0\overrightarrow{i}+0\overrightarrow{j}+1\overrightarrow{k}\\ =\overrightarrow{k}\end{array}$

The unit vector $\overrightarrow{n}$ for face 6 is $\left(0,1,0\right)$.

Substitute $0$ for $x$, $1$ for $y$, and $0$ for $z$ in the Equation (VI).

  $\begin{array}{c}\overrightarrow{n}=0\overrightarrow{i}+1\overrightarrow{j}+0\overrightarrow{k}\\ =\overrightarrow{j}\end{array}$

Substitute $0$ for $y$ and $-\overrightarrow{j}$ for $\overrightarrow{n}$ for face 1 in the Equation (VII).

  $\begin{array}{c}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_1=\left(4xz\overrightarrow{i}-\left(0\right)\overrightarrow{j}+\left(0\right)z\overrightarrow{k}\right)\overrightarrow{j}\\ =0\end{array}$

Substitute $1$ for $x$ and $\overrightarrow{i}$ for $\overrightarrow{n}$ for face 2 in the Equation (VII).

  $\begin{array}{c}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_2=\left(4\left(1\right)z\overrightarrow{i}-y^2\overrightarrow{j}+yz\overrightarrow{k}\right)\overrightarrow{i}\\ =4z\end{array}$

Substitute $0$ for $z$ and $-\overrightarrow{k}$ for $\overrightarrow{n}$ for face 3 in the Equation (VII).

  $\begin{array}{c}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_3=\left(4x\left(0\right)\overrightarrow{i}-y^2\overrightarrow{j}+y\left(0\right)\overrightarrow{k}\right)\left(-\overrightarrow{k}\right)\\ =0\end{array}$

Substitute $0$ for $x$ and $-\overrightarrow{i}$ for $\overrightarrow{n}$ for face 4 in the Equation (VII).

  $\begin{array}{c}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_4=\left(4\left(0\right)z\overrightarrow{i}-y^2\overrightarrow{j}+yz\overrightarrow{k}\right)\left(-\overrightarrow{i}\right)\\ =0\end{array}$

Substitute $1$ for $z$ and $\overrightarrow{k}$ for $\overrightarrow{n}$ for face 5 in the Equation (VII).

  $\begin{array}{c}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_5=\left(4x\left(1\right)\overrightarrow{i}-y^2\overrightarrow{j}+y\left(1\right)\overrightarrow{k}\right)\left(\overrightarrow{k}\right)\\ =1\end{array}$

Substitute $1$ for $y$ and $\overrightarrow{k}$ for $\overrightarrow{n}$ for face 6 in the Equation (VII).

  $\begin{array}{c}{\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_6=\left(4xz\overrightarrow{i}-{\left(1\right)}^2\overrightarrow{j}+yz\overrightarrow{k}\right)\left(\overrightarrow{j}\right)\\ =-1\end{array}$

Substitute $0$ for ${\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_1$, $4z$ for ${\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_2$, $0$ for ${\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_3$, $0$ for ${\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_4$, $1$ for ${\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_5$, and $-1$ for ${\left(\overrightarrow{G}\cdot \overrightarrow{n}\right)}_6$ in the Equation (VIII).

  $\begin{array}{c}{\displaystyle {\int }_A\overrightarrow{G}\cdot \overrightarrow{n}dA}=\left[\begin{array}{l}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}{\displaystyle \underset{x=0}{\overset{x=1}{\int }}\left(0\right)dxdz}}+{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}\left(4z\right)dzdy}}+{\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left(0\right)dydx}}\\ +{\displaystyle \underset{y=0}{\overset{y=1}{\int }}{\displaystyle \underset{z=0}{\overset{z=1}{\int }}\left(0\right)dzdy}}+{\displaystyle \underset{x=0}{\overset{x=1}{\int }}{\displaystyle \underset{y=0}{\overset{y=1}{\int }}\left(y\right)dydx}}+{\displaystyle \underset{z=0}{\overset{z=1}{\int }}{\displaystyle \underset{x=0}{\overset{x=1}{\int }}\left(-1\right)dxdz}}\end{array}\right]\\ ={\displaystyle \underset{y=0}{\overset{y=1}{\int }}2dy}+{\displaystyle \underset{x=0}{\overset{x=1}{\int }}\frac{1}{2}}dx+{\displaystyle \underset{z=0}{\overset{z=1}{\int }}\left(-1\right)}dz\\ ={\left[2y\right]}_{y=0}^{y=1}+{\left[\frac{1}{2}x\right]}_{x=0}^{x=1}+{\left[-z\right]}_{z=0}^{z=1}\\ =2+\frac{1}{2}-1\\ =\frac{3}{2}\end{array}$   ....... (X)

From the Equation (IX) and (X) integral of the divergence theorem is equal.

Conclusion:

Both integrals of the divergence theorem are equal.