Q30P - Introduction to Electrodynamics

Question

Calculate the surface integral of the function in Ex. 1.7, over the bottom of the box. For consistency, let "upward" be the positive direction. Does the surface integral depend only on the boundary line for this function? What is the total flux over the closed surface of the box (including the bottom)? [Note: For the closed surface, the positive direction is "outward," and hence "down," for the bottom face.]

Step-by-Step Solution

Answer

The value of surface integral is 32

Step 1: Define the surface integral

The surface integral of a vector V through a differential surface da  is defined as  $\int v.da$ . The surface integral of the function  $v=2xzi+(x+2)j+y(z^3-3)k$ which is defined as   . there are six differential surface in a cube as shown in following diagram: 

                         

          

Step 2: Compute the surface integral of the vector v at bottom plane

The bottom plane of the cube is xy plane. In xy plane $z=0$. Thus differential surface is defined as  $da=dxdy\hat{k}$

 

The surface integral of vector $\overrightarrow{v}$ , in the xy plane  is computed as:

$$\begin{gathered} \int v.da=\int \int (2xzi+(x+2)j+y(z^3-3\left)k\right)(dx dy k) \\ ={\int }_0^2{\int }_0^{2}y(z^2-3)dx dy \\ ={\int }_0^2{\int }_0^{2}y(0-3)dx dy \\ ={\int }_0^2{\int }_0^2-3y dx dy \end{gathered}$$

Solve further as,

$$\begin{gathered} {\int }_{\left(1\right)}v.da={\int }_0^2{\int }_0^2-3ydx dy \\ ={\int }_0^{2}3ydy{\int }_0^{2}dx \\ =\frac{3}{2}\times 4\times 2 \\ =12 \end{gathered}$$

Step 3: Compute the surface integral of the vector v at plane (i)

The bottom plane of the cube is yz plane. Thus differential surface is defined as  $da=dydz\hat{x}$

 

The surface integral of vector $\overrightarrow{v}$ , in the yz plane  is computed as:

$$\begin{gathered} \int v.da=\int \int (2xzi+(x+2)j+y(z^3-3\left)k\right)(dy dz i) \\ ={\int }_0^2{\int }_0^{2}2xz dy dz \\ ={\int }_0^2{\int }_0^{2}4z dy dz \end{gathered}$$

Solve further as,

$$\begin{gathered} \int v.da={\int }_0^2{\int }_0^{2}4z dydz \\ =a{\int }_0^{2}dy{\int }_0^{2}z dz \\ =16 \end{gathered}$$

Step 4: Compute the surface integral of the vector v at plane (ii)

The  plane (ii) of the cube is yz plane, where $x=0$  . Thus differential surface is defined as $da=-dydz \hat{x}$ 

 

The surface integral of vector $\overrightarrow{v}$ , in the yz plane  is computed as:

$$\begin{gathered} \int v.da=\int \int (2xzi+(x+2)j+y(z^3-3\left)k\right)(dy dz i) \\ ={\int }_0^2{\int }_0^{2}2xz dy dz \\ =0 \end{gathered}$$

Step 5: Compute the surface integral of the vector v at plane (iii)

The plane (iii) of the cube is xz plane , where  $y=2$ . Thus differential surface is defined as  $da=dxdz \hat{j}$

 

The surface integral of vector $\overrightarrow{v}$ , in the xz plane  is computed as:

$$\begin{gathered} \int v.da=\int \int (2xzi+(x+2)j+y(z^3-3\left)k\right)(dx dz j) \\ ={\int }_0^2{\int }_0^2(x+2)dx dz \\ ={\int }_0^2(x+2)dx{\int }_0^{2}dz \\ =12 \end{gathered}$$

Step 6 : Compute the surface integral of the vector v at plane (iv)

The plane (iv) of the cube is xz plane , where $y=0$  . Thus differential surface is defined as $da=-dxdz \hat{j}$ 

 

The surface integral of vector $\overrightarrow{v}$ , in the xz plane  is computed as:

$$\begin{gathered} \int v.da=\int \int (2xzi+(x+2)j+y(z^3-3\left)k\right)(-dx dz j) \\ =-{\int }_0^2{\int }_0^2(x+2)dx dz \\ =-{\int }_0^2(x+2)dx{\int }_0^{2}dz \\ =-12 \end{gathered}$$

Step 7 : Compute the surface integral of the vector v at plane (v)

The plane (v) of the cube is xy plane , where $z=2$ . Thus differential surface is defined as $da=-dxdy \hat{k}$ 

 

The surface integral of vector $\overrightarrow{v}$ , in the xy plane  is computed as:

$$\begin{gathered} \int v.da=\int \int (2xzi+(x+2)j+y(z^3-3\left)k\right)(-dx dz k) \\ ={\int }_0^2{\int }_0^{2}y(x+2)dx dy \\ ={\int }_0^{2}y(2^2-3)dx dy \\ ={\int }_0^2{\int }_0^{2}ydx dy \end{gathered}$$

Solve further as, 

$$\begin{gathered} \int v.da={\int }_0^{2}ydy{\int }_0^{2}dx \\ ={\left(\frac{y^2}{2}\right)}_0^2\left(2-0\right) \\ =4 \end{gathered}$$

Thus the net value of surface integral through the box is the sum of the surface integral of vector v through its open surfces, as follows:

$$\begin{gathered} \underset{ }{\sum {\int }_{6faces}v dl=16+0+12-12+4+12} \\ =32 \end{gathered}$$

 Thus the value of surface integral is 32.