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

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:

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:

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

Solve further as,

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

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:

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

Solve further as,

$$\begin{gathered} \int v.da={\int }_0^2{\int }_0^{2}4zdydz \\ =a{\int }_0^{2}dy{\int }_0^{2}zdz \\ =16 \end{gathered}$$

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:

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

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:

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

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:

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

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:

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

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:

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

Thus the value of surface integral is 32.