91Ó°ÊÓ

The surface integral of a vector V through a differential surface da is defined as $∫v.da$. 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 $\stackrel{→}{v}$, in the xy plane is computed as:

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

Solve further as,

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

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

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

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

Solve further as,

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

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 $\stackrel{→}{v}$, in the yz plane is computed as:

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

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 $\stackrel{→}{v}$, in the xz plane is computed as:

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

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 $\stackrel{→}{v}$, in the xz plane is computed as:

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

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 $\stackrel{→}{v}$, in the xy plane is computed as:

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

Solve further as,

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

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:

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

Thus the value of surface integral is 32.