8. Let F = xyzi + (y² + 1)ĵ + z³k and S be the surface of the cube 0 < x, y, z < 1. (a) Evaluate the surface integral S S;(V × F) · d§ using the divergence theorem (hint: is there a helpful vector identity?). (b) Evaluate the surface integral S Ss(V × F) · dS using Stokes' theorem (hint: is there a boundary?)

Given that

$\overrightarrow{F}=xyz\overrightarrow{i}+\left(y^2+1\right)\overrightarrow{j}+z^3\overrightarrow{k}$

S be the surface of the cube $0\le x,y,z\le 1$

Now,

$$\begin{gathered} \nabla \times \overrightarrow{F}=\left|\begin{array}{ccc}\overrightarrow{i}& \overrightarrow{j}& \overrightarrow{k}\\ \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\ xyz& y^2+1& z^3\end{array}\right| \\ =\overrightarrow{i}\left(\frac{\partial }{\partial y}\left(z^3\right)-\frac{\partial }{\partial z}\left(y^2+1\right)\right)-\overrightarrow{j}\left(\frac{\partial }{\partial x}\left(z^3\right)-\frac{\partial }{\partial z}\left(xyz\right)\right)+\overrightarrow{k}\left(\frac{\partial }{\partial x}\left(y^2+1\right)-\frac{\partial }{\partial y}\left(xyz\right)\right) \\ =0\overrightarrow{i}-j\left(0-xy\right)+\overrightarrow{k}\left(0-xz\right) \\ \nabla \times \overrightarrow{F}=xy\overrightarrow{j}-xz\overrightarrow{k} \end{gathered}$$

a)Using divergence theorem,

$\underset{S}{\iint }\overrightarrow{F.}d\overrightarrow{S}=\int \int \int \left(\nabla ·\overrightarrow{F}\right) d\overrightarrow{S}$

Here,

$$\begin{gathered} \underset{S}{\iint }\left(\nabla \times \overrightarrow{F}\right)d\overrightarrow{S}={\int }_0^1{\int }_0^1{\int }_0^1\nabla ·\left(\nabla \times \overrightarrow{F}\right) d\overrightarrow{S} \\ ={\int }_0^1{\int }_0^1{\int }_0^1\left(\frac{\partial }{\partial x}\overrightarrow{i}+\frac{\partial }{\partial y}\overrightarrow{j}+\frac{\partial }{\partial z}\overrightarrow{k}\right).\left(xy\overrightarrow{j}-xz\overrightarrow{k}\right)dx dy dz \\ ={\int }_0^1{\int }_0^1{\int }_0^1\left(\frac{\partial }{\partial y}\left(xy\right)-\frac{\partial }{\partial z}\left(xz\right)\right)dxdy dz \\ ={\int }_0^1{\int }_0^1{\int }_0^1(x-x)dx dy dz \\ \underset{S}{\iint }\left(\nabla \times \overrightarrow{F}\right)d\overrightarrow{S}=0 \end{gathered}$$

Thus, the integral value becomes 0.