Question

\({\bf{F}}(x,y,z) = {x^2}yz{\bf{i}} + x{y^2}z{\bf{j}} + xy{z^2}{\bf{k}}\), \(S\) is the surface of the box enclosed by the planes \(x = 0\), \({\bf{x}} = {\bf{a}}\), \({\bf{y}} = {\bf{0}},\,\,{\bf{y}} = {\bf{b}},\,\,{\bf{z}} = {\bf{0}},\,\,z = c\), where \(a,b\), and \(c\) are positive numbers.

Short answer

The flux of F across S is: \({\bf{F}}(0,y) = \frac{y}{{|y|}} = \pm 1\) and \({\bf{F}}(x,0) = \frac{{ - x}}{{|x|}}\).

Step-by-step solution

Statement of divergence theorem

The divergence theorem is a mathematical expression of the physical truth that, in the absence of matter creation or destruction, the density within a region of space can only be changed by allowing it to flow into or out of the region through its boundary.

Calculating the length of vectors

Let us solve the given problem.

It wants to draw the vector field\({\bf{F}}\)for\({\bf{F}}(x,y) = \frac{y}{{\sqrt {{x^2} + {y^2}} }}{\bf{i}} - \frac{x}{{\sqrt {{x^2} + {y^2}} }}{\bf{j}}\).

It has used the software Mathematica to plot the vector field.

Note that the length of the vector\(\frac{y}{{\sqrt {{x^2} + {y^2}} }}{\bf{i}} - \frac{x}{{\sqrt {{x^2} + {y^2}} }}{\bf{j}}\)is written as,

\(\begin{array}{c}\sqrt {{{\left( {\frac{y}{{\sqrt {{x^2} + {y^2}} }}} \right)}^2} + {{\left( {\frac{{ - x}}{{\sqrt {{x^2} + {y^2}} }}} \right)}^2}} = \sqrt {\frac{{{x^2} + {y^2}}}{{{x^2} + {y^2}}}} \\ = 1\end{array}\)

So, the length of the vector is 1.

Sketch the vector field \({\bf{F}}\)

\({\bf{F}}(0,y) = \frac{y}{{|y|}} = \pm 1\) and \({\bf{F}}(x,0) = \frac{{ - x}}{{|x|}}\)

All the vectors have same the length. Choose points and sketch diagram

Therefore, all the vector fields are tangent to the circle cantered at the origin and having radius \(\sqrt {{x^2} + {y^2}} \).