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Found in: Page 829

### Essential Calculus: Early Transcendentals

Book edition 2nd
Author(s) James Stewart
Pages 830 pages
ISBN 9781133112280

# $${\bf{F}}(x,y,z) = xy{e^z}{\bf{i}} + x{y^2}{z^3}{\bf{j}} - y{e^z}{\bf{k}}$$, $$S$$ is the surface of the box bounded by the coordinate planes and the planes $$x = 3,y = 2$$, and $$z = 1$$.

The flux of F across S is: $${\bf{F}}(x,y) = \frac{y}{{\sqrt {{x^2} + {y^2}} }}{\bf{i}} + \frac{x}{{\sqrt {{x^2} + {y^2}} }}{\bf{j}}$$.

See the step by step solution

## Step 1: Definition of coordinate plane

Two number lines constitute the coordinate plane, which is a two-dimensional surface. The horizontal x-axis is one of the number lines.

## Step 2: Draw the vector $${\bf{F}}$$

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.

## Step 3: Calculate the length of the vector

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

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

So, the length of the vector is $$1$$.

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

Therefore, along the axes the vectors are horizontal and vertical respectively on $$y$$ and $$x$$axes.