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Essential Calculus: Early Transcendentals
Found in: Page 829
Essential Calculus: Early Transcendentals

Essential Calculus: Early Transcendentals

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

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Short Answer

\({\bf{F}}(x,y,z) = xy{e^z}{\bf{i}} + x{y^2}{z^3}{\bf{j}} - y{e^z}{\bf{k}}\), \(S\) 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 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.

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