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Found in: Page 771

### Essential Calculus: Early Transcendentals

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

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# Use a calculator or CAS to evaluate the line integral correct to four decimal places.$$\int_{\rm{c}} {{\rm{z}}{{\rm{e}}^{{\rm{ - xy}}}}} {\rm{ds}}$$, where $${\rm{C}}$$ has parametric equations, $$\begin{array}{c}{\rm{x = t}}\\{\rm{y = }}{{\rm{t}}^{\rm{2}}}\\{\rm{z = }}{{\rm{e}}^{{\rm{ - t}}}}\end{array}$$$${\rm{0}} \le {\rm{t}} \le 1$$

Approximate value of integral is $$\approx {\rm{0}}{\rm{.8208}}$$.

See the step by step solution

## Step 1: Find the element of the arc length.

The respective derivatives of the parametric equations are:

$$\begin{array}{c}{{\rm{x}}^\prime }{\rm{(t) = 1,}}\\{{\rm{y}}^\prime }{\rm{(t) = 2t}}\\{{\rm{z}}^\prime }{\rm{(t) = - }}{{\rm{e}}^{{\rm{ - t}}}}\end{array}$$

Element of arc length,

$$\begin{array}{c}{\rm{ds = }}\sqrt {{{\left( {{{\rm{x}}^\prime }{\rm{(t)}}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {{{\rm{y}}^\prime }{\rm{(t)}}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {{{\rm{z}}^\prime }{\rm{(t)}}} \right)}^{\rm{2}}}} \\{\rm{ = }}\sqrt {{\rm{1 + 4}}{{\rm{t}}^{\rm{2}}}{\rm{ + }}{{\rm{e}}^{{\rm{ - 2t}}}}} \end{array}$$

## Step 2: Determine the value of integral.

$$\begin{array}{c}{\rm{e}}\int_{\rm{0}}^{\rm{1}} {{e^{ - t - {t^3}}}\sqrt {{{\left( {{x^\prime }\left( t \right)} \right)}^2} + {{\left( {{y^\prime }\left( t \right)} \right)}^2} + {{\left( {{z^\prime }\left( t \right)} \right)}^2}} } \\{\rm{ = }}\sqrt {{\rm{1 + 4}}{{\rm{t}}^{\rm{2}}}{\rm{ + }}{{\rm{e}}^{{\rm{ - 2t}}}}} \\ \approx {\rm{0}}{\rm{.8208}}\end{array}$$

Therefore, approximate value of integral is $$\approx {\rm{0}}{\rm{.8208}}$$.

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