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Q24E
Expert-verifiedUse 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 \( \approx {\rm{0}}{\rm{.8208}}\).
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,
\(\begin{array}{c}{\rm{ds = }}\sqrt {{{\left( {{{\rm{x}}^\prime }{\rm{(t)}}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {{{\rm{y}}^\prime }{\rm{(t)}}} \right)}^{\rm{2}}}{\rm{ + }}{{\left( {{{\rm{z}}^\prime }{\rm{(t)}}} \right)}^{\rm{2}}}} \\{\rm{ = }}\sqrt {{\rm{1 + 4}}{{\rm{t}}^{\rm{2}}}{\rm{ + }}{{\rm{e}}^{{\rm{ - 2t}}}}} \end{array}\)
\(\begin{array}{c}{\rm{e}}\int_{\rm{0}}^{\rm{1}} {{e^{ - t - {t^3}}}\sqrt {{{\left( {{x^\prime }\left( t \right)} \right)}^2} + {{\left( {{y^\prime }\left( t \right)} \right)}^2} + {{\left( {{z^\prime }\left( t \right)} \right)}^2}} } \\{\rm{ = }}\sqrt {{\rm{1 + 4}}{{\rm{t}}^{\rm{2}}}{\rm{ + }}{{\rm{e}}^{{\rm{ - 2t}}}}} \\ \approx {\rm{0}}{\rm{.8208}}\end{array}\)
Therefore, approximate value of integral is \( \approx {\rm{0}}{\rm{.8208}}\).
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