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### Essential Calculus: Early Transcendentals

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

# Evaluate the line integral$$\int_{\rm{C}} {\rm{F}} {\rm{ \times dr}}$$ where $${\rm{C}}$$is given by the vector function$${\rm{r(t)}}$$.$$\begin{array}{c}{\rm{F(x,y,z) = sinxi + cosyj + xzk,}}\\{\rm{r(t) = }}{{\rm{t}}^{\rm{3}}}{\rm{i - }}{{\rm{t}}^{\rm{2}}}{\rm{j + tk,}}\;\;\;{\rm{0}} \le {\rm{t}} \le {\rm{1}}\end{array}$$

The value of the line integral $$\int_{\rm{C}} {\rm{F}} {\rm{.dr = }}\frac{{\rm{6}}}{{\rm{5}}}{\rm{ - cos(1) - sin(1)}}.$$

See the step by step solution

## Step 1: Finding Derivative.

\begin{aligned}{\rm{r(t) = }}&{{\rm{t}}^{\rm{3}}}{\rm{i - }}{{\rm{t}}^{\rm{2}}}{\rm{j + tk}}\frac{{{\rm{dr}}}}{{{\rm{dt}}}}\\&{\rm{ = 3}}{{\rm{t}}^{\rm{2}}}{\rm{i - 2tj + kdr}}\\&{\rm{ = }}\left( {{\rm{3}}{{\rm{t}}^{\rm{2}}}{\rm{i - 2tj + k}}} \right){\rm{dt}}\end{aligned}

## Step 2: Integrating the Vector field.

\begin{align} & \text{F(x,y,z)=sinxi+cosyj+xzk} \\ & \text{F(r(t))=sin}\left( {{\text{t}}^{\text{3}}} \right)\text{i+cos}\left( \text{-}{{\text{t}}^{\text{2}}} \right)\text{j+}{{\text{t}}^{\text{3}}}\text{(t)k} \\ & \text{F(r(t))=sin}\left( {{\text{t}}^{\text{3}}} \right)\text{i+cos}\left( {{\text{t}}^{\text{2}}} \right)\text{j+}{{\text{t}}^{\text{4}}}\text{k} \end{align}

\begin{align} \int_{\text{C}}{\text{F}}\text{ }\!\!\times\!\!\text { dr=}&\left( \text{sin}\left( {{\text{t}}^{\text{3}}} \right)\text{i+cos}\left( {{\text{t}}^{\text{2}}} \right)\text{j+}{{\text{t}}^{\text{4}}}\text{k} \right)\cdot \left( \text{3}{{\text{t}}^{\text{2}}}\text{i-2tj+k} \right)\text{dt} \\ & \text{=}\int_{\text{0}}^{\text{1}}{\text{3}}{{\text{t}}^{\text{2}}}\text{sin}\left( {{\text{t}}^{\text{3}}} \right)\text{-2tcos}\left( {{\text{t}}^{\text{2}}} \right)\text{+}{{\text{t}}^{\text{4}}}\text{dt} \\ & \text{=}\left( \text{-cos}\left( {{\text{t}}^{\text{3}}} \right)\text{-sin}\left( {{\text{t}}^{\text{2}}} \right)\text{+}\frac{{{\text{t}}^{\text{5}}}}{\text{5}} \right)_{\text{0}}^{\text{1}} \\ & \text{=}\left( \text{-cos(1)-sin(1)+}\frac{\text{1}}{\text{5}} \right)\text{-(-1-0+0)} \\ & \text{=}\frac{\text{6}}{\text{5}}\text{-cos(1)-sin(1)} \end{align}

Thus,

$$\int_{\text{C}}{\text{F}}\text{.dr=}\frac{\text{6}}{\text{5}}\text{-cos(1)-sin(1)}.$$