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Essential Calculus: Early Transcendentals
Found in: Page 771
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

Evaluate the line integral, where C is the given curve. \(\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy}}\), \({\rm{C}}\)consists of line segments from \({\rm{(0,0)}}\)to\({\rm{(2,1)}}\)and from \({\rm{(2,1)}}\)to\({\rm{(3,0)}}\).

The line integral \(\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy = }}\frac{{\rm{5}}}{{\rm{2}}}\).

See the step by step solution

Step by Step Solution

TABLE OF CONTENTS :
TABLE OF CONTENTS

Step 1: Explanation of solution.

Find the line integral

\(\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy}}\),

Where \({\rm{C}}\) is the line segment from \({\rm{(0,0)}}\) to \({\rm{(2,1)}}\)and from \({\rm{(2,1)}}\) to \({\rm{(3,0)}}\). Note\({\rm{C = }}{{\rm{C}}_1}{\rm{ + }}{{\rm{C}}_2}\), where \({{\rm{C}}_1}\)is the line segment joining \((0,0)\) to \((2,1)\) and \({{\rm{C}}_2}\)is the line segment joining \({\rm{(2,1)}}\) to \({\rm{(3,0)}}\).

\(\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy = }}\int_{{{\rm{C}}_{\rm{1}}}{\rm{ + }}{{\rm{C}}_{\rm{2}}}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy}}\).

Step 2: Graph and Calculation.

On \({C_1},y = \frac{x}{2}\) which implies \(dy = \frac{{dx}}{2}\), and \({{\rm{C}}_{{\rm{2,y}}}}{\rm{ = - x + 3}} \Rightarrow {\rm{dy = - dx}}\).

\(\begin{array}{l}{{\rm{C}}_{\rm{1}}}{\rm{:y = }}\frac{{\rm{x}}}{{\rm{2}}}{\rm{,0}} \le x \le {\rm{2}}\\{{\rm{C}}_{\rm{2}}}{\rm{:y = - x + 3,2}} \le x \le 3\end{array}\)

Thus,

\(\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy = }}\int_{{{\rm{C}}_{\rm{1}}}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy + }}\int_{{{\rm{C}}_{\rm{2}}}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy}}\)

\(\begin{array}{l}{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\left( {\left( {{\rm{x + 2 \times }}\frac{{\rm{x}}}{{\rm{2}}}} \right){\rm{dx + }}{{\rm{x}}^{\rm{2}}}\frac{{{\rm{dx}}}}{{\rm{2}}}} \right)} {\rm{ + }}\int_{\rm{2}}^{\rm{3}} {\left( {{\rm{(x + 2 \times - x + 3)dx + }}{{\rm{x}}^{\rm{2}}}{\rm{ - dx}}} \right)} \\{\rm{ = }}\int_{\rm{0}}^{\rm{2}} {\left( {{\rm{2xdx + }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}{\rm{dx}}} \right)} {\rm{ + }}\int_{\rm{2}}^{\rm{3}} {{\rm{(6 - x)}}} {\rm{dx - }}{{\rm{x}}^{\rm{2}}}{\rm{dx}}\\{\rm{ = }}\left( {{{\rm{x}}^{\rm{2}}}{\rm{ + }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{6}}}} \right)_{\rm{0}}^{\rm{2}}{\rm{ + }}\left( {{\rm{6x - }}\frac{{{{\rm{x}}^{\rm{2}}}}}{{\rm{2}}}{\rm{ - }}\frac{{{{\rm{x}}^{\rm{3}}}}}{{\rm{3}}}} \right)_{\rm{2}}^{\rm{3}}\\{\rm{ = 4 + }}\frac{{\rm{4}}}{{\rm{3}}}{\rm{ + 6 - }}\frac{{\rm{5}}}{{\rm{2}}}{\rm{ - }}\frac{{{\rm{19}}}}{{\rm{3}}}\\{\rm{ = 10 - }}\frac{{\rm{5}}}{{\rm{2}}}{\rm{ - 5}}\\{\rm{ = }}\frac{{\rm{5}}}{{\rm{2}}}{\rm{.}}\end{array}\)

Therefore the line integral \(\int_{\rm{C}} {{\rm{(x + 2y)}}} {\rm{dx + }}{{\rm{x}}^{\rm{2}}}{\rm{dy = }}\frac{{\rm{5}}}{{\rm{2}}}\).

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