Chapter 16, Problem 1GYR

To determine

Explain the line integrals of scalar functions and evaluate them with examples.

Explanation of Solution

Calculation:

Definition:

If $f$ is defined on a curve $C$ given parametrically by $r\left(t\right)=g\left(t\right)i+h\left(t\right)j+k\left(t\right)k,a\le t\le b$, then the line integral of $f$ over $C$ is ${\displaystyle \underset{C}{\int }f\left(x,y,z\right)ds}=\underset{n\to \infty }{\text{lim}}{\displaystyle \sum _{k=1}^{n}f\left(x_k,y_k,z_k\right)\Delta s_k}$ provided this limit exists.

Evaluation of Line Integral:

To integrate a continuous function $f\left(x,y,z\right)$ over a curve $C$:

• Find a smooth parameterization of $C$, $r\left(t\right)=g\left(t\right)i+h\left(t\right)j+k\left(t\right)k,a\le t\le b$.
• Evaluate the integral as ${\displaystyle \underset{C}{\int }f\left(x,y,z\right)ds}={\displaystyle \underset{a}{\overset{b}{\int }}f\left(g\left(t\right),h\left(t\right),k\left(t\right)\right)\left|v\left(t\right)\right|dt}$.

Example:

Evaluate the Line integral ${\displaystyle {\int }_C\left(x+y\right)ds}$, over the space $C$, where $C$ is the straight-line segment $x=t$, $y=\left(1-t\right)$, $z=0$, from $\left(0,1,0\right)$ to $\left(1,0,0\right)$.

Solution:

Write the Line integral.

${\displaystyle {\int }_C\left(x+y\right)ds}$        (1)

Find the smooth parameterization of $C$.

$\begin{array}{c}r\left(t\right)=ti+\left(1-t\right)jk+\left(0\right)k\\ =ti+\left(1-t\right)j,\text{ 0}\le t\le \text{1}\end{array}$        (2)

The components have continuous first derivatives. Differentiate Equation (2) with respect to $t$.

$\begin{array}{c}\frac{dr}{dt}=\left(1\right)i+\left(0-1\right)j\\ =i-j\end{array}$        (3)

Find the value of $\left|\frac{dr}{dt}\right|$.

$\begin{array}{c}\left|\frac{dr}{dt}\right|=\sqrt{{\left(\frac{dx}{dt}\right)}^2+{\left(\frac{dy}{dt}\right)}^2+{\left(\frac{dz}{dt}\right)}^2}\\ =\sqrt{{\left(1\right)}^2+{\left(1\right)}^2}\\ =\sqrt{2}\end{array}$        (4)

Find the function $f\left(x,y,z\right)=f\left(x+y\right)$.

Substitute $t$ for $x$ and $1-t$ for $y$ in $f\left(x+y\right)$.

$\begin{array}{c}f\left(x+y\right)=t+\left(1-t\right)\\ =1\end{array}$        (5)

Evaluate the integral ${\displaystyle {\int }_{C}f\left(x,y,z\right)ds}={\displaystyle {\int }_C\left(x+y\right)ds}$.

Substitute Equation (4) and Equation (5) in ${\displaystyle {\int }_{C}f\left(x,y,z\right)ds}={\displaystyle {\int }_C\left(x+y\right)ds}$.

$\begin{array}{c}{\displaystyle {\int }_{C}f\left(x,y,z\right)ds}={\displaystyle \underset{0}{\overset{1}{\int }}f\left(t,1-t,0\right)\left|\frac{dr}{dt}\right|dt}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}f\left(x+y\right)\left|\frac{dr}{dt}\right|dt}\\ ={\displaystyle \underset{0}{\overset{1}{\int }}\left(1\right)\sqrt{2}dt}\\ =\sqrt{2}{\displaystyle \underset{0}{\overset{1}{\int }}dt}\end{array}$

$\begin{array}{c}{\displaystyle {\int }_{C}f\left(x,y,z\right)ds}=\sqrt{2}{\left[t\right]}_0^1\\ =\sqrt{2}\left[1-0\right]\\ =\sqrt{2}\left(1\right)\\ =\sqrt{2}\end{array}$

Therefore, the value of the Line integral ${\displaystyle {\int }_C\left(x+y\right)ds}$, over the space $C$, where $C$ is the straight-line segment $x=t$, $y=\left(1-t\right)$, $z=0$, from $\left(0,1,0\right)$ to $\left(1,0,0\right)$ is $\sqrt{2}$.