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### Essential Calculus: Early Transcendentals

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

# Use a graph of the vector field $${\bf{F}}$$ and the curve $$C$$ to guess whether the line integral of $${\bf{F}}$$ over $$C$$ is positive, negative, or zero. Then evaluate the line integral.$${\bf{F}}(x,y) = (x - y){\bf{i}} + xy{\bf{j}}$$, $$C$$ is the arc of the circle $${x^2} + {y^2} = 4$$ traversed counterclockwise from $$(2,0)$$ to $$(0, - 2)$$.

The line integral is $$3\pi + \frac{2}{3}$$.

See the step by step solution

## Step 1: Given data

The functions is $${\bf{F}}(x,y) = (x - y){\bf{i}} + xy{\bf{j}}$$ and $$C$$ is the arc of the circle $${x^2} + {y^2} = 4$$ traversed counterclockwise from $$(2,0)$$ to $$(0, - 2)$$.

## Step 2: Concept of line integral

Let $${\bf{F}}$$ be a continuous vector field defined on a smooth curve $$C$$given by vector function . Then the line integral of $${\bf{F}}$$ along $$C$$is given by as follows:

$$\int_C {\bf{F}} \cdot d{\bf{r}} = \int_a^b {\bf{F}} ({\bf{r}}(t)) \cdot {{\bf{r}}^\prime }(t)dt = \int_C {\bf{F}} \cdot {\bf{T}}ds$$

## Step 3: Check whether the line integral is positive or negative

A line integral of $$F$$ is positive if the angle between curve $$C$$ and the vector field $$F$$ is between $${0^\circ }$$to $${90^\circ }$$.

A line integral of $$F$$ is negative if the angle between curve $$C$$ and the vector field $$F$$ is greater than $${90^\circ }$$.

From the figure 1, the angle between the circle oriented counterclockwise and the vector field $$F$$ is between $${0^\circ }$$ and $${90^\circ }$$.

Thus, the line integral $$\int_{{C_1}} F dr$$ is positive.

Figure 1

The figure 1 shows a vector field $${\rm{F}}$$ and curve $$C$$.

If $$F$$ is a continuous vector field defined on a smooth curve $$C,r(t),a \le t \le b$$ is a vector function.

## Step 4: Calculate the equation for line integral

Then, calculate the line integral of $$F$$ along $$C$$ as follows:

$$\int_C F dr = \int_a^b F (r(t)) \cdot {r^\prime }(t)dt$$ ……. (1)

$$r(t)$$ is a vector function determined as follows:

$$r(t) = x(t){\bf{i}} + y(t){\bf{j}}$$

Parametric equations which represent the curve $$C$$ as follows:

$$x = 2\cos t\quad y = 2\sin t\quad 0 \le t \le 3\pi /2.$$

Since $$x = 2\cos t$$ and $$y = 2\sin t$$, it follows:

\begin{aligned}F(r(t)) &= 2(\cos t - \sin t){\bf{i}} + 4\cos t\sin t{\bf{j}}\\{r^\prime }(t) &= - 2\sin t{\bf{i}} + 2\cos t{\bf{j}}\end{aligned}

Using equation (1) as follows:

\begin{aligned}\int_C F dr &= \int_0^{\frac{{3\pi }}{2}} F (r(t)) \cdot {r^\prime }(t)dt\\ &= \int_0^{\frac{{3\pi }}{2}} {\left( {2\left( {\cos t - \sin t)i + 4\cos t\sin tj).( - 2\sin ti + 2\cos tj} \right)} \right)} dt\\ &= \int_0^{\frac{{3\pi }}{2}} {\left( { - 4(\cos t - \sin t)\sin t + 8{{\cos }^2}t\sin t} \right)} dt\\ &= \int_0^{\frac{{3\pi }}{2}} {\left( { - 4\cos t\sin t + 4{{\sin }^2}t + 8{{\cos }^2}t\sin t} \right)} dt\end{aligned}

Simplify further:

\begin{aligned}\int_C F dr &= - 4\int_0^{\frac{{3\pi }}{2}} {\cos t\sin t} dt + 4\int_0^{\frac{{3\pi }}{2}} {{{\sin }^2}t} dt + 8\int_0^{\frac{{3\pi }}{2}} {{{\cos }^2}t\sin t} dt\\ &= - 4{I_1} + 4{I_2} + 8{I_3}\end{aligned} ……. (2)

## Step 5: Calculate the line integral $${I_1}$$

Calculate the integral $${I_1}$$ as follows:

Let $$k = \cos t$$ then $$dk = - \sin tdt$$.

When, \begin{aligned}t &= 0\\k &= 1\end{aligned},

When, \begin{aligned}t &= 3\pi /2\\k &= 0\end{aligned}.

So the integral becomes as follows:

\begin{aligned}\int_0^{3\pi /2} {\cos } t\sin tdt &= - \int_1^0 k dk\\ &= \int_0^1 k dk\\ &= \left. {\frac{{{k^2}}}{2}} \right|_0^1\end{aligned}

Simplify further:

\begin{aligned}\int_0^{3\pi /2} {\cos } t\sin tdt &= \frac{{{1^2}}}{2} - \frac{{{{(0)}^2}}}{2}\\ &= \frac{1}{2}\end{aligned}

## Step 6: Calculate the line integral $${I_2}$$

Calculate the integral $${I_2}$$ as follows:

Write $${\sin ^2}t$$ as $$1/2 - 1/2\cos (2t)$$.

So the integral becomes as follows:

$$\int_0^{3\pi /2} {{{\sin }^2}} tdt = \int_0^{3\pi /2} {\left( {\frac{1}{2} - \frac{1}{2}\cos 2t} \right)} dt$$

Let $$k = 2t$$ then $$dk = 2dt$$.

When,\begin{aligned}t &= 0\\k &= 0\end{aligned},

When, \begin{aligned}t &= 3\pi /2\\k &= 3\pi \end{aligned}.

So the integral becomes as follows:

\begin{aligned}\int_0^{3\pi /2} {\left( {\frac{1}{2} - \frac{1}{2}\cos 2t} \right)} dt &= \frac{1}{2}\int_0^{3\pi /2} d t - \frac{1}{2} \cdot \frac{1}{2}\int_0^{3\pi } {\cos } kdk\\ &= \left. {\frac{1}{2}t} \right|_0^{3\pi /2} - \left. {\frac{1}{4}\sin k} \right|_0^{3\pi }\\ &= \frac{1}{2} \cdot \frac{{3\pi }}{2} - \frac{1}{4}(\sin 3\pi - \sin 0)\\ &= \frac{{3\pi }}{4}\end{aligned}

## Step 7: Calculate the line integral $${I_3}$$

Calculate the integral $${I_3}$$ as follows:

Let $$k = \cos t$$ then $$dk = - \sin tdt$$.

When $$t = 0,\;k = 1$$, when $$t = 3\pi /2,\;k = 0$$.

So the integral becomes as follows:

\begin{aligned}\int_0^{3\pi /2} {{{\cos }^2}} t\sin tdt &= - \int_1^0 {{k^2}} dk\\ &= \int_0^1 {{k^2}} dk &= \left. {\frac{{{k^3}}}{3}} \right|_0^1\\ &= \frac{{{1^3}}}{3} - \frac{{{{(0)}^3}}}{3}\\ &= \frac{1}{3}\end{aligned}

Substitute these values in equation (2) as follows:

\begin{aligned} - 4{I_1} + 4{I_2} + 8{I_3} &= - 4 \cdot \frac{1}{2} + 4 \cdot \frac{{3\pi }}{4} + 8 \cdot \frac{1}{3}\\ &= 3\pi + \frac{2}{3}\end{aligned}

Thus, the line integral is $$3\pi + \frac{2}{3}$$.