Question

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 \((0, - 2)\).

Short answer

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

Step-by-step solution

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 \((0, - 2)\).

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:

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

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.

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)

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:

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

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:

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

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 \(3\pi + \frac{2}{3}\).