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### Section 16.4 : Line Integrals of Vector Fields

6. Evaluate $$\displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}$$ where $$\vec F\left( {x,y} \right) = xy\,\vec i + \left( {1 + 3y} \right)\vec j$$ and $$C$$ is the line segment from $$\left( {0, - 4} \right)$$ to $$\left( { - 2, - 4} \right)$$ followed by portion of $$y = - {x^2}$$ from $$x = - 2$$ to $$x = 2$$ which is in turn followed by the line segment from $$\left( {2, - 4} \right)$$ to $$\left( {5,1} \right)$$. See the sketch below.

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To help with the problem let’s label each of the curves as follows,

The parameterization of each curve is,

$${C_1}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {0, - 4} \right\rangle + t\left\langle { - 2, - 4} \right\rangle = \left\langle { - 2t, - 4} \right\rangle \hspace{0.25in}0 \le t \le 1$$
$${C_2}:\,\,\vec r\left( t \right) = \left\langle {t, - {t^2}} \right\rangle \hspace{0.25in} - 2 \le t \le 2$$
$${C_3}:\,\,\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {2, - 4} \right\rangle + t\left\langle {5,1} \right\rangle = \left\langle {2 + 3t, - 4 + 5t} \right\rangle \hspace{0.25in}0 \le t \le 1$$

Show Step 2

Now we need to compute the line integral for each of the curves. In the first few problems in this section we evaluated the vector function along the curve, took the derivative of the parameterization and computed the dot product separately. For this problem we’ll be doing all that work in the integral itself.

Here is the line integral for each of the curves.

\begin{align*}\int\limits_{{{C_1}}}{{\vec F\centerdot d\vec r}} & = \int_{0}^{1}{{\left\langle {\left( { - 2t} \right)\left( { - 4} \right),1 + 3\left( { - 4} \right)} \right\rangle \centerdot \left\langle { - 2,0} \right\rangle \,dt}}\\ & = \int_{0}^{1}{{ - 16t\,dt}} = \left. {\left( { - 8{t^2}} \right)} \right|_0^1 = \underline { - 8} \end{align*} \begin{align*}\int\limits_{{{C_2}}}{{\vec F\centerdot d\vec r}} & = \int_{{ - 2}}^{2}{{\left\langle {\left( t \right)\left( { - {t^2}} \right),1 + 3\left( { - {t^2}} \right)} \right\rangle \centerdot \left\langle {1, - 2t} \right\rangle \,dt}}\\ & = \int_{{ - 2}}^{2}{{5{t^3} - 2t\,dt}} = \left. {\left( {\frac{5}{4}{t^4} - {t^2}} \right)} \right|_{ - 2}^2 = \underline 0 \end{align*} \begin{align*}\int\limits_{{{C_3}}}{{\vec F\centerdot d\vec r}} & = \int_{0}^{1}{{\left\langle {\left( {2 + 3t} \right)\left( { - 4 + 5t} \right),1 + 3\left( { - 4 + 5t} \right)} \right\rangle \centerdot \left\langle {3,5} \right\rangle \,dt}}\\ & = \int_{0}^{1}{{45{t^2} + 69t - 79\,dt}} = \left. {\left( {15{t^3} + \frac{{69}}{2}{t^2} - 79t} \right)} \right|_0^1 = \underline { - \frac{{59}}{2}} \end{align*} Show Step 3

Okay to finish this problem out all we need to do is add up the line integrals over these curves to get the full line integral.

$\int\limits_{C}{{\vec F\centerdot d\vec r}} = \left( { - 8} \right) + \left( 0 \right) + \left( { - \frac{{59}}{2}} \right) = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{75}}{2}}}$

Note that we put parenthesis around the result of each individual line integral simply to illustrate where it came from and they aren’t needed in general of course.