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### Section 5-4 : Line Integrals of Vector Fields

1. Evaluate $$\displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}$$ where $$\vec F\left( {x,y} \right) = {y^2}\,\vec i + \left( {3x - 6y} \right)\vec j$$ and $$C$$ is the line segment from $$\left( {3,7} \right)$$ to $$\left( {0,12} \right)$$.

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Start Solution

Here is a quick sketch of $$C$$ with the direction specified in the problem statement shown.

Show Step 2

Next, we need to parameterize the curve.

$\vec r\left( t \right) = \left( {1 - t} \right)\left\langle {3,7} \right\rangle + t\left\langle {0,12} \right\rangle = \left\langle {3 - 3t,7 + 5t} \right\rangle \,\,\,\,\,\,\,\,0 \le t \le 1$ Show Step 3

In order to evaluate this line integral we’ll need the dot product of the vector field (evaluated at the along the curve) and the derivative of the parameterization.

Here is the vector field evaluated along the curve (i.e. plug in $$x$$ and $$y$$ from the parameterization into the vector field).

$\vec F\left( {\vec r\left( t \right)} \right) = {\left( {7 + 5t} \right)^2}\,\vec i + \left( {3\left( {3 - 3t} \right) - 6\left( {7 + 5t} \right)} \right)\vec j = {\left( {7 + 5t} \right)^2}\,\vec i + \left( { - 33 - 39t} \right)\vec j$

The derivative of the parameterization is,

$\vec r'\left( t \right) = \left\langle { - 3,5} \right\rangle$

Finally, the dot product of the vector field and the derivative of the parameterization.

$\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) = - 3{\left( {7 + 5t} \right)^2} - 5\left( {33 + 39t} \right)$ Show Step 4

Now all we need to do is evaluate the integral.

\begin{align*}\int\limits_{C}{{\vec F\centerdot d\vec r}} & = \int_{0}^{1}{{ - 3{{\left( {7 + 5t} \right)}^2} - 5\left( {33 + 39t} \right)\,dt}}\\ & = \left. {\left[ { - \frac{1}{5}{{\left( {7 + 5t} \right)}^3} - 165t - \frac{{195}}{2}{t^2}} \right]} \right|_0^1 = \require{bbox} \bbox[2pt,border:1px solid black]{{ - \frac{{1079}}{2}}}\end{align*}