Next, we need to parameterize the curve.

\[\vec r\left( t \right) = \left\langle {t,{{\left( {t - 1} \right)}^2}} \right\rangle \,\,\,\,\,\,\,\,0 \le t \le 3\] 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).

\[\vec F\left( {\vec r\left( t \right)} \right) = {\left[ {{{\left( {t - 1} \right)}^2}} \right]^2}\,\vec i + \left( {{{\left( t \right)}^2} - 4} \right)\vec j = {\left( {t - 1} \right)^4}\,\vec i + \left( {{t^2} - 4} \right)\vec j\]

The derivative of the parameterization is,

\[\vec r'\left( t \right) = \left\langle {1,2\left( {t - 1} \right)} \right\rangle \]

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

\[\begin{align*}\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) & = {\left( {t - 1} \right)^4}\left( 1 \right) + \left( {{t^2} - 4} \right)\left( {2t - 2} \right)\\ & = {\left( {t - 1} \right)^4} + 2{t^3} - 2{t^2} - 8t + 8\end{align*}\]

Make sure that you simplify the dot product with an eye towards doing the integral!

Show Step 4

Now all we need to do is evaluate the integral.

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\vec r}} & = \int_{0}^{3}{{{{\left( {t - 1} \right)}^4} + 2{t^3} - 2{t^2} - 8t + 8\,dt}}\\ & = \left. {\left[ {\frac{1}{5}{{\left( {t - 1} \right)}^5} + \frac{1}{2}{t^4} - \frac{2}{3}{t^3} - 4{t^2} + 8t} \right]} \right|_0^3 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{171}}{{10}}}}\end{align*}\]