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Section 16.2 : Line Integrals - Part I

3. Evaluate \( \displaystyle \int\limits_{C}{{6x\,ds}}\) where \(C\) is the portion of \(y = {x^2}\) from \(x = - 1\) to \(x = 2\). The direction of \(C\) is in the direction of increasing \(x\).

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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

In this case we can just use the equation of the curve for the parameterization because the specified direction is going in the direction of increasing \(x\) which will give us integral limits from smaller value to larger value as needed. Here is a parameterization for this curve.

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

We could also break this up into parameter form as follows.

\[\begin{array}{*{20}{l}}{x = t}\\{y = {t^2}}\end{array}\hspace{0.25in} - 1 \le t \le 2\]

Either form of the parameterization will work for the problem but we’ll use the vector form for the rest of this problem.

Show Step 3

We’ll need the magnitude of the derivative of the parameterization so let’s get that.

\[\vec r'\left( t \right) = \left\langle {1,2t} \right\rangle \hspace{0.25in}\hspace{0.25in}\left\| {\vec r'\left( t \right)} \right\| = \sqrt {{{\left( 1 \right)}^2} + {{\left( {2t} \right)}^2}} = \sqrt {1 + 4{t^2}} \]

We’ll also need the integrand “evaluated” at the parameterization. Recall all this means is we replace the \(x\)/\(y\) in the integrand with the \(x\)/\(y\) from parameterization. Here is the integrand evaluated at the parameterization.

\[6x = 6t\] Show Step 4

The line integral is then,

\[\int\limits_{C}{{6x\,ds}} = \int_{{ - 1}}^{2}{{6t\sqrt {1 + 4{t^2}} \,dt}} = \left. {\frac{1}{2}{{\left( {1 + 4{t^2}} \right)}^{\frac{3}{2}}}} \right|_{ - 1}^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{1}{2}\left( {{{17}^{\frac{3}{2}}} - {5^{\frac{3}{2}}}} \right)}}\]