Calculus III - Line Integrals of Vector Fields

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Section 16.4 : Line Integrals of Vector Fields

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2. Evaluate \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) where \(\vec F\left( {x,y} \right) = \left( {x + y} \right)\,\vec i + \left( {1 - x} \right)\vec j\) and \(C\) is the portion of \(\displaystyle \frac{{{x^2}}}{4} + \frac{{{y^2}}}{9} = 1\) that is in the 4^th quadrant with the counter clockwise rotation.

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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.

\[\vec r\left( t \right) = \left\langle {2\cos \left( t \right),3\sin \left( t \right)} \right\rangle \,\,\,\,\,\,\,\,\frac{3}{2}\pi \le t \le 2\pi \] 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).

\[\vec F\left( {\vec r\left( t \right)} \right) = \left( {2\cos \left( t \right) + 3\sin \left( t \right)} \right)\,\vec i + \left( {1 - 2\cos \left( t \right)} \right)\vec j\]

The derivative of the parameterization is,

\[\vec r'\left( t \right) = \left\langle { - 2\sin \left( t \right),3\cos \left( t \right)} \right\rangle \]

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

\[\begin{align*}\vec F\left( {\vec r\left( t \right)} \right)\centerdot \vec r'\left( t \right) & = \left( {2\cos \left( t \right) + 3\sin \left( t \right)} \right)\left( { - 2\sin \left( t \right)} \right) + \left( {1 - 2\cos \left( t \right)} \right)\left( {3\cos \left( t \right)} \right)\\ & = - 4\cos \left( t \right)\sin \left( t \right) - 6{\sin ^2}\left( t \right) + 3\cos \left( t \right) - 6{\cos ^2}\left( t \right)\\ & = - 4\cos \left( t \right)\sin \left( t \right) - 6\left[ {{{\sin }^2}\left( t \right) + {{\cos }^2}\left( t \right)} \right] + 3\cos \left( t \right)\\ & = - 2\sin \left( {2t} \right) + 3\cos \left( t \right) - 6\end{align*}\]

Make sure that you simplify the dot product with an eye towards doing the integral! In this case that meant using the double angle formula for sine to “simplify” the first term for the integral.

Show Step 4

Now all we need to do is evaluate the integral.

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\vec r}} & = \int_{{\frac{3}{2}\pi }}^{{2\pi }}{{ - 2\sin \left( {2t} \right) + 3\cos \left( t \right) - 6\,dt}}\\ & = \left. {\left[ {\cos \left( {2t} \right) + 3\sin \left( t \right) - 6t} \right]} \right|_{\frac{3}{2}\pi }^{2\pi } = \require{bbox} \bbox[2pt,border:1px solid black]{{5 - 3\pi }}\end{align*}\]