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Section 16.4 : Line Integrals of Vector Fields

4. Evaluate \( \displaystyle \int\limits_{C}{{\vec F\centerdot d\vec r}}\) where \(\vec F\left( {x,y,z} \right) = {{\bf{e}}^{2x}}\,\vec i + z\left( {y + 1} \right)\vec j + {z^3}\,\vec k\) and \(C\) is given by \(\vec r\left( t \right) = {t^3}\,\vec i + \left( {1 - 3t} \right)\vec j + {{\bf{e}}^t}\,\vec k\) for \(0 \le t \le 2\).

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Start Solution

Okay, for this problem we’ve been given the parameterization of the curve in the problem statement so we don’t need to worry about that for this problem and we can jump right into the work needed to evaluate the line integral. This means that 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).

\[\vec F\left( {\vec r\left( t \right)} \right) = {{\bf{e}}^{2{t^{\,3}}}}\,\vec i + {{\bf{e}}^t}\left( {1 - 3t + 1} \right)\vec j + {\left( {{{\bf{e}}^t}} \right)^3}\,\vec k = {{\bf{e}}^{2{t^{\,3}}}}\,\vec i + {{\bf{e}}^t}\left( {2 - 3t} \right)\vec j + {{\bf{e}}^{3t}}\,\vec k\]

The derivative of the parameterization is,

\[\vec r'\left( t \right) = 3{t^2}\,\vec i - 3\vec j + {{\bf{e}}^t}\,\vec k\]

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

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

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

Show Step 2

Now all we need to do is evaluate the integral.

\[\begin{align*}\int\limits_{C}{{\vec F\centerdot d\vec r}} & = \int_{0}^{2}{{3{t^2}{{\bf{e}}^{2{t^{\,3}}}} - 3{{\bf{e}}^t}\left( {2 - 3t} \right) + {{\bf{e}}^{4t}}\,dt}}\\ & = \left. {\left[ {\frac{1}{2}{{\bf{e}}^{2{t^{\,3}}}} - 3{{\bf{e}}^t}\left( {5 - 3t} \right) + \frac{1}{4}{{\bf{e}}^{4t}}\,dt} \right]} \right|_0^2 = \require{bbox} \bbox[2pt,border:1px solid black]{{\frac{{57}}{4} + 3{{\bf{e}}^2} + \frac{1}{4}{{\bf{e}}^8} + \frac{1}{2}{{\bf{e}}^{16}}}}\end{align*}\]

Note that the second term in the integral involved integration by parts (you do recall how to do that right?). We’ll leave the integration by parts details to you to verify and note that we did simplify the results a little bit.

Also, do not get excited about the “messy” answer here! You will get these kinds of answers on occasion.