Stefan's question about a line integral.

[Prove It]

Evaluate the line integral

$\displaystyle \begin{align*} \int_{(0,0,0)}^{ \left( \frac{1}{3}, \frac{\pi}{2}, 1 \right) }{ \left[ 12\,z^2 + 4\,\mathrm{e}^{4x}\cos{(5y)} \right] \,\mathrm{d}x - \left[ 5\,\mathrm{e}^{4x}\sin{(5y)} \right] \,\mathrm{d}y + 24\,x\,z\,\mathrm{d}z } \end{align*}$

The line starting at $\displaystyle \begin{align*} \left( 0, 0, 0 \right) \end{align*}$ and ending at $\displaystyle \begin{align*} \left( \frac{1}{3}, \frac{\pi}{2}, 1 \right) \end{align*}$ can be expressed in a parametric (vector) form as $\displaystyle \begin{align*} \left( x, y, z \right) = \left( \frac{1}{3}t , \frac{\pi}{2}t, t \right) \end{align*}$ with $\displaystyle \begin{align*} 0 \leq t \leq 1 \end{align*}$. Thus

$\displaystyle \begin{align*} x &= \frac{1}{3}t \implies \mathrm{d}x = \frac{1}{3}\mathrm{d}t \\ y &= \frac{\pi}{2}t \implies \mathrm{d}y = \frac{\pi}{2}\mathrm{d}t \\ z &= t \implies \mathrm{d}z = \mathrm{d}t \end{align*}$

and so the line integral becomes

$\displaystyle \begin{align*} &= \int_0^1{ \left\{ 12t^2 + 4\mathrm{e}^{ 4 \left( \frac{1}{3} t \right) } \cos{ \left[ 5 \left( \frac{\pi}{2}t \right) \right] } \right\} \frac{1}{3}\mathrm{d}t - \left\{ 5\mathrm{e}^{4 \left( \frac{1}{3}t \right) } \sin{ \left[ 5 \left( \frac{\pi}{2}t \right) \right] } \right\} \frac{\pi}{2}\mathrm{d}t + 24 \left( \frac{1}{3}t \right) t \, \mathrm{d}t } \\ &= \int_0^1{ 4\,t^2 + \frac{4}{3}\,\mathrm{e}^{ \frac{4}{3}t } \cos{ \left( \frac{5\pi}{2}t \right)} - \frac{5\pi}{2}\,\mathrm{e}^{\frac{4}{3}t}\sin{ \left( \frac{5\pi}{2}t \right) } + 8\,t^2 \, \mathrm{d}t } \\ &= \int_0^1{ 12\,t^2 + \frac{4}{3}\,\mathrm{e}^{\frac{4}{3}t}\cos{ \left( \frac{5\pi}{2}t \right) } - \frac{5\pi}{2}\,\mathrm{e}^{\frac{4}{3}t} \sin{ \left( \frac{5\pi}{2} t\right) } \, \mathrm{d}t } \\ \end{align*}$

Now to integrate these, the product functions either require integration by parts, or use of the rules $\displaystyle \begin{align*} \int{ \mathrm{e}^{b\,x}\sin{(a\,x)} \, \mathrm{d}x } = \frac{1}{a^2 + b^2} \, \mathrm{e}^{b\,x} \, \left[ b\sin{(a\,x)} - a\cos{(a\,x)} \right] + C \end{align*}$ and $\displaystyle \begin{align*} \int{ \mathrm{e}^{b\,x}\cos{(a\,x)} \,\mathrm{d}x } = \frac{1}{a^2 + b^2}\,\mathrm{e}^{b\,x} \, \left[ a\sin{(a\,x)} + b\cos{(a\,x)} \right] + C \end{align*}$.

 

[jvicens]

Using these rules, the integral evaluates to

$\displaystyle \begin{align*} &= \left[ 4\,t^3 + \frac{4}{3}\,\mathrm{e}^{\frac{4}{3}t}\cos{\left( \frac{5\pi}{2}t \right) } - \frac{5\pi}{2}\,\mathrm{e}^{\frac{4}{3}t}\sin{\left( \frac{5\pi}{2}t \right) } + \frac{8}{3}\,t^3 \right]_0^1 \\ &= \frac{13}{3} + \frac{4}{3}\,\mathrm{e}^{\frac{4}{3}}\cos{\frac{5\pi}{2}} - \frac{5\pi}{2}\,\mathrm{e}^{\frac{4}{3}}\sin{\frac{5\pi}{2}} - \frac{8}{3} \\ &= \frac{5\pi}{2}\,\mathrm{e}^{\frac{4}{3}} \\ &\approx 26.9 \end{align*}$

In conclusion, the line integral evaluates to approximately 26.9. This integral represents the work done by the vector field $\displaystyle \begin{align*} \mathbf{F} (x,y,z) = \left( 12\,z^2 + 4\,\mathrm{e}^{4x}\cos{(5y)} , -5\,\mathrm{e}^{4x}\sin{(5y)} , 24\,x\,z \right) \end{align*}$ over the given line segment. This could have practical applications in physics or engineering, where the work done by a force field over a specific path is of interest.