How do you evaluate the following line integral (x^2)zds, where c is the line segment from the point (0, 6, -1) to the point (4,1,5)?

Mar 17, 2015

The answer is: $\setminus \frac{56 \setminus \sqrt{77}}{3} \setminus \approx 163.8$

First, parameterize the line segment. The quickest way to do this is to let $P = \left(0 , 6 , - 1\right)$ and $Q = \left(4 , 1 , 5\right)$ and, thinking of these as vectors, find a vector going from $P$ to $Q$ by subtracting the components/coordinates of $P$ from the corresponding components/coordinates of $Q$ to get $\setminus \vec{v} = \left(4 , - 5 , 6\right)$. Now let $\setminus \vec{c} \left(t\right) = P + t \setminus \vec{v} = \left(4 t , 6 - 5 t , - 1 + 6 t\right)$ for $0 \setminus \le q t \setminus \le q 1$.

The velocity vector is $\setminus \vec{c} ' \left(t\right) = \setminus \vec{v}$, and its length (the speed) is $| | \setminus \vec{v} | | = \setminus \sqrt{16 + 25 + 36} = \setminus \sqrt{77}$. Letting $f \left(x , y , z\right) = {x}^{2} z$, the function we have to integrate as $t$ goes from $t = 0$ to $t = 1$ is $f \left(4 t , 6 - 5 t , - 1 + 6 t\right) | | \setminus \vec{v} | | = \setminus \sqrt{77} {\left(4 t\right)}^{2} \setminus \cdot \left(- 1 + 6 t\right) = \setminus \sqrt{77} \left(96 {t}^{3} - 16 {t}^{2}\right) .$

(Note that $\mathrm{ds} = \setminus \frac{\mathrm{ds}}{\mathrm{dt}} \mathrm{dt} = | | \setminus \vec{v} | | \mathrm{dt}$.)

Here's the integral calculation:

$\setminus {\int}_{0}^{1} \setminus \sqrt{77} \left(96 {t}^{3} - 16 {t}^{2}\right) \mathrm{dt} = \setminus \sqrt{77} \left(24 {t}^{4} - \setminus \frac{16}{3} {t}^{3}\right) \setminus {|}_{t = 0}^{t = 1}$

$= \setminus \sqrt{77} \setminus \cdot \setminus \frac{72 - 16}{3} = \setminus \frac{56 \setminus \sqrt{77}}{3} \setminus \approx 163.8$