# How do you evaluate the line integral, where c is the line segment from (0,8,4) to (6,7,7)?

Jul 15, 2015

Let $\vec{c} \left(t\right) = \left(6 t , - t + 8 , 3 t + 4\right)$ and compute ${\int}_{0}^{1} \vec{F} \left(\vec{c} \left(t\right)\right) \setminus \cdot \vec{c} ' \left(t\right) \setminus \mathrm{dt}$, where $\vec{F} \left(\vec{c} \left(t\right)\right) \setminus \cdot \vec{c} ' \left(t\right)$ is a dot product of two vectors.

#### Explanation:

The parametric equation $\vec{c} \left(t\right) = \left(6 t , - t + 8 , 3 t + 4\right)$ will parameterize the line segment $c$ from $\left(0 , 8 , 4\right)$ to $\left(6 , 7 , 7\right)$ as $t$ increases from $t = 0$ to $t = 1$.

If $\vec{F} \left(x , y , z\right)$ is the vector field you want to integrate over this line segment, the way to calculate the line integral is to calculate ${\int}_{0}^{1} \vec{F} \left(\vec{c} \left(t\right)\right) \setminus \cdot \vec{c} ' \left(t\right) \setminus \mathrm{dt}$, where $\vec{F} \left(\vec{c} \left(t\right)\right) \setminus \cdot \vec{c} ' \left(t\right)$ is the dot product of $\vec{F} \left(\vec{c} \left(t\right)\right)$ with $\vec{c} ' \left(t\right)$.

For example, if $\vec{F} \left(x , y , z\right) = \left(x + y , {y}^{2} + z , x y z\right)$, then $\vec{F} \left(\vec{c} \left(t\right)\right) = \vec{F} \left(6 t , - t + 8 , 3 t + 4\right)$

$= \left(5 t + 8 , {\left(- t + 8\right)}^{2} + 3 t + 4 , 6 t \left(- t + 8\right) \left(3 t + 4\right)\right)$

$= \left(5 t + 8 , {t}^{2} - 13 t + 68 , - 18 {t}^{3} + 120 {t}^{2} + 192 t\right)$.

Hence,

$\vec{F} \left(\vec{c} \left(t\right)\right) \setminus \cdot \vec{c} ' \left(t\right)$

$= \left(5 t + 8 , {t}^{2} - 13 t + 68 , - 18 {t}^{3} + 120 {t}^{2} + 192 t\right) \setminus \cdot \left(6 , - 1 , 3\right)$

$= 30 t + 48 - {t}^{2} + 13 t - 68 - 54 {t}^{3} + 360 {t}^{2} + 576 t$

$= - 54 {t}^{3} + 359 {t}^{2} + 619 t - 20$,

and

${\int}_{0}^{1} \vec{F} \left(\vec{c} \left(t\right)\right) \setminus \cdot \vec{c} ' \left(t\right) \setminus \mathrm{dt} = {\int}_{0}^{1} \left(- 54 {t}^{3} + 359 {t}^{2} + 619 t - 20\right) \setminus \mathrm{dt}$

$= - \frac{27}{2} {t}^{4} + \frac{359}{3} {t}^{3} + \frac{619}{2} {t}^{2} - 20 t {|}_{t = 0}^{t = 1}$

$= - \frac{27}{2} + \frac{359}{3} + \frac{619}{2} - 20 = \frac{1187}{3} = 395.666666 \ldots$

You might wonder whether you would get the same answer if you used a different parameterization, such as $\vec{c} \left(t\right) = \left(3 t , - \frac{1}{2} t + 8 , \frac{3}{2} t + 4\right)$ for $0 \setminus \le q t \setminus \le q 2$. The answer is "yes", and you can do the calculation to check it.

If you want to prove this fact in the general case, you have to think about what happens when you "change variables" in the general case. The formula for changing variables in a general case can be written as ${\int}_{a}^{b} f \left(g \left(x\right)\right) g ' \left(x\right) \setminus \mathrm{dx} = {\int}_{g \left(a\right)}^{g \left(b\right)} f \left(u\right) \setminus \mathrm{du}$, where $u = g \left(x\right)$ is the substitution, $\mathrm{du} = g ' \left(x\right) \setminus \mathrm{dx}$, $u = g \left(a\right)$ when $x = a$, and $u = g \left(b\right)$ when $x = b$.