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)?

Question

Luke Alvarez · Accepted Answer

The answer is: #\frac{56\sqrt{77}}{3}\approx 163.8# First, parameterize the line segment.  The quickest way to do this is to let #P=(0,6,-1)# and #Q=(4,1,5)# 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 #\vec{v}=(4,-5,6)#.  Now let #\vec(c)(t)=P+t\vec{v}=(4t,6-5t,-1+6t)# for #0\leq t\leq 1#.   The velocity vector is #\vec{c}'(t)=\vec{v}#, and its length (the speed) is #||\vec{v}||=\sqrt{16+25+36}=\sqrt{77}#.  Letting #f(x,y,z)=x^2z#, the function we have to integrate as #t# goes from #t=0# to #t=1# is #f(4t,6-5t,-1+6t)||\vec{v}||=\sqrt(77)(4t)^2\cdot (-1+6t)=\sqrt{77}(96t^3-16t^2).#  (Note that #ds=\frac{ds}{dt}dt=||\vec{v}||dt#.) Here's the integral calculation: #\int_{0}^{1}\sqrt{77}(96t^3-16t^2)dt=\sqrt(77)(24t^{4}-\frac{16}{3}t^{3})\|_{t=0}^{t=1}# #=\sqrt{77}\cdot \frac{72-16}{3}=\frac{56\sqrt{77}}{3}\approx 163.8#

Luke Alvarez · Answer

undefined To evaluate the line integral $ \int_C x^2 z \, ds $, where $ C $ is the line segment from the point $ (0, 6, -1) $ to the point $ (4, 1, 5) $, follow these steps:

1. Parametrize the curve $ C $ with a vector-valued function $ \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle $ where $ t $ varies from $ t = t_1 $ to $ t = t_2 $, and $ \mathbf{r}(t_1) $ corresponds to the initial point and $ \mathbf{r}(t_2) $ corresponds to the final point on the curve.

2. Calculate the derivative of the parametric equations to find $ ds $.

3. Substitute the parametric equations for $ x $, $ y $, and $ z $ into the integrand $ x^2 z $ to express it in terms of $ t $.

4. Compute the integral $ \int_{t_1}^{t_2} (x(t))^2 z(t) \, ds $ using the limits $ t = t_1 $ to $ t = t_2 $.

Let's proceed with the solution:

1. Parametrize the curve $ C $:
$$ \mathbf{r}(t) = \langle x(t), y(t), z(t) \rangle = \langle 4t, 6 - 5t, -1 + 6t \rangle $$
where $ 0 \leq t \leq 1 $.

2. Calculate the derivative of $ \mathbf{r}(t) $ to find $ ds $:
$$ |\mathbf{r}'(t)| = \sqrt{(4)^2 + (-5)^2 + (6)^2} = \sqrt{16 + 25 + 36} = \sqrt{77} $$
$$ ds = |\mathbf{r}'(t)| \, dt = \sqrt{77} \, dt $$

3. Substitute the parametric equations into the integrand:
$$ x(t) = 4t, \quad y(t) = 6 - 5t, \quad z(t) = -1 + 6t $$
$$ (x(t))^2 z(t) = (4t)^2 (-1 + 6t) = 16t^2 (-1 + 6t) $$

4. Compute the integral:
$$ \int_{0}^{1} 16t^2 (-1 + 6t) \sqrt{77} \, dt $$

This integral can be evaluated using standard techniques, such as integration by parts or u-substitution.

After integrating, you will obtain the value of the line integral.