Question

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) $ ?

Answer 1

Hint: An integral in which the function to be integrated is determined along a curve in the coordinate system is known as a line integral. The function to be integrated may be a scalar or a vector field. A scalar-valued function or a vector-valued function may be integrated along a curve. The line integral's value can be calculated by adding the values of all the points on the vector field.

Complete step by step solution:
We should parameterize the line segment first. The easiest way to do this is to represent $ P $ and $ Q $ as vectors. Consider a vector $ \vec v $ such that it passes from P to Q. To calculate the value we subtract the coordinates of $ P $ to the coordinates of $ Q $ and we get,
 $ \vec v = (4, - 5,6) $
Now, let us consider a vector $ \vec c(t) $ such that
 $
  \vec c(t) = P + t\vec v \\
  \vec c(t) = (0,6, - 1) + t(4, - 5,6) \\
  \vec c(t) = (4t,6 - 5t, - 1 + 6t) \;
  $
Where $ 0 \leqslant t \leqslant 1 $ .
The velocity vector is $ \vec c'(t) = \vec v $ and its length is
 $
  \left\| {\vec v} \right\| = \sqrt {16 + 25 + 36} \\
  \left\| {\vec v} \right\| = \sqrt {77} \;
  $
Now, let us assume a function such that $ f(x,y,z) = {x^2}z $ where we have to integrate as t goes from $ t = 0 $ to $ t = 1 $ is
 $
  f(4t,6 - 5t, - 1 + 6t)\left\| {\vec v} \right\| = \sqrt {77} {(4t)^2}.( - 1 + 6t) \\
  f(4t,6 - 5t, - 1 + 6t)\left\| {\vec v} \right\| = \sqrt {77} (16{t^2}).( - 1 + 6t) \\
  f(4t,6 - 5t, - 1 + 6t)\left\| {\vec v} \right\| = \sqrt {77} (96{t^3} - 16{t^2}) \;
  $
Now, we know that $ ds = \dfrac{{ds}}{{dt}}dt = \left\| {\vec v} \right\|dt $
Using integral calculation:
 $
  \int\limits_0^1 {\sqrt {77} } (96{t^3} - 16{t^2})dt = \sqrt {77} [24{t^4} - \dfrac{{16}}{3}{t^3}]_{t = 0}^{t = 1} \\
   = \sqrt {77} .\dfrac{{72 - 16}}{3} = 56\dfrac{{\sqrt {77} }}{3} \approx 163.8 \;
  $
Hence, the answer is $ 163.8 $ .
So, the correct answer is “ $ 163.8 $ ”.

Note: The following are few examples of line integral applications of vector calculus. The mass of wire is calculated using a line integral. It aids in the calculation of the wire's moment of inertia and centre of mass. It is used to calculate the magnetic field around a conductor in Ampere's Law. A line integral is used in Faraday's Law of Magnetic Induction to calculate the voltage produced in a circle.