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Question: How do you evaluate the following line integral \( ({x^2})zds \) ,where C is the line segment from t...

How do you evaluate the following line integral (x2)zds({x^2})zds ,where C is the line segment from the point (0,6,−1)(0,6, - 1) to the point (4,1,5)(4,1,5) ?

Explanation

Solution

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 PP and QQ as vectors. Consider a vector v⃗\vec v such that it passes from P to Q. To calculate the value we subtract the coordinates of PP to the coordinates of QQ and we get,
v⃗=(4,−5,6)\vec v = (4, - 5,6)
Now, let us consider a vector c⃗(t)\vec c(t) such that
c⃗(t)=P+tv⃗ c⃗(t)=(0,6,−1)+t(4,−5,6) c⃗(t)=(4t,6−5t,−1+6t)   \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⩽t⩽10 \leqslant t \leqslant 1 .
The velocity vector is c⃗′(t)=v⃗\vec c'(t) = \vec v and its length is
∥v⃗∥=16+25+36 ∥v⃗∥=77   \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)=x2zf(x,y,z) = {x^2}z where we have to integrate as t goes from t=0t = 0 to t=1t = 1 is
f(4t,6−5t,−1+6t)∥v⃗∥=77(4t)2.(−1+6t) f(4t,6−5t,−1+6t)∥v⃗∥=77(16t2).(−1+6t) f(4t,6−5t,−1+6t)∥v⃗∥=77(96t3−16t2)   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=dsdtdt=∥v⃗∥dtds = \dfrac{{ds}}{{dt}}dt = \left\| {\vec v} \right\|dt
Using integral calculation:
∫0177(96t3−16t2)dt=77[24t4−163t3]t=0t=1 =77.72−163=56773≈163.8   \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 163.8163.8 .
So, the correct answer is “ 163.8163.8 ”.

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.