Line Integral of a Vector Field: Lang (5.2.2)

Here’s a problem (5.2.2, p. 58) from Lang’s “A Second Course in Calculus”:

Compute the line integral of the vector field over the indicated curve. (x, y, xz - y) over the line segment from (0, 0, 0) to (1, 2, 4).

x(t) = x₀ + (x₁ – x₀)t

y(t) = y₀ + (y₁ – y₀)t

z(t) = z₀ + (z₁ – z₀)t

Here, t varies between 0 and 1. When t is 0, you get (0, 0, 0); when t is 1, you get (1, 2, 4). Let’s get the parametrization now:

x(t) = 0 + (1 - 0)t = t

y(t) = 0 + (2 - 0)t = 2t

z(t) = 0 + (4 - 0)t = 4t

Therefore,

r(t) = (t, 2t, 4t)

r’(t) = (1, 2, 4)

            Now, we will substitute t for x, 2t for y, and 4t for z in the vector field:

F(r, t) = (x, y, xz - y)

F(r, t) = (t, 2t, 4t² – 2t)

            Next, we have to take the dot product of the vector field along the curve with the derivative of the parametrization:

F(r(t)) *  r’(t)

(t, 2t, 4t² – 2t) * (1, 2, 4)

Using the definition of the dot product, we get:

t + 4t + (4t² – 2t)(4)

5t+ 16t² – 8t

16t² – 3t

We can now compute the integral of this dot product over the interval [0, 1].

Therefore, the answer is 23/6.

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