Question

evaluate the line integral, where c is the given space curve. xeyz ds, c is the line segment from (0, 0, 0) to (3, 2, 4)

Answer 1

Evaluating the line integral, where c is the given space curve. xeyz ds, c is the line segment from (0, 0, 0) to (3, 2, 4) is; 30008.8

How to evaluate the Line Integral?

We need the derivative of position vector to solve this integral. The first step will be to evaluate the value of the function F(x,y,z) in terms of the given parametric curve r(t) then multiply it with the magnitude of its derivative i.e. |r'(t)| and then integrate it with the given limits of t.

f(x, y, z) = xe^(yz)

Now the integral;. $\int\limits^._c {f(x,y.z)} \, ds = \int\{f(rt)} \, * |r'(t)| dt$

As C is line passing through (0,0,0) and (3,2,4) vector for parametrization is;⟨3 − 0, 2 − 0, 4 − 0⟩v = ⟨3, 2,4⟩

Parametrization is;x = 0 + 3t, y = 0 + 2t,z = 0 + 4t

Thus;r(t) = ⟨3t, 2t, 4t⟩taking the derivative r′(t) = ⟨3, 2 ,4⟩ |r'(t)| = √(9 + 4 + 16) |r'(t)| =√29f(r(t)) = 3te^(8t²)

Integrating with this formula . $\int\limits^._c {f(x,y.z)} \, ds = \int\{f(rt)} \, * |r'(t)| dt$

Using online derivative calculator gives us;

$\int\limits^._c {f( xe^{yz})} \, ds

= 30008.8$

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