Question

Evaluate the line integral, where C is the given curve. C (x yz) dx 2x dy xyz dz, C consists of line segments (3, 0, 1) to (4, 3, 1) and from (4, 3, 1) to (4, 5, 4)

Answer 1

Split up C into two component paths C₁ and C₂, where each line segment is respectively parameterized by

r₁(t) = (1 - t ) (3i + k) + t (4i + 3j + k) = (t + 3) i + 3t j + k

r₂(t) = (1 - t ) (4i + 3j + k) + t (4i + 5j + 4k) = 4i + (2t + 3) j + (3t + 1) k

both with 0 ≤ t ≤ 1.

It's a bit unclear what function you're supposed to integrate (looks like xyz ?) so I'll give a more general result. The line integral of a scalar function f(x, y, z) along the given path C is

`\displaystyle \int_C f(x,y,z)\,\mathrm ds = \int_(C_1)f(\mathbf r_1(t))\left\|(\mathrm d\mathbf r_1)/(\mathrm dt)\right\|\,\mathrm dt + \int_(C_1)f(\mathbf r_2(t))\left\|(\mathrm d\mathbf r_2)/(\mathrm dt)\right\|\,\mathrm dt`

We have

dr₁/dt = i + 3j ==> || dr₁/dt || = √(1² + 3²) = √10

dr₂/dt = 2j + 3k ==> || dr₂/dt || = √(2² + 3²) = √13

Then the integrals reduce to

`\displaystyle \int_0^1 \left(√(10)\,f(t+3,3t,1) + √(13)\,f(4,2t+3,3t+1)\right)\,\mathrm dt`

If indeed f(x, y, z) = xyz, then we have

`\displaystyle \int_0^1 \left(3√(10)\,t(t+3) + 4√(13)\,(2t+3)(3t+1)\right)\,\mathrm dt = 11√(\frac52)+42√(13)`