Question

In Exercises 17 and 18 , evaluate \(\int_{C}\left(2 x+y^{2}-z\right) d s\) along the given path. \(C:\) line segments from (0,0,0) to (1,0,0) to (1,0,1) to (1,1,1)

Short answer

The value of the line integral \(\int_{C}\left(2 x+y^{2}-z\right) d s\) along the given path is zero. This comes from the result of summing the computed integrals for each line segment: 1, -1, and 0 respectively.

Step-by-step solution

Parameterize the Line Segments

Define three parameterized vectors for each line segment of the path. For the segment from (0,0,0) to (1,0,0), let \(r_1(t) = t\hat{i}\) where \(0 \leq t \leq 1\). For the segment from (1,0,0) to (1,0,1), let \(r_2(t) = \hat{i} + t\hat{k}\) where \(0 \leq t \leq 1\). For the segment from (1,0,1) to (1,1,1), let \(r_3(t) = \hat{i} + t\hat{j} + \hat{k}\) where \(0 \leq t \leq 1\).

Compute the Line Integral for Each Segment

Calculate the line integral for each segment. For the first segment, substitute \(x = t, y = 0, z = 0\) into the integrand, and integrate from 0 to 1. For the second segment, substitute \(x = 1, y = 0, z = t\) into the integrand, and integrate from 0 to 1. For the third segment, substitute \(x = 1, y = t, z = 1\) into the integrand, and integrate from 0 to 1.

Sum the Results

Add the results of the integrals computed in the previous step. This gives the total line integral along the given path.