Question

Evaluate line integral of (x+y) ds where C is the straight-line segment x=t, y=(1-t), z=0, from (0,1,0) to (1,0,0)

Answer 1

In order to evaluate the line integral, we have to express everything in terms of the parameter t. Since we have x, y, and z in terms of t already, we specifically need to worry about ds.

The small piece of the curve C associated with the small changes dx, dy, and dz has length
`ds = √(dx^2 + dy^2 + dz^2)`. Using this, we can represent the length of ds with the associated change in t as
`ds = \sqrt{((dx)/(dt))^2 + ((dy)/(dt))^2 + ((dz)/(dt))^2 } \ dt`.

What are the limits of integration in terms of t? By drawing the curve and/or plugging in the given points into the equations for x, y, and z in terms of t, we can see that the curve C is traversed by t as it goes from 0 to 1.

Putting all this together and evaluating, we get


`\int\limits_C {x(t) + y(t)} \, ds`

`= \int\limits_C {(x(t) + y(t)) \sqrt{((dx)/(dt))^2 + ((dy)/(dt))^2 + ((dz)/(dt))^2} \, dt`

`= \int\limits^1_0 {(t + (1 - t)) \sqrt{(1)^2 + ({-1})^2 + (0)^2}} \, dt`

`= \int\limits^1_0 {√(2)} \, dt`

`= \bf √(2)`