Question

Evaluate the line integral over C ∫(x + y)ds where C is the line segment from (0,1,1) to (3,2,2). I've parameterized C as 3ti + (1+t)j + (1+t)k, and I get the result 3√(11). Is that correct? Thanks!!

Answer 1

Yes, you are correct!!!

The result is
`\int\limits^(1)_(0) {4t+1dt} \,=3√(11)`

Explanation:

The given line integral is

`\int\limits_C {x+y} \, ds`, where C is the line segment from
`(0,1,1)` to
`(3,2,2)`.

First we need to get the direction vector,

`d=\:<\:3,2,2\:>-\:<\:0,1,1\:>`

`d=\:<\:3,1,1\:>`

The parametric equation now becomes,

`r(t)=\:<\:0,1,1\:>+\:t<\:3,1,1\:>`.

`\Rightarrow r(t)=3ti+(1+t)j+(1+t)k`.

We parameterize the curve with the equations,

`x=3t` and
`y=t+1`

`ds=|r'(t)|dt`

`\Rightarrow ds=|3i+j+k|dt`

`\Rightarrow ds=√(3^2+1^2+1^2)dt`

`\Rightarrow ds=√(11)dt`.

The line integral now becomes,

`\int\limits^(1)_(0) [{3t+(t+1)]√(11)dt} \,`

`√(11)\int\limits^(1)_(0) {4t+1dt} \,=3√(11)`