Triangle line integral

[xmflea]

Let C be the triangle in the plane from (0,0) to (1,1) to (0,1) back to (0,0). evaluate the line integral of f along C if f(x,y)=(x+2y).

attempt:
C1: x=t y=t ds=root(1)dt, integrated t+2t from 0 to 1 to get 3/2
C2: x=-t ds=root(1)dt, integrated -t from -1 to 0 to get 1/2
C3: y=-t, ds=root(1)dt integrated -2t from -1 to 0 to get 1.
added up 3/2, 1/2, and 1 to get the wrong answer.
the correct answer is (1/2)(3root2 + 7)

looks like I am parametizing completely wrong. so how do i go about this? thanks.

 

[Cyosis]

For starters regarding path C1, c'(t)=(1,1). What is the length of the vector (1,1)? The rest of C1 is correct.

For path C2 c(1) should yield (1,1). In your case c(1)=(1,k), with k an undetermined parameter since you didn't parametrize y for this side of the triangle.

For path C3 a similar problem occurs.

You want to find yourself a parametrization of the following form:

$$c(t) = \left\{ \begin{matrix} (t,t) & 0 \leq t \leq 1 \\ (..,..) & 1 \leq t \leq 2 \\ (..,..) & 2 \leq t \leq 3 \end{matrix} \right$$

What you have been doing so far is making some weird jumps in time for your C2 and C3 paths. I suggest you draw the triangle and seeing as all sides are straight lines use y=at+b to parametrize each segment.

Example for the side (1,1)->(0,1)

We know that the y coordinate remains constant and equals 1. Therefore our parametrization will be of the form (x(t),1). We now need to find x(t). x(t) decreases linearly between 1 and 0 so its equation will look like x=at+b. For t=1 we know that x=1 therefore 1=a+b. For t=2 we know that x=0 therefore 0=2a+b. Solving for a and b will yield x(t)=2-t therefore c(t)=(2-t,1).