# Computing a line integral

1. Jan 11, 2013

### Zondrina

1. The problem statement, all variables and given/known data

Let C be the arc $x=t^2, \space y=2t, \space z= \sqrt{4+3t}$ for $t \in [-1,0]$

Evaluate the line integral :

$\int_{C} z^2dx + \sqrt{x}dy - 4xyz dz$

2. Relevant equations

$\int_{C} f(P) dx = \int_{a}^{b} f(P(t)) x'(t) dt$ for $t \in [a,b]$

3. The attempt at a solution

So I have not computed one of these before. I know I'm supposed to parametrize ( choose a parameter ) and then compute my according integral. I really only need help setting it up, so here goes nothing :

This question already has a nice parameter t picked out for me on an appropriate interval. Since the curve is an arc, we know it is one to one, i.e there are no self intersections.

So we compute :

$dx = 2tdt, \space dy = 2dt, \space dz = \frac{1}{2} (4+3t)^{-1/2} dt$

*I'm not sure I should be concerned with smoothness here*.

Thus our integral becomes :

$\int_{C} z^2dx + \sqrt{x}dy - 4xyz dz = \int_{-1}^{0} (6t^2+8t)dt + (2t)dt - (4t^3)dt$

So after computing everything and setting it all up that's what I think I should get. Then computing the integral is a simple matter. I just want to make sure I'm not jumping too quickly missing anything and if anyone could verify it would be much appreciated.

2. Jan 11, 2013

### Dick

You've got the right procedure. You don't need to check that it's 1-1. You do want it to be differentiable, but it clearly is. You are missing some details. Check dz again, and I don't think sqrt(x) is t. Remember t is negative.

3. Jan 11, 2013

### Zondrina

Silly me, I forgot my chain rule for a second there.

Also, x = t2, so sqrt(x) = t.

4. Jan 11, 2013

### haruspex

You expression for dz is not quite right. Other than that it looks ok.

5. Jan 11, 2013

### Dick

t is NEGATIVE. I don't think the square root of t^2 can be negative.

6. Jan 11, 2013

### Zondrina

The most common mistake in math I believe... I see why it's -t now.

7. Jan 12, 2013

### HallsofIvy

Staff Emeritus
If "t in [-1, 0]" bothers you let s= -t. Then $x= t^2= s^2$, $y= 2t= -2s$, and $z= \sqrt{4+ 3t}= \sqrt{4- 3s}$ with s in [0, 1].

8. Jan 12, 2013

### Zondrina

Ohh I understand now. I had a question to prove this in general in my tutorial that I never had a chance to do.