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

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.