# Evaluating a line integral

1. Apr 9, 2013

### dumbperson

1. The problem statement, all variables and given/known data
P is the part of the curve 9y²=4x³ between the points (1,-2/3) and (1,2/3).

Evaluate the integral $$\int_P x ds$$

2. Relevant equations

3. The attempt at a solution

$$\int_P x ds = \int_P x |r'(t)| dt$$

My problem is that I cannot find a right parametrization r(t) for this.

I tried y=t, then x=((3/2)t)^(2/3)

but thent goes from -2/3 to 2/3, but that gives a complex x for t=-2/3

Tried it with x=3t² and y=2sqrt(3)t³, I get the same problem.

Any tips ? Thanks!

2. Apr 9, 2013

### Staff: Mentor

How about x = t and y = ±(2/3)t3/2?

3. Apr 9, 2013

### dumbperson

so split the integral up in 2 parts?

First along the path

$$r(t)=t\hat{x} -\frac{2}{3}t^{\frac{3}{2}}\hat{y}$$ from t=1 to t=0, then

$$r(t)=t\hat{x} + \frac{2}{3}t^{\frac{3}{2}}\hat{y}$$ from t=0 to t=1

$$|r'(t)| = \sqrt{1+t}$$

So the integral becomes

$$\int^1_0 t \sqrt{1+t} dt + \int^0_1 t \sqrt{1+t} dt = 0$$

Is this correct? Zero?

4. Apr 9, 2013

### SammyS

Staff Emeritus
I would be inclined to use the parametrization, $\displaystyle \ y=t\,,\ \text{ and }\ x=\sqrt[3]{\frac{9t^2}{4}}\ .$

... but I'm not sure how messy the resulting integral is.

5. Apr 9, 2013

### dumbperson

Could zero be the answer? It doesn't feel right,but is it possible for a line integral like that to be zero?

6. Apr 9, 2013

### LCKurtz

Yes, in general it is possible for ds line integrals to come out to zero. But not in this case. Your integrand is nonnegative and $ds$ is always positive. The problem is that when you use the formula $ds =|\vec r'(t)|dt$ you are assuming the curve is parameterized in the positive direction. Since that is not the case in your second integral, it needs a negative sign in front or, what is the same thing, the limits reversed.

7. Apr 9, 2013

### dumbperson

Ahh, thanks.

So the integral becomes

$$2\int^1_0 t\sqrt{1+t} dt$$

u= 1+t, du= dt, t=u-1

$$2\int^{u(1)}_{u(0)} (u^{\frac{3}{2}}-u^{\frac{1}{2}}) du = [\frac{4}{5}u^{\frac{5}{2}}-\frac{4}{3}u^{\frac{3}{2}}]^{{2}}_{1} =$$

Is this alright then?

8. Apr 9, 2013

### LCKurtz

Yes, except the final simplified answer seems to be missing...