Evaluating the line integral for a specific curve

[dumbperson]

Homework Statement

P is the part of the curve 9y²=4x³ between the points (1,-2/3) and (1,2/3).

Evaluate the integral $$\int_P x ds $$

Homework Equations

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!

 

[Mark44]

Homework Statement

P is the part of the curve 9y²=4x³ between the points (1,-2/3) and (1,2/3).

Evaluate the integral $$\int_P x ds $$

Homework Equations

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!

How about x = t and y = ±(2/3)t^3/2?

 

[dumbperson]

How about x = t and y = ±(2/3)t^3/2?

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?

 

[SammyS]

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.

 

[dumbperson]

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

 

[LCKurtz]

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?

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

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.

 

[dumbperson]

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.

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?

 

[LCKurtz]

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?

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