Line Integral of a parametric curve

1. May 11, 2012

chrisy2012

1. The problem statement, all variables and given/known data
Evaluate the line integral over the curve C
$$\int_{C}^{}e^xdx$$
where C is the arc of the curve
$$x=y^3$$
from (-1,-1) to (1,1)
2. Relevant equations
$$\int_{C}^{}f(x,y)ds=\int_{a}^{b}f(x(t),y(t))\sqrt((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)dt$$
3. The attempt at a solution
I tried parametrizing the curve to y=t and x=t^3
therefore dy/dt = 1 and dx/dt = 3t^2
plug this back into the formula, we get
∫ from -1 to 1 (e^t^3)sqrt(3t^2+1)dt
but this is an insolvable integral, anything I did wrong or is there another way?

2. May 11, 2012

sharks

Your line integral is obviously wrong. You might have typed it incorrectly. It should be:
$$\int_{C}^{}e^xds$$
$$\frac{dy}{dx}=\frac{1}{3y}$$
$$\int_{C}^{}f(x,y)ds=\int_{x=a}^{x=b}f(x,y)\sqrt {1+(\frac{dy}{dx})^2}\,.dx$$
Afterwards, use the substitution $\frac{1}{3x^{1/3}}=\tan \theta$ and evaluate your line integral.

Last edited: May 11, 2012
3. May 11, 2012

jackmell

Keep in mind there are three basic ways of integrating over a curve (line integral), you can integrate over the arc-length (ds), but also over the shadow of the curve along the x and y axis (by dx or dy). Now you wrote it as dx, so that's just a regular integral:

$$\int_C f(x,y)dx=\int_C f(x,y(x))dx=\int_{-1}^1 e^x dx$$

If it were ds, then you'd need to use the formula you posted.

4. May 11, 2012

chrisy2012

But I am suppose to integrate over the arclength(ds). I'm suppose to parametrize the curve with respect to t so that the curve imoves along 1 unit of length per unit of time. That's how I got the bounds for the integral

5. May 11, 2012

sharks

Post the entire question correctly.

6. May 11, 2012

chrisy2012

The entire question is posted correctly.

7. May 11, 2012

sharks

In that case, you won't have any use for the relevant equations that you've posted.