Line Integral of a parametric curve

[chrisy2012]

Homework Statement

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)

Homework Equations

$$\int_{C}^{}f(x,y)ds=\int_{a}^{b}f(x(t),y(t))\sqrt((\frac{dx}{dt})^2+(\frac{dy}{dt})^2)dt$$

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?

 

[DryRun]

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.

 

[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.

 

[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

 

[DryRun]

Post the entire question correctly.

 

[chrisy2012]

The entire question is posted correctly.

 

[DryRun]

The entire question is posted correctly.

Homework Statement

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)

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