What is the potential function for a line integral with a vector field?

[tomelwood]

Homework Statement

I have to calculate the following line integral

\int_{\gamma}y^{2}cos(xy^{2})dx + 2xycos(xy^{2})dy where \gamma is the path defined by the equations x(t) = t^{4} and y(t)=sin^{3}(\frac{t\pi}{2}) t between 0 and 1

Homework Equations

Now I know that the formula for calculating this is the integral over where gamma is defined (ie 0 to 1) of F(\gamma(t))\bullet\gamma'(t)dt where \bullet is the scalar product.

The Attempt at a Solution

Therefore wherever I see an x in the original integral, I can substitute it for t^{4} and similarly for the y subbing with the sin expression. And then instead of dx I put x'(t) and instead of dy I put y'(t) and integrate everything between 0 and 1.

The problem is is that this yields the following horrendous expression, which I don't know how to integrate.
\int^{1}_{0}(sin^{6}(\frac{t\pi}{2})cos(t^{4}sin^{6}(\frac{t\pi}{2}))4t^{3}+2t^{4}sin^{3}(\frac{t\pi}{2})cos(t^{4}sin^{6}(\frac{t\pi}{2}))\frac{3\pi}{2}sin^{2}(\frac{t\pi}{2})cos(\frac{t\pi}{2})) dt

Where have I gone wrong?
Hopefully the Latex works!

 

[hunt_mat]

Write it out carefully:

<br /> dx=x&#039;(t)dt=3t^{3}dt,\quad dy=y&#039;(t)dt=\frac{3\pi}{2}\sin^{2}\left(\frac{t\pi}{2}\right)\cos\left(\frac{t\pi}{2}\right) dt<br />

Also:

<br /> y^{2}=\sin^{6}\left(\frac{t\pi}{2}\right) ,\quad 2xy=2t^{4}\sin^{3}\left(\frac{t\pi}{2}\right)<br />

 

[grey_earl]

You have done nothing wrong! The indefinite integral of your monster expression is
\sin\left(t^4 \sin^6\left(\frac{t \pi}{2}\right)\right), although I don't know any easy way to get to this. However, the proof that it is an indefinite integral is easy, but I'm sure you know this :)
BTW: In LaTeX, you write \sin instead of sin, and using \left( and \right) makes your parentesis the right height.

@hunt_mat: x'(t) = 4 t^3

 

[hunt_mat]

.

@hunt_mat: x'(t) = 4 t^3

Well spotted! My bad.

 

[tmiddlet]

This problem illustrates how nice it can be to notice that the vector field in this problem is a vector field. If you notice that the potential function is f(x,y) = \sin(xy^2) + c
then it is easy to find an answer. The answer found this way would agree with @grey_earl's indefinite integral.