First, let's check to see if it's a conservative vector field -- if it is, we don't have to mess with parametrizing that slightly nasty curve. Comparing the partial with respect to y of i's coefficient, which is 2x, to the partial with respect to x of j's coefficient, which is 2x, we see they're the same. Yay! We can do it the short way.

Now we look for a potential function. The antiderivative of i's coefficient with respect to x is x²y. This alone won't produce the whole coefficient of j -- it gets the x² part, but we need the -6y part as well. We'll get that if our potential function also has a -3y² term, so try f(x,y) = x²y - 3y² as a potential function. If you double-check, it does have our original integrand as its gradient, so it works.

All we need to do now is (because of the Fun. Thrm. for Line Integrals): f(0,-2) - f(3,0) = -12 - 0 = -12, so the value of the original integral is -12.
 
 
2. Compute  for a path C given by r = <t³, t⁴> for 0  t  1.

This is problem 17 from Stewart section 14.2, p. 906, but with a sign changed.

In this case there's no potential function (the mixed partials, x² and y, don't match) so we do it the long way.

I. The path is already parametrized for us.

II. We build F(r(t)) = <(t³)²(t⁴), (t³)(t⁴)>.

III. We build r'(t) = <3t², 4t³>.

IV. We put them together and use the bounds from our parametrization as limits of integration to form .