In order to use the fundamental theorem of line integrals, you need to find a scalar potential function - that is, a scalar function f(x, y) for which
grad f(x, y) = F(x, y)
This amounts to solving for f such that
∂f/dx = x + 3
∂f/∂y = 6y + 3
Integrating both sides of the first equation with respect to x gives
f = 1/2 x ^2 + 3x + g(y)
Differentiating with respect to y gives
∂f/∂y = dg/dy = 6y + 3
Solving for g gives
g = ∫ (6y + 3) dy = 3y ^2 + 3y + C
and hence
f(x, y) = 1/2 x ^2 + 3x + 3y ^2 + 3y + C
(a) By the fundamental theorem, the integral of F along any path starting at the point P (1, 0) and ending at Q (3, 3) is
∫ F(x, y) • dr = f (3, 3) - f (1, 0) = 99/2 - 7/2 = 46
(b) Now we're talking about a closed path, so the integral is simply 0. We can verify this by checking the integral over the origin-containing paths:
• From the origin to P :
∫ F(x, y) • dr = f (1, 0) - f (0, 0) = 7/2 - 0 = 7/2
• From Q back to the origin:
∫ F(x, y) • dr = f (0, 0) - f (3, 3) = 0 - 99/2 = -99/2
Then the total integral is 7/2 + 46 - 99/2 = 0, as expected.
