Determine if the vector filed Fxy is conservative If it is f

Solution

Let P(x, y) = 1 + 2xy + ln x and Q(x, y) = x2. Then Py = 2x and Qx = 2x. Since Py = Qx, and the domain of F is {(x, y)| x > 0} which is open and simply connected, F is conservative. This means there is a function f such that F = f, or, fx(x, y) = 1 + 2xy + ln x and fy(x, y) = x2 Integrate 1 + 2xy + ln x with respect to x to get f(x, y) = x + x2y + x ln x - x + g(y) = x2y + x ln x + g(y), where g(y) is the constant of integration with respect to x. (Recall that the integral of ln x is x ln x - x. Use integration by parts.) To find g(y), take the derivative of f(x, y) = x2y + x ln x + g(y) with respect to y: fy(x, y) = x2 + g\'(y) Comparing this to fy(x, y) = x2 above, we see that g\'(y) = 0 so g(y) is K, a constant. The potential function for F is f(x, y) = x2y + x ln x + K.