LAST NAME : FIRST NAME : QUIZ 7, Version A : MATH 251, Section 506 last name : . . . . . . . . first name : . . . . . . . . GRADE : . . . . . . . . ”An Aggie does not lie, cheat or steal, or tolerate those who do” signature : . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Write up your result, detail your calculations if necessary and BOX your final answer 1. Given F (x, y) = hsin y, x cos y + sin yi. (a) [25pts] Show that F is conservative. (b) [25pts] Find a potential function of F . 2. [50pts] Use the Green’s Theorem to compute (0, 1) and (1, 1). H C −xydx+ydy where C is the triangle of vertices (0, 0), 1. F =< P, Q > is a vector field in R2 whose every component function is defined in R2 and has also continuous first partial derivatives. (a) Since the domain of F is R2 which is simply connected . We need to check that Qx (x, y) = Py (x, y) for every (x, y) in R2 : Py (x, y) = cos y and Qx (x, y) = cos y. So, F is conservative. (b) Let f be a potential function which satisfies fx (x, y) = sin y, and fy (x, y) = x cos y + sin y. We integrate first fx with respect to x and obtain f (x, y) = x sin y+g(y) where g is a differentiable function. So, now we differentiate this function f w.r.t y and we obtain fy (x, y) = x cos y + g 0 (y) which needs to be equal to Q(x, y) = x cos y + sin y. By identification, we obtain g 0 (y) = sin y which implies g(y) = − cos y + Cst. Then, a potential function is f (x, y) = x sin y − cos y , (by choosing Cst = 0). 2. Let P (x, y) = −xy and Q(x, y) = y. By Green’s theorem, by denoting D = {(x, y) | 0 ≤ x ≤ 1, x ≤ y ≤ 1} the domain bounded by C, we have Z Z Z Z Z −xydx + ydy = (Qx − Py )dA = xdA, C Z D 1Z 1 = xdydx = 0 = D Z 1 . 6 x 0 1 x2 x3 x(1 − x)dx = − 2 3 1 , 0