LAST NAME : FIRST NAME : QUIZ 7, Version A : MATH 251, Section 505 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) = hy cos x, sin x + cos yi. (a) [25pts] Show that F is conservative. (b) [25pts] Find a potential function of F (show your work). H 2. [50pts] Use the Green’s Theorem to compute C −xydx + y 2 dy where C is the triangle of vertices (0, 0), (0, 1) and (1, 1). (a) F is a vector field defined on R2 , which is a simply connected domain . And Py (x, y) = Qx (x, y) = cos x where P (x, y) = y cos x and Q(x, y) = sin x + cos y, so F is conservative. (b) Since F is conservative, there exists a potential function f such that F (x, y) = ∇f (x, y) i.e fx (x, y) = y cos x and fy (x, y) = sin x + cos y. Let’s begin by integrating fx (x, y) = y cos x w.r.t x, we obtain fx (x, y) = y sin x + g(y) where g is a differentiable function in R. Now, differentiate the function previously obtained w.r.t y and identify it to Q(x, y) : ∂ ∂ f (x, y) = (y sin x + g(y)) = fy (x, y) = Q(x, y) = sin x + cos y. ∂y ∂y We have sin x+g 0 (y) = sin x+cos y so g 0 (y) = cos y and by integrating w.r.t y, g(y) = sin y +Cst. A potential function f for F is f (x, y) = y sin x + sin y (by taking Cst = 0). 1. Since C is a simple closed curve enclosing a domain D = {(x, y) | 0 ≤ x ≤ 1, Theorem where P (x, y) = −xy and Q(x, y) = y 2 , I Z Z Z Z 2 −xydx + y dy = Qx − P ydA = xdA C Z D 1Z 1 = D Z xdydx = 0 1 = . 6 x x ≤ y ≤ 1}, by Green’s 0 1 [xy]1x dx Z = 0 1 x2 x3 x − x dx = − 2 3 2 1 0