LAST NAME :
FIRST NAME :
QUIZ 5, 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 + y cos x, sin x + x cos yi.
(a) [25pts] Show that F is conservative.
(b) [25pts] Find a potential function of F (show your work).
I
2. [50pts] Use the Green’s Theorem to compute
−xydx + y 2 dy where C is the triangle of vertices
C
(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 y + cos x where P (x, y) = sin y + y cos x and Q(x, y) = sin x +
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) = sin y + y cos x and fy (x, y) = sin x + x cos y. Let’s begin by integrating fx (x, y) =
sin y + y cos x w.r.t x, we obtain fx (x, y) = x sin 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) =
(x sin y+y sin x+g(y)) = x cos y+sin x+g 0 (y) = fy (x, y) = Q(x, y) = sin x+x cos y.
∂y
∂y
We have x cos y +sin x+g 0 (y) = sin x+x cos y so g 0 (y) = 0 and by integrating w.r.t y, g(y) = Cst.
A potential function f for F is f (x, y) = x sin y + y sin x (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