EXAM 3 MATH 251
SPRING 2016
INSTRUCTOR: KEVIN KORDEK
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INSTRUCTOR: KEVIN KORDEK
(3) (10 points) Suppose C is the path in R2 parametrized by r(t) = hsin7051 (t), cos899 (t)i with
π
0 ≤ t ≤ . Compute the line integral
2
Z
F · dr
C
where F =
hesin(y) , x cos(y)esin(y) i.
Solution The large numbers in the problem were meant to deter you from trying to do the integral using a parametrization. What we need to observe is that the vector field F is conservative.
To check this, we compute
∂ x cos(y)esin(y) = cos(y)esin(y)
∂x
and notice that it equals
∂ sin(y) e
= cos(y)esin(y) .
∂y
This implies that F has a potential function, i.e. there is a function f such that ∇f = F. To
find it, we set up the equations
fx = esin(y)
fy = x cos(y)esin(y) .
If we anti-differentiate the first equation with respect to x, we find that f = xesin(y) + h(y), for
some function h(y) that depends only on y. If we insert this expression for f into the equation
fy = x cos(y)esin(y) , we find that
dh
∂f
= x cos(y)esin(y) +
= x cos(y)esin(y) .
∂y
dy
dh
This implies that
= 0. In other words, h = C, where C is some constant. Now we can
dy
conclude that f = xesin(y) + C.
We can now solve the problem using the Fundamental Theorem of Line Integration. Notice that
r(t) parametrizes a path that starts at (0, 1) and ends at (1, 0). The theorem then tells us that
Z
Z
F · dr =
∇f · dr = f (1, 0) − f (0, 1)
C
C
= (1 + C) − (0 + C)
= 0.
EXAM 3 MATH 251
SPRING 2016
3
Z Z
curl(F) · ndS, where S is the part of the
(8) (15 points) Use Stokes’ Theorem to evaluate
S
parabaloid z = 1 − x2 − y 2 that satisfies z ≥ 0 and F = h2xy, y 2 , z 2 i.
Solution
The surface S is the part of the parabaloid z = 1 − x2 − y 2 that lies above the xy-plane. Its
boundary C is equal to the intersection of the plane z = 0 with z = 1 − x2 − y 2 .
This is a circle in the xy-plane with equation 0 = 1 − x2 − y 2 , i.e. x2 + y 2 = 1. Since we
assume that S has the outward orientation, C is oriented counterclockwise. The boundary C
is therefore parametrized by r(t) = hcos(t), sin(t), 0i, where 0 ≤ t ≤ 2π.
Z
Z Z
F · dr. Notice that
curl(F) · n dS =
Stokes’ Theorem says that
S
C
dr = h− sin(t), cos(t), 0idt.
We now have everything we need to solve the problem.
Let’s compute:
Z
Z
2π
h2 cos(t) sin(t), sin2 (t), 0i · h− sin(t), cos(t), 0idt
F · dr =
0
C
Z
=
2π
− sin2 (t) cos(t)dt
0
= −
=0
1
sin3 (t)|2π
0
3