Quiz 5
Math 0240
Spring 2014
Solutions
Problem: Show that F̄ = h2ye2x , e2x + 2yi is a conservative vector field and find its potential
function.
Solution:
P (x, y) = 2ye2x ,
concervative.
Q(x, y) = e2x + 2y,
Py = 2e2x = Qx . Hence the field is
fx = 2ye2x , f = ye2x + g(y), fy = e2x + g 0 (y) = e2x + 2y, g 0 (y) = 2y, g(y) = y 2 + C,
f (x, y) = ye2x + y 2 + C.
Z
Problem: Show that the line integral I =
tan y dx + x sec2 y dy is independent of path and
C
evaluate the integral if C is any path from (1, 0) to (2, π/4).
Solution:
P (x, y) = tan y,
concervative.
Q(x, y) = x sec2 y,
Py = sec2 y = Qx . Hence the field is
fx = tan y, f = x tan y + g(y), fy = x sec2 y + g 0 (y) = x sec2 y, g 0 (y) = 0, g(y) = C,
f (x, y) = x tan y. Then I = f (2, π/4) − f (1, 0) = 2 − 0 = 2.
I
(tan x − 2y) dx + (x − 4y sec y) dy,
Problem: Evaluate the integral I =
C
2
2
where C is the circle x + y = 4.
P = tan x − 2y, Q = x − 4y sec y, Py = −2, Qx = 1, Qx − Py = 3.
ZZ
By Green’s Theorem I = 3
dA = 3A(D) = 3 · 4π = 12π,
Solution:
D
2
2
where D is the circle x + y ≤ 4 and A(D) = 4π is its area.
I
F̄ · dr̄,
Problem: Evaluate the integral I =
C
2
2
where C is the circle (x − 3) + y = 4 oriented counterclockwise and
F̄ = htan x − 2y, x − 4y sec yi.
1
P = tan x − 2y, Q = x − 4y sec y, Py = −2, Qx = 1, Qx − Py = 3.
ZZ
dA = 3A(D) = 3 · 4π = 12π,
By Green’s Theorem I = 3
Solution:
D
where D is the circle (x − 3)2 + y 2 ≤ 4 with center at (3, 0) and radius 2. A(D) = 4π is its area.
Bonus problem: Find the surface area of the part of the cone S: z =
between the planes z = 0 and z = a.
Solution:
p
x2 + y 2 that lies
Let x = r cos θ, y = r sin θ, z = r.
Then the cone is defined parametrically as r̄(r, θ) = hr cos θ, r sin θ, ri.
ī
j̄
k̄
√
sin θ 1 = h−r cos θ, −r sin θ, ri,
r̄r × r̄θ = cos θ
|r̄r × r̄θ | = r 2.
−r sin θ r cos θ 0 ZZ √
ZZ
r 2 dArθ , where D = {(r, θ) | 0 ≤ r ≤ a, 0 ≤ θ ≤ 2π}
dS =
Then A(S) =
S
D
2π a
√ Z Z
√
r dr dθ = πa2 2.
A(S) = 2
0
0
2