The University of Kansas
Asst. Instructor: Cuong Ngo
Department of Mathematics
Math 127: Calculus III
Quiz # 7 - FALL 2016: Sections 13.2, 13.3, and 13.4 — Solutions!
Instruction: Please show in detail all necessary steps that lead to each of your answers.
1. Given the vector field F = hcos(xy)y, cos(xy)xi.
(a) Find a function such that ∇f = F .
Ans.:
We need to find a scalar function f such that
∇f = F
hfx , fy i = hcos(xy)y, cos(xy)xi ,
which implies that
(
fx = cos(xy)y
fy = cos(xy)x .
(1)
From here, we can choose either the first or the second equation in (1) to take antiderivative.
Let us pick the first one; with this we have
fx = cos(xy)y
Z
fx dx = cos(xy)y dx
Z
f = sin(xy) + G(y) ,
(since G(y) is a constant in the context of antiderivative with respect to x). Now we use the
other equation in (1) to find G(y), namely
f = sin(xy) + G(y)
fy = cos(xy)x + G0 (y)
cos(xy)x = cos(xy)x + G0 (y)
(from (1))
0
0 = G (y)
C = G(y) .
Hence
f (x, y) = sin(xy) + C .
Z
F · dr where C is part of the circle given by the parametrization
(b) Evaluate
C
r(t) = hcos(t), sin(t)i ,
h πi
t ∈ 0,
.
4
Ans.:
From part (a), we know that F is a conservative
Z vector field, and therefore, line integral in this
field is path-independent. Hence, to compute
F · dr, we can apply the fundamental theorem
C
of Calculus for conservative field:
Z
Z
F · dr =
∇f (r(t)) · r 0 (t) dt = f (r(π/4)) − f (r(0))
C
C
√
√
= f ( 2/2, 2/2) − f (1, 0)
= sin(1/2) − sin(0) = sin(1/2) .
2. Given the vector field F = −yx2 , y 2 x . Compute
I
F · dr, where C is a closed curve with positive
C
orientation given in the figure below
Ans.:
Since the curve C is closed and positively-oriented, it is convenient for us to use the Green’s theorem
to compute the line integral. Denote
F = h−yx2 , y 2 x i .
| {z } |{z}
P
By the Green’s Theorem, we have
I
ZZ
F · dr =
(∇ × F ) · k̂ dA
C
D
ZZ ∂Q ∂P
=
−
dA
∂x
∂y
Z ZD
=
(y 2 + x2 ) dA
D
Z πZ 2
=
(r2 ) rdrdθ
0
=
1
15π
.
4
Q
(polar coordinates)