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Quiz 10
Name:
SOLUTIONS
Maths 114 - Calculus II
November 30, 2010
Note: In order to receive full credit, you must show work that justifies your answer.
2 2
x
3
1. Show that the vector field F(x, y)
R = h3x y + e , 2x yi is conservative, and
hence evaluate
the line integral C F · dr, where C is the curve parameterized by
r(t) = sin2 (2πt)et , cos(2πt) − t , 0 ≤ t ≤ 1.
To show that the vector field is conservative, we need to show that the partial y
derivative of the x-component of F(x, y) is equal to the partial x derivative of
the y-component of F(x, y) (and also note that the function is defined on a
∂
∂
(3x2 y 2 + ex ) = 6x2 y and ∂x
(2x3 y) = 6x2 y. The
simply connected region). ∂y
domain of F(x, y) is all of R2 , so it is defined on a simply connected region.
Therefore F(x, y) is a conservative vector field.
Now that we know the vector field is conservative, we are able to apply the
fundamental theorem of line integrals. In order to do so, we first need to find a
potential function
for F(x, y). If fx = 3x2 y 2 + ex and fy = 2x3 y, then
R
2 2
f (x, y) = (3x y + ex ) dx = x3 y 2 + ex + C(y), so
∂
2x3 y = fy = ∂y
(x3 y 2 + ex + C(y)) = 2x3 y + C 0 (y). Therefore C 0 (y) = 0, so
C(y) is a constant. Any choice of constant will give us a potential function, so
we may as well choose the constant to be zero. Thus a potential function is
f (x, y) = x3 y 2 + ex .
R
The fundamental theorem of line integrals says that C F · dr = f (b) − f (a),
where a is the starting point of the curve C, and b is the end point. From the
parameterization, we see that a = r(0) = h0, 1i and b = r(1) = h0, 0i. Therefore
Z
F · dr = f (0, 0) − f (0, 1) = 1 − 1 = 0.
C
2. Calculate the work done by the force field F(x, y) = hx3 − y, x + ey i on a
particle which moves once around the circle x2 + y 2 = 1 in the anticlockwise
direction.
R
2
2
The work done is C F · dr, where C is the circle
RR x + y = 1, positively
oriented. By Green’s theorem, this is equal to D Qx − Py dA, where D is the
disk bounded by C and P (x, y) = x3 − y, Q(x, y) = x + ey are the components
of F(x, y). Therefore the work done is
ZZ
ZZ
∂
∂
y
3
(x + e ) −
x − y dA =
1 − (−1) dA = 2 Area(D) = 2π.
∂y
D
D ∂x
Quiz 10
Name:
SOLUTIONS
Maths 114 - Calculus II
December 2, 2010
Note: In order to receive full credit, you must show work that justifies your answer.
1. Consider F(x, y) =
D
1
, 1
x+y x+y
E
+ y and C : r(t) = ht, et i, 0 ≤ t ≤ 1.
(a) Show that F is a conservative vector field.
To show that the vector field is conservative, we need to show that the
partial y derivative of the x-component of F(x,
y)is equal to the partial x
∂
1
−1
derivative of the y-component of F(x, y). ∂y
= (x+y)
2 and
x+y
1
∂
−1
1
+ y = (x+y)
2 . Therefore F(x, y) is a conservative vector field .
∂x x+y
R
(b) Evaluate C F · dr.
Because we know that F is conservative, we can apply the fundamental
theorem of line integrals. To do so, we first need to find a potential
1
1
and Φy = x+y
+ y, then
function for F. If Φx = x+y
R 1
Φ(x, y) = x+y dx = ln |x + y| + C(y), so
1
∂
1
+ y = Φy = ∂y
(ln |x + y| + C(y)) = x+y
+ C 0 (y). Therefore C 0 (y) = y,
x+y
2
so C(y) = y2 + c, where c is a constant. Any choice of constant will give us
a potential function, so we may as well put c = 0. Thus a potential
2
function is Φ(x, y) = ln |x + y| + y2 .
The
fundamental theorem of line integrals says that
R
F · dr = Φ(b) − Φ(a), where a is the starting point of the curve C, and
C
b is the end point. From the parameterization, we see that
a = r(0) = h0, 1i and b = r(1) = h1, ei. Therefore
Z
e2 1
F · dr = Φ(1, e) − Φ(0, 1) = ln(1 + e) + − .
2
2
C
You could also do this problem without using the fundamental theorem of
line integrals, just by using the definition of a line integral.
1
Strictly speaking, we should also show that the function is defined on a simply connected region.
The domain of F(x, y) is R2 minus the points where x + y = 0, which is not connected since it consists
of two pieces, namely the half-planes x + y < 0 and x + y > 0. However, the curve C is contained in
the half-plane x + y > 0, and this is simply connected, so we are OK.
R
2. Evaluate C F · dr using Green’s Theorem, where C is the circle x2 + y 2 = 9
oriented clockwise, and where
F = hex + sin(x), x + yi .
Because
C is oriented
clockwise, Green’s theorem says that
R
RR
F
·
dr
=
−
Q
−
Py dA, where D is the disk bounded by C and
x
C
D
x
P (x, y) = e + sin(x), Q(x, y) = x + y are the components of F. Therefore
ZZ
Z
∂ x
∂
(x + y) −
(e + sin(x)) dA
F · dr = −
∂x
∂y
C
D
ZZ
= −
1 − 0 dA
D
= −Area(D)
= −π32
= −9π.