MATH 263
Solutions: TEST III
Instructions: Answer any 8 of the 10 questions. You may answer more than 8 to earn
extra credit.
Whoever despises the high wisdom of mathematics nourishes himself
on delusion and will never still the sophistic sciences whose only
product is an eternal uproar.
- Leonardo da Vinci (1452-1519)
1. (a) Let f and g be (scalar-valued) functions of x, y and z and let F and G be vector fields defined on R3. For
each of the following expressions, determine if the result is a scalar, a vector, or meaningless. (No need to give
reasons.)
(b) Prove that div(curl F) = 0.
Solution:
Let F(x, y, z) = P(x, y, z) i + Q(x, y, z) j + R(x, y, z) k.
Then curl F = (Qz – Ry) i – (Rx – Pz) j + (Qx – Py) k.
Thus div(curl F) = (Qz – Ry)x – (Rx – Pz)y + (Qx – Py)z = Qzx – Ryx – Rxy + Pzy + Qxz – Pyz = 0 by Clairaut’s
theorem.
2
2. [Stewart] Let R be the trapezoidal region with vertices
(1, 0), (2, 0), (0, -2), and (0, -1). Evaluate the integral
∬ 𝑒 (𝑥+𝑦)/(𝑥−𝑦) 𝑑𝐴
𝑅
𝐻𝑖𝑛𝑡: 𝑀𝑎𝑘𝑒 𝑡ℎ𝑒 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝑣𝑎𝑟𝑖𝑎𝑏𝑙𝑒𝑠 𝑢 = 𝑥 + 𝑦, 𝑣 = 𝑥 − 𝑦.
Note: Be certain to sketch R in the xy-plane and its pullback in the uv-plane.
(Give a numerical answer.)
u = x + y and v = x – y
by solving for x and y.
So x = ½ (u + v) and y = ½ (u – v).
3
3. Find the path integral of f(x, y, z) = x + y + z over the straight-line segment from
(1, 2, 3) to (0, -1, 1). (Give a numerical answer.)
Solution: Define a parametrization as follows:
 (t) = t(0, -1, 1) + (1 – t)(1, 2, 3) for 0 ≤ t ≤ 1 .
So x(t) = 1 – t, y(t) = -t + 2 (1 – t) = 2 – 3t, z(t) = t + 3(1 – t)= 3 – 2t
Now (t) =(-1, -3, -4), so || (t) || = (1 + 9 + 4)1/2 =(14)1/2.
Also f( (t)) = 1 – t – t + 2 (1 – t) + t + 3(1 – t)= 6 – 6t = 6(1 – t)
Thus
1
 f ( (t )) ||  ' (t ) || ds   6(1  t )
C
1
14 dt  6 14  (1  t ) dt  3 14
0
0
4. If a circle C with radius 1 rolls along the outside of the circle x2 + y2 = 16, a fixed point P on C traces out a
curve called an epicycloid, with parametric equations
x = 5 cos(t) − cos(5t), y = 5 sin(t) − sin(5t).
Below is a graph of the epicycloid. Find the area it encloses.
Solution: Let (t) = (x(t), y(t)) = (5 cos t – cos 5t, 5 sin t – sin(5t), for 0 ≤ t ≤ 2.
Let E be the region in question. Now, invoking Green’s theorem for area:
𝑎𝑟𝑒𝑎 𝑜𝑓 𝐸 = ∮ 𝑥 𝑑𝑦 − 𝑦 𝑑𝑥
Now x dy = (5 cos t – cos 5t)(5cos t – 5 cos 5t) = 25 cos2 t – 25 cos t cos 5t + 5cos2 5t
4
and, similarly:
– y dx = 25 sin2 t – 25 sin t sin 5t + 5 sin2 5t
Now x dy – y dx = 30 – 25(cos t cos 5t – sin t sin 5t) = 30 – cos 6t.
Hence Area of E =
2𝜋
2𝜋
0
0
1
1
1
∮ 𝑥 𝑑𝑦 − 𝑦 𝑑𝑥 = ∫ (30 − 𝑐𝑜𝑠 6𝑡) 𝑑𝑡 =
∫ 30 𝑑𝑡 = 𝟑𝟎𝝅
2
2
2
5.
Let G(x, y) = 3x i + y2 j be given.
(a)
Find div G.
P Q (3x) ( y 2 )



 3 2y
Solution: div F    F 
x y
x
y
(b) Find curl G.
Solution: curl F    F  (Qx  Py )k  (2 y  3)k
(c)
Find the circulation of G around the closed curve C that is the triangle joining the points (0, 0), (2, 0),
and (0, 4). Assume that the curve is oriented in the counterclockwise direction and that the curve begins
and ends at the origin. Show all work!
Solution: Let T denote the triangle including its interior.
Using Green’s theorem,
2
4−2𝑥
𝑐𝑖𝑟𝑐𝑢𝑙𝑎𝑡𝑖𝑜𝑛 = ∬(3 + 2𝑦)𝑑𝐴 = ∫ ∫
𝑇
0
0
2
(3 + 2𝑦)𝑑𝑦 𝑑𝑥 = ∫ (3(4 − 2𝑥) + (4 − 2𝑥)2 )𝑑𝑥 = 132
0
5
6.
Find a potential function for the vector field

  2
1 
1 

F ( x, y , z )   y 2 
 i  (2 xy  5) j   3z 
k
x z
x z


Solution:
Assume that F = grad f = fx i + fy j + fz k
Then fx (x, y, z) = y 2 −
1
;
x+z
Next, fy (x, y, z) = 2xy +
so f(x, y, z) = xy 2 − ln(x + z) + g(y, z)
∂g(x,z)
∂y
2xy +
and so
∂g(x, z)
= 2xy + 5 which implies that g(x, z) = 5y + h(z)
∂y
Now f(x, y, z) = xy 2 − ln(x + z) + 5y + h(z)
Next, fz (x, y, z) = −
1
dh
+ and so
x+z
dz
3z 2 −
1
dh
+
x+z
dz
= −
1
x+z
which implies that
dh
dz
= 0.
Finally, a potential function is
f(x, y, z) = xy 2 − ln(x + z) + 5y
7. Find the work done by the vector field
F(x, y, z) = (x2 + y) i + (y2 + x) j + zez k
over the path given by the curve
(t) = (2 cos t) i + (2 sin t) j + (2t/) k, 0 ≤ t ≤ 2.
Solution: Note that this vector field is conservative: F = g,
where g(x, y, z) = (x3 + y3)/3 + xy + zez – ez.
Invoking the Fundamental theorem of line integrals:

F

ds


g

d
s
g ( (2))  g ( (0))  g (2, 0, 4)  g (2, 0, 0)  3e 4  1


C
8.
C
Let C be the helix given by (t) = (3cos t, 3sin t, 4t) , defined for 0  t  2 and let
T(x, y, z) = ln (1 + x + y + 2z) be the temperature (in degrees Fahrenheit) at a point (x, y, z).
Calculate the
average temperature of the curve C. Express your answer as a Riemann integral. (No need to evaluate.)
2𝜋
𝑑𝑥
𝑑𝑦
𝑑𝑧
Solution: Computing the length of the curve: 𝐿 = ∫𝐶 1𝑑𝑠 = ∫0 √( 𝑑𝑡 )2 + ( 𝑑𝑡 )2 + (𝑑𝑡 )2 𝑑𝑡 =
2𝜋
∫0 √(−3 𝑠𝑖𝑛 𝑡)2 + (3 𝑐𝑜𝑠 𝑡)2 + (4)2 𝑑𝑡 = 5(2p) = 10p
6
Thus the average temperature is
1
∫
10𝜋 𝐶
𝑇 𝑑𝑠 =
2𝜋
1
∫ 𝑇(𝜎(𝑡))|𝜎 ′ (𝑡)| 𝑑𝑡
10𝜋 0
=
1
10𝜋
2𝜋
∫0 𝑙𝑛(1 + 3 𝑐𝑜𝑠 𝑡 + 3 𝑠𝑖𝑛 𝑡 + 4𝑡) 𝑑𝑡 ℉
9. (Stewart) Evaluate the following integral, where C is the circle x2 + y2 = 9.
∮ (2𝑦 − 𝑒 sin 𝑥 )𝑑𝑥 + (5𝑥 + √𝑦 4 + 1 ) 𝑑𝑦
𝐶
(Give a numerical answer.)
Solution: First observe that the scalar curl of the vector field is 5 – 2 = 3.
Using Green’s theorem,
∮𝐶 (2𝑦 − 𝑒 sin 𝑥 )𝑑𝑥 + (5𝑥 + √𝑦 4 + 1 )𝑑𝑦 = ∬𝐵 𝑑𝑖𝑣 𝐺 𝑑𝐴 = 3(𝑎𝑟𝑒𝑎 𝑜𝑓 𝐶) = 3(𝜋9) = 27𝜋
10.
Let F(x, y) = 2y i + x j, and let Ca be the circle of radius a centered at the origin, oriented in a counterclockwise direction.
(a)
Compute

F
Ca
Solution: Note that curl F = 1 – 2 = -1. Using Green’s theorem,
∮ 𝐹 ∙ 𝑑𝑠 = ∬ 𝑑𝑖𝑣 𝐺 𝑑𝐴 = −1(𝑎𝑟𝑒𝑎 𝑜𝑓 𝐶) = −𝜋𝑎2
𝐶𝑎
(b)
𝐷
Show that
lim a  0

Ca
2
Solution: lim −𝜋𝑎 = 0
𝑎→0
F 0
7
Extra Extra Credit:
Let R be the region in the first-quadrant that lies inside x2 + y2 = 1 but outside x2 + y2 = 2y.
(a)
Sketch this region.
Solution:
Here is the region of integration.
xy plane
xy plane
y
2.0
y
2.0
1.5
1.5
1.0
1.0
0.5
0.5
0.2 0.4 0.6 0.8 1.0
x
0.2 0.4 0.6 0.8 1.0
x
Water is fluid, soft and yielding. But water will wear away rock, which is rigid and cannot
yield. As a rule, whatever is fluid, soft and yielding will overcome whatever is rigid and
hard. This is another paradox: what is soft is strong.
-
Lao-Tzu (600 B.C.)
8
Klein bottle (from virtual math museum)