Solutions #6
~ : R3 → R3 , show that ∇
~ · (∇
~ ×F
~ ) = 0; in other words, the curl of
1(a). For a vector field F
a vector field is incompressible.
~ (x, y, z) = F1 (x, y, z)~ı + F2 (x, y, z)~
Solution. If F
 + F3 (x, y, z)~
k, then we have
∂ ∂ ∂
∂x
∂y
∂z
~ · (∇
~ ×F
~ ) = det  ∂ ∂ ∂ 
∇
∂x
∂y
∂z
F1 F2 F3
∂ ∂F3 ∂F2
∂ ∂F3 ∂F1
∂ ∂F2 ∂F1
=
−
−
−
+
−
∂x ∂y
∂z
∂y ∂x
∂z
∂z ∂x
∂y
2
2
2
2
2
2
∂ F3
∂ F2
∂ F3
∂ F1
∂ F2
∂ F1
=
−
−
+
+
−
= 0.
∂x∂y ∂x∂z ∂x∂y ∂y∂z ∂x∂z ∂y∂x
~ : R3 → R3 is the vector field defined by G
~ = f ∇g,
~
1(b). If f, g : R3 → R be scalar fields. If G
~
~
~
then show ∇ × G is everywhere perpendicular to G.
~ × (f F
~ ) = ∇f
~ ×F
~ + f (∇
~ ×F
~ ) for any vector field F
~ , we have
Solution. Since ∇
~ ×G
~ =∇
~ × (f ∇g)
~ = ∇f
~ × ∇g
~ + f (∇
~ × ∇g)
~ .
∇
~ × (∇g)
~ = ~0, we have ∇
~ ×G
~ = ∇f
~ × ∇g.
~ Hence,
Since all gradient fields are irrotational ∇
 ∂g



∂g
∂g
∂g
∂g
f ∂x f ∂y
f ∂g
∂z
∂x
∂y
∂z
 ∂f ∂f ∂f 
∂f
∂f 
∂f
~ · (∇
~ × G)
~ = f ∇g
~ · (∇f
~ × ∇g)
~ = det 
G
=
f
det

 ∂x

∂y
∂z
∂x
∂y
∂z  = 0
∂g
∂x
∂g
∂y
∂g
∂z
∂g
∂x
∂g
∂y
∂g
∂z
~ ×G
~ is everywhere perpendicular to G.
~
which implies that ∇
~ : R3 → R3 such that ∇
~ ×H
~ = 2~ı − 3~
2. Find a vector field H
 + 4~
k.
Solution. If ~
r (x, y, z) = x~ı + y~
 + z~
k and ~
v = a~ı + b~
 + c~
k for a, b, c ∈ R then we have


~ı ~
 ~
k

~
v×~
r = det a b c  = (bz − cy)~ı − (az − cx)~
 + (ay − bx)~
k
x y z
and

~ı
~

~
k

∂
∂
∂
~ × (~

∇
v×~
r ) = det  ∂x
∂y
∂z
bz − cy −az + cx ay − bx
h
i
h
i
∂
∂
∂
∂
= ∂y (ay − bx) − ∂z (cx − az) ~ı − ∂x (ay − bx) − (bz − cy) ~

∂z
h
i
∂
∂
(cx − az) − ∂y
(bz − cy) ~
k
+ ∂x
= 2a~ı + 2b~
 + 2c~
k = 2~
v.
MATH 280: page 1 of 2
Solutions #6
MATH 280: page 2 of 2
~ =~
Hence, if a = 1, b = −3/2 and c = 2 then H
v ×~
r = − 32 z − 2y ~ı + (2x − z)~ı + y + 32 x ~
k
~
~
~
satisfies ∇ × H = 2~ı − 3~
 + 4k.
~ is not uniquely determined. Indeed, two distinct vector
Remark. The vector potential F
potentials differ by an element of the kernel of the curl operator. Since any gradient field
~ + ∇f
~ is also a vector potential for any scalar field
is annihilated by the curl operator, F
3
f : R → R.
3(a). If p(x, y) gives the pollution density, in micrograms per square
R meter, and x and y are
measured in meters, give the units and practical interpretation of R p(x, y) dA.
Solution. Consider a small piece ∆A := ∆x∆y of the region R, small enough so that the
pollution density remains close to constant over the piece. Since p(x, y) is measured in
µg · m−2 and both x and y are measured in m, the amount of pollution on ∆A measured in
micrograms is approximately p(x, y)∆x∆y. To get the total amount
of polution, we add the
R
masses on each piece and take the limit as ∆A → 0. Therefore, R p dA gives the total mass
the pollution measured in micrograms in the region R.
3(b). Using Riemann sums with four equal subdivisions in each direction, find upper and lower
bounds for the volume under the graph of h(x, y) = 2 + xy over the rectangle R = [0, 2] × [0, 4].
Solution. We divide R into 16 subrectangles by dividing each edge into 4 equal parts. Since
~
∇h(x,
y) = y~ı+x~
 and R = [0, 2]×[0, 4], it follows that h(x, y) increases as we move away from
the origin in the region R. Hence, to get an upper sum we evaluate h on each subrectangle at
the corner farthest from the origin. For example, in the rectangle [0, 0.5] × [0, 1], we evaluate
h at (0.5, 1).
x\y
0
0.5
1
1.5
2
0
1 2
3 4
2
2 2
2 2
2 2.5 3 3.5 4
2
3 4
5 6
2 3.5 5 6.5 8
2
4 6
8 10
Values of h(x, y) on the rectangle R.
Using the table, we find that
Upper sum = (2.5 + 3 + 3.5 + 4 + 3 + 4 + 5 + 6 + 3.5 + 5 + 6.5 + 8 + 4 + 6 + 8 + 10)(0.5)(1) = 41 .
To get a lower sum, we evaluate h at the corner nearest the origin:
Lower sum = (2 + 2 + 2 + 2 + 2 + 2.5 + 3 + 3.5 + 2 + 3 + 4 + 5 + 2 + 3.5 + 5 + 6.5)(0.5)(1) = 25 .
R
Therefore, we have 25 ≤ R h dA ≤ 41.