SME
Calculus III (Multivariable) · Problem type
The Divergence theorem on a box
Use this for “Use the Divergence theorem to find the flux of F = ⟨x², y², z²⟩ out of the box 0 ≤ x ≤ 2, 0 ≤ y ≤ 1, 0 ≤ z ≤ 3.”
Diagram
a = ⟨2, 0, 0⟩, with length |a| = 2.
b = ⟨0, 1, 0⟩, with length |b| = 1.
c = ⟨0, 0, 3⟩, with length |c| = 3.
a·(b × c) = 6, so the box’s volume is |6| = 6.
This is an example. Type your own values to start a new problem.
∬
⟨x²,
y²,
z²⟩
·
n
dS
=
∭
(2x
+
2y
+
2z)
dV
=
Φ
0
≤
x
≤
a
,
0
≤
y
≤
b
0
≤
z
≤
c
V
Volume of the box
worked out (not typed)
Φ₁
Flux through the x-faces
worked out (not typed)
Φ₂
Flux through the y-faces
worked out (not typed)
Φ₃
Flux through the z-faces
worked out (not typed)
I
Triple integral
worked out (not typed)
Formulas
V
=
a
×
b
×
c
6 = 2 × 1 × 3
Φ₁
=
a
2
×
b
×
c
12
=
2
2
×
1
×
3
Φ₂
=
b
2
×
a
×
c
6
=
1
2
×
2
×
3
Φ₃
=
c
2
×
a
×
b
18
=
3
2
×
2
×
1
Φ
=
Φ₁
+
Φ₂
+
Φ₃
36 = 12 + 6 + 18
I
=
(
a
+
b
+
c
)
×
V
36 = (2 + 1 + 3) × 6
Assumptions
•
F = ⟨x², y², z²⟩ on the box 0 ≤ x ≤ a, 0 ≤ y ≤ b, 0 ≤ z ≤ c, with n pointing out of each face.
•
The Divergence theorem: the flux out through the closed surface, ∬ F · n dS, equals ∭ div F dV over the solid inside.
•
div F = 2x + 2y + 2z is the outflow per unit volume at each point.
Step-by-step
WE KNOW
a = 2, b = 1, c = 3
FIND
volume of the box (V), flux through the x-faces (Φ₁), flux through the y-faces (Φ₂), flux through the z-faces (Φ₃), flux out of the box (Φ), triple integral (I)
Step 1 · Find volume of the box (V)
Use V = a × b × c
The box’s volume: length times width times height.
V
=
2
×
1
×
3
V
=
6
Step 2 · Find flux through the x-faces (Φ₁)
Use Φ₁ = a² × b × c
On
the
face
x
=
a
,
F
·
n
=
x
2
=
a
2
over
an
area
bc;
on
the
face
x
=
0,
F
·
n
=
−
x
2
=
0.
Φ₁
=
2
2
×
1
×
3
Φ₁
=
4
×
3
Φ₁
=
12
Step 3 · Find flux through the y-faces (Φ₂)
Use Φ₂ = b² × a × c
On
the
face
y
=
b
,
F
·
n
=
y
2
=
b
2
over
an
area
ac;
on
the
face
y
=
0,
F
·
n
=
−
y
2
=
0.
Φ₂
=
1
2
×
2
×
3
Φ₂
=
2
×
3
Φ₂
=
6
Step 4 · Find flux through the z-faces (Φ₃)
Use Φ₃ = c² × a × b
On
the
face
z
=
c
,
F
·
n
=
z
2
=
c
2
over
an
area
ab;
on
the
face
z
=
0,
F
·
n
=
−
z
2
=
0.
Φ₃
=
3
2
×
2
×
1
Φ₃
=
9
×
2
Φ₃
=
18
Step 5 · Find flux out of the box (Φ)
Use Φ = Φ₁ + Φ₂ + Φ₃
Six faces in three pairs: add the pairs for the flux out through the whole surface.
Φ
=
12
+
6
+
18
Φ
=
36
Step 6 · Find triple integral (I)
Use I = (a + b + c) × V
div
F
=
∂(
x
2
)/∂x
+
∂(
y
2
)/∂y
+
∂(
z
2
)/∂z
=
2x
+
2y
+
2z.
Each
part
over
the
box
gives
one
pair’s
flux:
the
area
of
the
face
times
∫
2x
dx
across.
Together
they
make
(
a
+
b
+
c
)
×
V
.
I
=
(2
+
1
+
3)
×
6
1
×
3
×
2
∫
0
2
x
dx
=
12
2
×
3
×
1
∫
0
2
y
dy
=
6
2
×
1
×
3
∫
0
2
z
dz
=
18
I
=
6
×
6
I
=
36
(the
flux
again:
the
Divergence
theorem)
Check
Put the numbers back in. Both sides should match.
6 = 2 × 1 × 3 ✓
12
=
2
2
×
1
×
3
✓
6
=
1
2
×
2
×
3
✓
18
=
3
2
×
2
×
1
✓
36 = 12 + 6 + 18 ✓
36 = (2 + 1 + 3) × 6 ✓
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