SME
Calculus III (Multivariable) · Problem type
Flux out of a sphere
Use this for “Find the outward flux of F = ⟨3x, 3y, 3z⟩ through the sphere x² + y² + z² = 4.”
Diagram
On the sphere of radius 2, n = ⟨x, y, z⟩ ÷ 2, so F·n = 1 × 2 = 2 everywhere.
Φ = 2 × 4π × 2² = 100.5.
This is an example. Type your own values to start a new problem.
F
=
k
⟨x,
y,
z⟩
x²
+
y²
+
z²
=
R
2
Φ
=
∬
F
·
n
dS
=
Fₙ
×
S
=
Formulas
Fₙ
=
k
×
R
2 = 1 × 2
S
=
4
×
π
×
R
2
16π
=
4
×
π
×
2
2
Φ
=
Fₙ
×
S
32π = 2 × 16π
Φ
=
4
×
π
×
k
×
R
3
32π
=
4
×
π
×
1
×
2
3
Assumptions
•
On the sphere of radius R about the origin, the outward unit normal is n = ⟨x, y, z⟩ ÷ R.
•
F = k⟨x, y, z⟩ points straight out, so F · n = k(x² + y² + z²) ÷ R = kR at every point of the sphere.
•
The flux is Φ = ∬ F · n dS. With F · n the same everywhere, it is F · n times the area 4πR².
•
Check by the divergence theorem: div F = 3k, and 3k times the volume (4 ÷ 3)πR³ is 4πkR³ too.
Step-by-step
WE KNOW
k = 1, R = 2
FIND
outward part of F (Fₙ), area of the sphere (S), outward flux (Φ)
Step 1 · Find outward part of F (Fₙ)
Use Fₙ = k × R
On
the
sphere
x
2
+
y
2
+
z
2
=
R
2
,
so
Fₙ
=
F
·
n
=
k
(
x
2
+
y
2
+
z
2
)
R
=
kR.
Fₙ = 1 × 2
Fₙ = 2
Step 2 · Find area of the sphere (S)
Use S = 4 × π × R²
The
area
of
a
sphere
of
radius
R
is
4πR².
S
=
4
×
π
×
2
2
S
=
4
×
4
×
π
S
=
16
×
π
S
=
16π
(≈
50.27)
Step 3 · Find outward flux (Φ)
Use Φ = Fₙ × S
Fₙ is the same at every point, so the surface integral ∬ F · n dS is Fₙ times the area.
Φ
=
2
×
16π
Φ
=
32π
(≈
100.53)
Check
Put the numbers back in. Both sides should match.
2 = 1 × 2 ✓
16π
=
4
×
π
×
2
2
✓
32π = 2 × 16π ✓
32π
=
4
×
π
×
1
×
2
3
✓
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