SME
Calculus III (Multivariable) · Problem type
Mass of a wire along a segment
Use this for “A wire runs straight from (1, 0) to (4, 4) with density f(x, y) = x + 2y. Find its mass ∫ f ds.”
Chart
M
=
(
0 + 3
2,
0 + 4
2
)
=
(
3
2
, 2)
This is an example. Type your own values to start a new problem.
f(x,
y)
=
α
x
+
β
y
+
γ
m
=
∫
f
ds
=
a₁
Start A, x
example
a₂
Start A, y
example
b₁
End B, x
example
b₂
End B, y
example
L
Length of the wire
worked out (not typed)
m₁
Midpoint M, x
worked out (not typed)
m₂
Midpoint M, y
worked out (not typed)
f(M)
Density at the midpoint
worked out (not typed)
Formulas
L
=
√(
(
b₁
−
a₁
)
2
+
(
b₂
−
a₂
)
2
)
5
=
√(
(3 − 0)
2
+
(4 − 0)
2
)
m₁
=
a₁
+
b₁
2
1.5
=
0 + 3
2
m₂
=
a₂
+
b₂
2
2
=
0 + 4
2
f
(M)
=
α
×
m₁
+
β
×
m₂
+
γ
3.5 = 1 × 1.5 + 1 × 2 + 0
m
=
L
×
f
(M)
17.5 = 5 × 3.5
Assumptions
•
The wire is r(t) = A + t(B − A) for t from 0 to 1, so ds = |B − A| dt = L dt, with L the length of the wire.
•
The mass is ∫ f ds = L × ∫ from 0 to 1 of f(r(t)) dt. f is linear, so its average along the wire is its value at the midpoint M.
•
The density f is mass per unit length and is not negative anywhere on the wire.
Step-by-step
WE KNOW
α = 1, β = 1, γ = 0, a₁ = 0, a₂ = 0, b₁ = 3, b₂ = 4
FIND
length of the wire (L), midpoint M, x (m₁), midpoint M, y (m₂), density at the midpoint (f(M)), mass of the wire (m)
Step 1 · Find length of the wire (L)
Use L = √((b₁ − a₁)² + (b₂ − a₂)²)
The
length
of
the
wire
is
the
distance
from
A
to
B,
so
ds
=
L
dt.
L
=
√(
(3 − 0)
2
+
(4 − 0)
2
)
L
=
√
9 + 16
L
=
√
25
L
=
5
Step 2 · Find midpoint M, x (m₁)
Use m₁ = (a₁ + b₁) ÷ 2
The midpoint is halfway across: average the two x-coordinates.
m₁
=
0 + 3
2
m₁
=
1.5
Step 3 · Find midpoint M, y (m₂)
Use m₂ = (a₂ + b₂) ÷ 2
Halfway up: average the two y-coordinates.
m₂
=
0 + 4
2
m₂
=
2
Step 4 · Find density at the midpoint (f(M))
Use f(M) = α × m₁ + β × m₂ + γ
Put the midpoint into f. f changes evenly along the wire, so this is its average there.
f
(M)
=
1
×
1.5
+
1
×
2
+
0
f
(M)
=
3.5
Step 5 · Find mass of the wire (m)
Use m = L × f(M)
The
mass
is
the
length
times
the
average
density:
∫
f
ds
=
L
×
f
(M).
m
=
5
×
3.5
f
(r(t))
=
7t
m
=
5
×
1
∫
0
(7
t
) dt
=
5
×
3.5
m
=
17.5
Check
Put the numbers back in. Both sides should match.
5
=
√(
(3 − 0)
2
+
(4 − 0)
2
)
✓
1.5
=
0 + 3
2
✓
2
=
0 + 4
2
✓
3.5 = 1 × 1.5 + 1 × 2 + 0 ✓
17.5 = 5 × 3.5 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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