SME
Calculus III (Multivariable)
Line and surface integrals
Use this for “Find the work done by F = ⟨2y, x − y⟩ along the segment from (0, 1) to (3, 2).”
Diagram
F = ⟨−y, x⟩
Work along the path W = ∫F·dr = 1
This is an example. Type your own values to start a new problem.
F
=
⟨
α
x
+
(
β
)
y,
γ
x
+
δ
y⟩
from
(
a₁
,
a₂
)
to
(
b₁
,
b₂
)
W
=
∫
F
·
dr
=
m₁
Midpoint M, x
worked out (not typed)
m₂
Midpoint M, y
worked out (not typed)
P(M)
Field at M, x part
worked out (not typed)
Q(M)
Field at M, y part
worked out (not typed)
Formulas
m₁
=
a₁
+
b₁
2
0.5
=
1 + 0
2
m₂
=
a₂
+
b₂
2
0.5
=
0 + 1
2
P(M)
=
α
×
m₁
+
β
×
m₂
−0.5 = 0 × 0.5 + (−1) × 0.5
Q(M)
=
γ
×
m₁
+
δ
×
m₂
0.5 = 1 × 0.5 + 0 × 0.5
W
=
P(M)
×
(
b₁
−
a₁
)
+
Q(M)
×
(
b₂
−
a₂
)
1 = −0.5 × (0 − 1) + 0.5 × (1 − 0)
Assumptions
•
The segment is r(t) = A + t(B − A) for t from 0 to 1, so dr = (B − A) dt and W = ∫ from 0 to 1 of F(r(t)) · (B − A) dt.
•
F is linear, so F(r(t)) · (B − A) is a line in t, and its average over 0 ≤ t ≤ 1 is its value at t = ½, the midpoint M.
•
Running the path backward, from B to A, flips the sign of the work.
Step-by-step
WE KNOW
α = 0, β = −1, γ = 1, δ = 0, a₁ = 1, a₂ = 0, b₁ = 0, b₂ = 1
FIND
midpoint M, x (m₁), midpoint M, y (m₂), field at M, x part (P(M)), field at M, y part (Q(M)), work along the segment (W)
Step 1 · Find midpoint M, x (m₁)
Use m₁ = (a₁ + b₁) ÷ 2
The midpoint is halfway across: average the two x-coordinates.
m₁
=
1 + 0
2
m₁
=
0.5
Step 2 · Find midpoint M, y (m₂)
Use m₂ = (a₂ + b₂) ÷ 2
Halfway up: average the two y-coordinates.
m₂
=
0 + 1
2
m₂
=
0.5
Step 3 · Find field at M, x part (P(M))
Use P(M) = α × m₁ + β × m₂
Put the midpoint into P = αx + βy, the x part of F.
P(M) = 0 × 0.5 + (−1) × 0.5
P(M) = −0.5
Step 4 · Find field at M, y part (Q(M))
Use Q(M) = γ × m₁ + δ × m₂
Put the midpoint into Q = γx + δy, the y part of F.
Q(M) = 1 × 0.5 + 0 × 0.5
Q(M) = 0.5
Step 5 · Find work along the segment (W)
Use W = P(M) × (b₁ − a₁) + Q(M) × (b₂ − a₂)
Dot F at the midpoint with the step B − A. This is the integral of F · dr, since a line in t averages to its middle value.
W
=
−0.5
×
(0
−
1)
+
0.5
×
(1
−
0)
On the segment, x = −t + 1 and y = t
F(r(t)) · r′(t) = 1
W
=
1
∫
0
1 dt
=
1
W
=
1
Check
Put the numbers back in. Both sides should match.
0.5
=
1 + 0
2
✓
0.5
=
0 + 1
2
✓
−0.5 = 0 × 0.5 + (−1) × 0.5 ✓
0.5 = 1 × 0.5 + 0 × 0.5 ✓
1 = −0.5 × (0 − 1) + 0.5 × (1 − 0) ✓
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