SME
Calculus III (Multivariable)
Green's, Stokes' and Divergence theorems
Use this for “Use Green’s theorem to find the circulation of F = ⟨−y², x²⟩ around the rectangle 0 ≤ x ≤ 3, 0 ≤ y ≤ 2.”
Diagram
F = ⟨−y², x²⟩
Counterclockwise, side by side: C₁ 0, C₂ 4, C₃ 2, C₄ 0; the total is 6
This is an example. Type your own values to start a new problem.
∮
(−y²
dx
+
x²
dy)
=
∬
(2x
+
2y)
dA
=
∮ F · dr
0
≤
x
≤
a
,
0
≤
y
≤
b
C₁
Integral along the bottom
worked out (not typed)
C₂
Integral up the right side
worked out (not typed)
C₃
Integral along the top
worked out (not typed)
C₄
Integral down the left side
worked out (not typed)
I
Double integral
worked out (not typed)
Formulas
C₁
=
a
∫
0
0 dx
0
=
2
∫
0
0 dx
C₂
=
b
∫
0
a
2
dy
4
=
1
∫
0
2
2
dy
C₃
=
∫
from
a
to
0
of
−(
b
2
)
dx
2
=
∫
from
2
to
0
of
−(
1
2
)
dx
C₄
=
0
∫
b
0 dy
0
=
0
∫
1
0 dy
∮
F
·
dr
=
C₁
+
C₂
+
C₃
+
C₄
6 = 0 + 4 + 2 + 0
I
=
a
×
b
×
(
a
+
b
)
6 = 2 × 1 × (2 + 1)
Assumptions
•
F = ⟨P, Q⟩ = ⟨−y², x²⟩, and the boundary runs counterclockwise: bottom, right side, top, left side.
•
Green’s theorem: ∮ F · dr = ∬ (Qₓ − Pᵧ) dA. Here Qₓ − Pᵧ = 2x + 2y, the circulation per unit area.
•
Along a side where y is fixed dy = 0, so only P dx counts; where x is fixed only Q dy counts.
Step-by-step
WE KNOW
a = 2, b = 1
FIND
integral along the bottom (C₁), integral up the right side (C₂), integral along the top (C₃), integral down the left side (C₄), circulation (∮ F · dr), double integral (I)
Step 1 · Find integral along the bottom (C₁)
Use C₁ = ∫ from 0 to a of 0 dx
Along
the
bottom
y
=
0,
so
P
=
−
y
2
=
0,
and
dy
=
0
there:
nothing
adds
up.
C₁
=
2
∫
0
0 dx
C₁
=
0
Step 2 · Find integral up the right side (C₂)
Use C₂ = ∫ from 0 to b of a² dy
Up
the
right
side
x
=
a
,
so
dx
=
0
and
Q
=
x
2
=
a
2
the
whole
way
up.
C₂
=
1
∫
0
2
2
dy
1
∫
0
4 dy
=
4
×
(1
−
0)
C₂
=
4
Step 3 · Find integral along the top (C₃)
Use C₃ = ∫ from a to 0 of −(b²) dx
Leftward
along
the
top
y
=
b
:
P
=
−
b
2
,
and
x
runs
back
from
a
to
0,
so
the
two
minus
signs
cancel.
C₃
=
∫
from
2
to
0
of
−(
1
2
)
dx
0
∫
2
(−1) dx
=
(−1)
×
(0
−
2)
C₃
=
2
Step 4 · Find integral down the left side (C₄)
Use C₄ = ∫ from b to 0 of 0 dy
Down
the
left
side
x
=
0,
so
Q
=
x
2
=
0,
and
dx
=
0
there:
nothing
adds
up.
C₄
=
0
∫
1
0 dy
C₄
=
0
Step 5 · Find circulation (∮ F · dr)
Use ∮ F · dr = C₁ + C₂ + C₃ + C₄
Add the four sides, taken counterclockwise: the circulation round the whole rectangle.
∮ F · dr = 0 + 4 + 2 + 0
∮ F · dr = 6
Step 6 · Find double integral (I)
Use I = a × b × (a + b)
Qₓ − Pᵧ = 2x − (−2y) = 2x + 2y. Integrate in y first with x fixed, then in x.
I
=
2
×
1
×
(2
+
1)
1
∫
0
(2x + 2
y
) dy
=
2x
+
1
2
∫
0
(2
x
+ 1) dx
=
4
+
2
I
=
6
(the
circulation
again:
Green’s
theorem)
Check
Put the numbers back in. Both sides should match.
0
=
2
∫
0
0 dx
✓
4
=
1
∫
0
2
2
dy
✓
2
=
∫
from
2
to
0
of
−(
1
2
)
dx
✓
0
=
0
∫
1
0 dy
✓
6 = 0 + 4 + 2 + 0 ✓
6 = 2 × 1 × (2 + 1) ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
More problem types
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.”
Stokes’ theorem round a circle
Use this for “Use Stokes’ theorem to find the circulation of F = ⟨−2y, 2x, 0⟩ round the circle x² + y² = 4, z = 0.”
Which theorem fits?
Use this for “Which theorem turns the flux of F out of a closed surface into a triple integral?”