SME
Calculus III (Multivariable) · Problem type
Area under a linear change of variables
Use this for “With x = 2u + v, y = u + 3v, find the area of the image of the rectangle 0 ≤ u ≤ 1, 0 ≤ v ≤ 2.”
Diagram
A = [[2, 1], [1, 3]] sends (1, 0) to (2, 1) and (0, 1) to (1, 3): its columns.
Area of the image = |det A| × area of the figure: |2 × 3 − 1 × 1| × 2 = 10.
This is an example. Type your own values to start a new problem.
x
=
a
u
+
b
v,
y
=
c
u
+
d
v
∂(x,
y)
∂(u,
v)
=
=
J
A
=
|
J
|
×
U
×
W
=
A₀
Area in the uv-plane
worked out (not typed)
Formulas
J
=
a
×
d
−
b
×
c
5 = 2 × 3 − 1 × 1
A₀
=
U
×
W
2 = 1 × 2
A
=
|
J
|
×
A₀
10 = |5| × 2
Assumptions
•
x = au + bv and y = cu + dv take the rectangle 0 ≤ u ≤ U, 0 ≤ v ≤ W to a parallelogram.
•
The Jacobian is J = ∂(x, y)/∂(u, v) = ad − bc, and dx dy = |J| du dv.
•
J = 0 squashes the rectangle flat onto a line; a negative J only flips its orientation.
Step-by-step
WE KNOW
a = 2, b = 1, c = 1, d = 3, U = 1, W = 2
FIND
Jacobian (J), area in the uv-plane (A₀), area of the image (A)
Step 1 · Find Jacobian (J)
Use J = a × d − b × c
J
is
the
determinant
of
the
partial
derivatives:
∂x/∂u
=
a
,
∂x/∂v
=
b
,
∂y/∂u
=
c
and
∂y/∂v
=
d
.
J
=
2
×
3
−
1
×
1
J
=
6
−
1
J
=
5
Step 2 · Find area in the uv-plane (A₀)
Use A₀ = U × W
The
rectangle
in
the
uv-plane:
side
U
times
side
W
.
A₀
=
1
×
2
A₀
=
2
Step 3 · Find area of the image (A)
Use A = |J| × A₀
dx
dy
=
|
J
|
du
dv:
every
patch
of
the
rectangle
is
stretched
by
the
same
factor
|
J
|,
so
the
whole
area
is
too.
A
=
|5|
×
2
A
=
5
×
2
A
=
10
Check
Put the numbers back in. Both sides should match.
5 = 2 × 3 − 1 × 1 ✓
2 = 1 × 2 ✓
10 = |5| × 2 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.