Multivariable-Calculus · Unit 16 · Video 1 · Interactive Practice

Change of Variables: How a Substitution Rescales Area

IKey Formulas

Formula	Meaning	When it applies
$u = \dfrac{x}{a},\ v = \dfrac{y}{b} \ \Rightarrow\ dx\,dy = ab\,du\,dv$	Two one-variable substitutions multiply	Each new variable depends on one old variable
$\text{Area}\big\{(\tfrac{x}{a})^2 + (\tfrac{y}{b})^2 < 1\big\} = \pi ab$	Ellipse = stretched unit disk	Semi-axes $a$ and $b$
$u = px + qy,\ v = rx + sy \ \Rightarrow\ du\,dv = \|ps - qr\|\,dx\,dy$	Linear substitution: constant area scale factor	Any linear change of variables
$dx\,dy = \dfrac{1}{\|ps - qr\|}\,du\,dv$	The factor used when switching an integral to $u, v$	Same substitution, read the other way

Key Insight: for $u = 3x - 2y,\ v = x + y$ every piece of the $uv$ -plane has $5$ times the area of the matching piece of the $xy$ -plane, so $du\,dv = 5\,dx\,dy$ and $dx\,dy = \tfrac15\,du\,dv$ .

IIUndoing the Squish

The substitution $u = x/a,\ v = y/b$ squeezes the ellipse onto the unit disk, shrinking every cell by the factor $ab$ .

a 3 b 2 xy → uv xy

IIISame Rectangle Anywhere, Same Ratio

Under $u = 3x - 2y,\ v = x + y$ every rectangle becomes a parallelogram with exactly $5$ times its area.

x 0.5 y 0.5 Δx 1 Δy 0.5

IVThe Unit Square and the Determinant

The unit square's image has sides given by the coefficient matrix's columns, so its area is $|\det|$ .

u = 3 x + −2 y v = 1 x + 1 y

VQuiz Questions

Problem 1 · Area of an Ellipse

Given: the region $\dfrac{x^2}{16} + \dfrac{y^2}{9} < 1$ — find its area using the substitution $u = x/4,\ v = y/3$ .

✅ Correct! The semi-axes are

a = 4

and

b = 3

, so the area is

\pi ab = 12\pi

❌ Not quite. The two one-variable factors multiply:

dx\,dy = (4\,du)(3\,dv) = 12\,du\,dv

, not

4 + 3

❌ Not quite.

16

and

9

are

a^2

and

b^2

; the semi-axes are

a = 4

and

b = 3

, and the factor is

ab = 12

❌ Not quite. The area scale factor is the product

ab = 4 \cdot 3

, not

a^2 + b^2

Show solution

With $u = x/4$ and $v = y/3$ the region becomes $u^2 + v^2 < 1$ , the unit disk. From $du = \tfrac14\,dx$ and $dv = \tfrac13\,dy$ we get $dx\,dy = 12\,du\,dv$ , so

\iint_{x^2/16 + y^2/9 < 1} dx\,dy = \iint_{u^2 + v^2 < 1} 12\,du\,dv = 12 \cdot \pi = 12\pi.

Problem 2 · Which Way Does the Factor Go?

Given: $u = 3x - 2y,\ v = x + y$ maps a region $R$ in the $xy$ -plane onto a region $S$ in the $uv$ -plane, and $S$ has area $20$ — find the area of $R$ .

✅ Correct! Pieces of the

uv

-plane are

5

times larger, so

R

has area

20/5 = 4

❌ Wrong direction.

du\,dv = 5\,dx\,dy

: the

uv

-region is the larger one, so divide by

5

❌ Not quite. A linear substitution with determinant

5

does not preserve area;

S

5

times as large as

R

❌ Not quite. That is

5/20

; the area of

R

20/5

Show solution

The coefficient matrix $\begin{pmatrix} 3 & -2 \\ 1 & 1 \end{pmatrix}$ has determinant $3 \cdot 1 - (-2)\cdot 1 = 5$ , so $du\,dv = 5\,dx\,dy$ and $dx\,dy = \tfrac15\,du\,dv$ :

\text{Area}(R) = \iint_R dx\,dy = \iint_S \tfrac15\,du\,dv = \tfrac15 \cdot 20 = 4.

Problem 3 · A New Linear Substitution

Given: $u = x + 2y,\ v = 3x - y$ — find the factor $k$ in $dx\,dy = k\,du\,dv$ .

✅ Correct!

|\det| = 7

gives

du\,dv = 7\,dx\,dy

, so

dx\,dy = \tfrac17\,du\,dv

❌ Wrong direction.

7

is the factor in

du\,dv = 7\,dx\,dy

; solving for

dx\,dy

divides by

7

❌ Check the sign. The determinant is

-7

, but an area is positive: the factor uses

|\det| = 7

❌ Not quite. The determinant is

ps - qr = (1)(-1) - (2)(3) = -7

, a difference, not

(1)(-1) + (2)(3) = 5

Show solution

The unit square goes to the parallelogram with sides $\langle 1, 3\rangle$ (image of $(1,0)$ ) and $\langle 2, -1\rangle$ (image of $(0,1)$ ). Its area is

\left|\det\begin{pmatrix} 1 & 2 \\ 3 & -1 \end{pmatrix}\right| = |{-1} - 6| = 7.

So $du\,dv = 7\,dx\,dy$ , and therefore $dx\,dy = \tfrac17\,du\,dv$ .

Problem 4 · Simplifying the Bounds

Given: $R$ is the parallelogram $0 \le x + y \le 2,\ 0 \le x - 2y \le 3$ — find its area using $u = x + y,\ v = x - 2y$ .

✅ Correct! The

uv

-region is a

2 \times 3

rectangle and

dx\,dy = \tfrac13\,du\,dv

, so the area is

6/3 = 2

❌ Wrong direction.

du\,dv = 3\,dx\,dy

, so the

xy

-area is the

uv

-area divided by

3

❌ Not quite.

6

is the area of the rectangle in the

uv

-plane; the area element must be converted too.

❌ Not quite. That is

3/6

; the area of

R

6/3

Show solution

In $u, v$ the region is the rectangle $0 \le u \le 2,\ 0 \le v \le 3$ , of area $6$ . The coefficient matrix has

\det\begin{pmatrix} 1 & 1 \\ 1 & -2 \end{pmatrix} = -2 - 1 = -3,

so $du\,dv = 3\,dx\,dy$ and $dx\,dy = \tfrac13\,du\,dv$ :

\text{Area}(R) = \int_0^3 \int_0^2 \tfrac13\,du\,dv = \tfrac13 \cdot 6 = 2.

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