Multivariable-Calculus · Unit 16 · Video 3 · Interactive Practice

Change of Variables Start to Finish: x²y with u = x, v = xy

IKey Formulas

Step	Result for $u = x,\ v = xy$	How
1. Area element	$\dfrac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 1 & 0 \\ y & x \end{vmatrix} = x,\quad dx\,dy = \dfrac1x\,du\,dv$	Jacobian, then its reciprocal
2. Integrand	$x^2 y \cdot \dfrac1x = xy = v$	Conversion factor included
3. Bounds	$\displaystyle\int_0^1\!\!\int_0^1 x^2 y\,dx\,dy = \int_0^1\!\!\int_v^1 v\,du\,dv = \frac16$	Slice by $v =$ const

Key Insight: holding $v = xy$ fixed moves you along a hyperbola in the square, and along a horizontal line in the triangle; both pictures give $v \le u \le 1$ , $0 \le v \le 1$ .

IIOne Slice, Two Pictures

The slice $xy = v$ enters the square through the top side at $u = v$ and leaves through the right side at $u = 1$ .

v 0.4 position along slice 0.5

IIIMapping the Square onto the Triangle

Under $(u, v) = (x, xy)$ the left side collapses to the origin and the area drops from $1$ to $\tfrac12$ .

xy → uv 0%

IVQuiz Questions

Problem 1 · Rewriting a New Integrand

Given: $u = x,\ v = xy$ , so $dx\,dy = \frac1x\,du\,dv$ — find what $x y^2\,dx\,dy$ becomes in the new variables.

✅ Correct!

xy^2 = u \cdot (v/u)^2 = v^2/u

, and the factor

1/u

makes it

v^2/u^2

❌ Missing the conversion factor.

xy^2 = v^2/u

is right, but

dx\,dy

also brings

\frac1x = \frac1u

❌ Not quite.

xy^2 = (xy)\,y = v \cdot \frac{v}{u}

, so a

u

is left over even before the factor

\frac1u

❌ Not quite. With

y = v/u

y^2 = v^2/u^2

and

x = u

cancels one power:

xy^2 = v^2/u

; the factor

\frac1u

gives

v^2/u^2

, not

v^2/u^3

Show solution

Express everything through $u, v$ : $x = u$ and $y = v/u$ . Then

x y^2\,dx\,dy = u \cdot \frac{v^2}{u^2} \cdot \frac1u\,du\,dv = \frac{v^2}{u^2}\,du\,dv.

Problem 2 · Where the Slice Enters

Given: the unit square $0 \le x, y \le 1$ , the substitution $u = x,\ v = xy$ , and the order $du$ first — find the bounds of the inner integral for a fixed $v$ with $0 < v < 1$ .

✅ Correct! The hyperbola

xy = v

enters through the top side

y = 1

, where

x = v

, and leaves through

x = 1

❌ Not quite. Those are the old bounds for

x

; on the slice

xy = v

, a point with

x < v

would need

y > 1

❌ Not quite. For

u < v

the slice point

(u, v/u)

has

y > 1

, outside the square; the slice lives on

v \le u \le 1

❌ Check the algebra. On the top side

y = v/u = 1

gives

u = v

, not

u = 1/v

(which would exceed

1

Show solution

Fixing $v$ means moving along the hyperbola $y = v/x$ . It meets the top side $y = 1$ at $x = v$ , i.e. $u = v$ , and the right side $x = 1$ , i.e. $u = 1$ . So

\int_0^1\!\!\int_0^1 f\,dx\,dy = \int_0^1\!\!\int_v^1 (\cdots)\,du\,dv.

Problem 3 · A Full Computation

Given: $\displaystyle\int_0^1\!\!\int_0^1 x^3 y\,dx\,dy$ — find its value by changing to $u = x,\ v = xy$ .

✅ Correct!

\int_0^1\!\int_v^1 uv\,du\,dv = \tfrac18

, matching

\big(\int_0^1 x^3dx\big)\big(\int_0^1 y\,dy\big) = \tfrac14 \cdot \tfrac12

❌ Missing the conversion factor. Integrating

x^3y = u^2v

over the triangle without the

\frac1u

gives

\tfrac1{10}

; the integrand must be

u^2 v \cdot \frac1u = uv

❌ Check the bounds.

\tfrac14

comes from

u

running over

[0, 1]

; the region in the

uv

-plane is the triangle

v \le u \le 1

❌ Not quite.

\tfrac16

is the value for

x^2 y

from the video; here the integrand is

x^3 y

Show solution

Integrand: $x^3 y \cdot \frac1x = x^2 y = u^2 \cdot \frac{v}{u} = uv$ . Bounds as in the video: $v \le u \le 1$ , $0 \le v \le 1$ .

\int_0^1\!\!\int_v^1 uv\,du\,dv = \int_0^1 v\,\frac{1 - v^2}{2}\,dv = \frac12\left(\frac12 - \frac14\right) = \frac18.

Directly: $\int_0^1 x^3\,dx \cdot \int_0^1 y\,dy = \frac14 \cdot \frac12 = \frac18$ ✓

Problem 4 · The Image of a Rectangle

Given: the rectangle $0 \le x \le 2,\ 0 \le y \le 1$ and the map $u = x,\ v = xy$ — find its image region in the $uv$ -plane and that region's area.

✅ Correct! Its area also equals

\iint x\,dx\,dy = \int_0^2 x\,dx = 2

, since

du\,dv = x\,dx\,dy

❌ Not quite.

v = xy

, not

y

: the top side

y = 1

maps to the line

v = u

, and the left side collapses to the origin.

❌ Check the top side. On

y = 1

(u, v) = (x, x)

, so it maps to the segment from

(0,0)

(2,2)

, not to

(0,2)

❌ Not quite. The corner

(2, 1)

maps to

(u, v) = (2, 2 \cdot 1) = (2, 2)

Show solution

Map each side: $x = 2$ goes to $u = 2$ , $0 \le v \le 2$ ; $y = 1$ goes to $v = u$ , $0 \le u \le 2$ ; $y = 0$ goes to $v = 0$ ; and $x = 0$ collapses to the origin. The image is the triangle with vertices $(0,0)$ , $(2,0)$ , $(2,2)$ , of area $\tfrac12 \cdot 2 \cdot 2 = 2$ .

Check with the area element: $\iint du\,dv = \int_0^1\!\int_0^2 x\,dx\,dy = 2$ ✓

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