Designing Substitutions from Curved Boundaries

Why Build Coordinates Around a Boundary?

A useful substitution is often suggested by the way a region is bounded. When the upper boundary is a curve, ordinary rectangular coordinates may leave one variable with a bound that depends on the other. A carefully chosen coordinate can convert that curved limit into a constant one.

Consider the planar region

D=\{(x,y):0\le x\le 1,\;0\le y\le x^2\}.

For each fixed $x$, the coordinate $y$ runs from the lower boundary $0$ to the upper boundary $x^2$. When $x>0$, the ratio $y/x^2$ measures the fraction of the way from the lower boundary to the upper boundary. This suggests using that fraction as a new coordinate.

Boundary-adapted idea: Keep $x$ as one coordinate and express the position between the two $y$-boundaries as a fraction between $0$ and $1$. This turns the curved upper boundary into a constant coordinate bound.

Constructing the Substitution

Set $u=x$, and define $v=y/x^2$ when $x>0$. Solving for $x$ and $y$ gives the map from the new coordinates to the original ones:

x=u,\qquad y=u^2v.

The bounds follow directly from the original region. Since $0\le x\le1$, we need $0\le u\le1$. For $x>0$, the condition $0\le y\le x^2$ becomes $0\le y/x^2\le1$, so $0\le v\le1$. Thus the parameter rectangle is the unit square.

Definition: The boundary-adapted map is $F(u,v)=(u,u^2v)$, with parameter rectangle $R=[0,1]\times[0,1]$. For $u>0$, its inverse is $u=x,\;v=y/x^2$.

There is a boundary detail worth stating precisely. The rectangle $0<u\le1,\;0\le v\le1$ maps onto the part of $D$ with $x>0$; the open interior of the unit square maps onto the corresponding interior of that region. At $u=0$, every value of $v$ maps to the single point $(0,0)$. Consequently, the map is not one-to-one on the entire closed square, and its inverse formula involving division by $x^2$ is not defined at that point.

This collapse occurs only on the boundary. The interior map is one-to-one, and the omitted boundary has area zero. For continuous integrands on this bounded region, the change-of-variables theorem can be applied on the nonsingular interior; including or excluding its boundary does not alter the integral.

Deriving and Checking the Jacobian

The Jacobian determinant measures the local area-scaling factor for the map from $(u,v)$ to $(x,y)$. Differentiate each component of $F$ with respect to both new coordinates:

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v}\\[4pt] \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{vmatrix} = \begin{vmatrix} 1 & 0\\ 2uv & u^2 \end{vmatrix} =u^2.

For $u>0$, the determinant is positive, so the absolute Jacobian is also $u^2$. At $u=0$, it is zero, in agreement with the collapse of that edge to one point. Therefore the area element transforms as $dA=dx\,dy=u^2\,du\,dv$ on the nonsingular part of the map.

The inverse also verifies the determinant. For $x>0$, $u=x$ and $v=y/x^2$. Its Jacobian is

\frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} 1 & 0\\[4pt] -\frac{2y}{x^3} & \frac{1}{x^2} \end{vmatrix} =\frac{1}{x^2}.

On corresponding points, $x=u$, so the inverse determinant is $1/u^2$, the reciprocal of the forward determinant $u^2$. This reciprocal check is available wherever the map is nonsingular, namely when $u>0$.

Area transformation: For an integrable function $f$ on $D$, this substitution gives $$ \iint_D f(x,y)\,dA = \int_0^1\int_0^1 f(u,u^2v)\,u^2\,dv\,du. $$ The factor $u^2$ is the absolute Jacobian. The collapsed edge $u=0$ does not contribute area.

Why the Bounds Become a Rectangle

The construction can be justified by checking the inequalities in both directions. If $(x,y)\in D$ and $x>0$, then $u=x$ satisfies $0<u\le1$, and $v=y/x^2$ satisfies $0\le v\le1$. Substitution back gives $x=u$ and $y=u^2v$, so the point is reached by the map.

Conversely, if $0<u\le1$ and $0\le v\le1$, then $x=u$ satisfies $0<x\le1$, while $y=u^2v$ satisfies $0\le y\le u^2=x^2$. Hence $(x,y)\in D$. These two implications show that the rectangle covers exactly the portion with $x>0$, with a unique parameter pair for each point there: $u=x$ and $v=y/x^2$.

The same reasoning explains the usefulness of the substitution. In the original coordinates, the height of the vertical slice depends on $x$. In the new coordinates, $v$ always traverses the same interval, from the lower boundary at $v=0$ to the upper boundary at $v=1$. The variation in slice height is instead represented by the Jacobian factor $u^2$.

Worked Examples

Worked Example: Find the Area of the Curved Region

The area is $\iint_D 1\,dA$. Under the substitution, the integrand remains $1$, while the area factor is $u^2$:

\begin{aligned} \operatorname{Area}(D) &=\int_0^1\int_0^1 u^2\,dv\,du\\ &=\int_0^1\left[u^2v\right]_{v=0}^{v=1}\,du\\ &=\int_0^1 u^2\,du\\ &=\left[\frac{u^3}{3}\right]_{u=0}^{u=1}\\ &=\frac13. \end{aligned}

Check in the original coordinates by integrating the vertical slice heights:

\operatorname{Area}(D) =\int_0^1\int_0^{x^2}1\,dy\,dx =\int_0^1 x^2\,dx =\left[\frac{x^3}{3}\right]_0^1 =\frac13.

Both calculations give area $1/3$.

Worked Example: Integrate a Function Involving Both Coordinates

Evaluate $\iint_D (x^2+y)\,dA$. First replace $x$ and $y$ by their expressions in $u,v$:

$$ x^2+y=u^2+u^2v=u^2(1+v). $$

Multiplying by the Jacobian $u^2$, the transformed integrand is $u^4(1+v)$. Therefore

\begin{aligned} \iint_D (x^2+y)\,dA &=\int_0^1\int_0^1 u^4(1+v)\,dv\,du\\ &=\int_0^1 u^4\left[v+\frac{v^2}{2}\right]_{0}^{1}\,du\\ &=\int_0^1 u^4\left(1+\frac12\right)\,du\\ &=\frac32\left[\frac{u^5}{5}\right]_{0}^{1}\\ &=\frac{3}{10}. \end{aligned}

Check directly using the original bounds:

\begin{aligned} \iint_D (x^2+y)\,dA &=\int_0^1\int_0^{x^2}(x^2+y)\,dy\,dx\\ &=\int_0^1\left[x^2y+\frac{y^2}{2}\right]_{0}^{x^2}\,dx\\ &=\int_0^1\left(x^4+\frac{x^4}{2}\right)\,dx\\ &=\frac32\left[\frac{x^5}{5}\right]_{0}^{1} =\frac{3}{10}. \end{aligned}

The direct and transformed integrals agree.

Worked Example: Integrate the Vertical Coordinate

Evaluate $\iint_D y\,dA$. Since $y=u^2v$, multiplication by the Jacobian $u^2$ gives $u^4v$:

\begin{aligned} \iint_D y\,dA &=\int_0^1\int_0^1 u^4v\,dv\,du\\ &=\int_0^1 u^4\left[\frac{v^2}{2}\right]_{0}^{1}\,du\\ &=\frac12\int_0^1u^4\,du\\ &=\frac12\left[\frac{u^5}{5}\right]_{0}^{1}\\ &=\frac{1}{10}. \end{aligned}

To check, integrate directly in $y$ first:

\begin{aligned} \iint_D y\,dA &=\int_0^1\int_0^{x^2}y\,dy\,dx\\ &=\int_0^1\left[\frac{y^2}{2}\right]_{0}^{x^2}\,dx\\ &=\frac12\int_0^1x^4\,dx\\ &=\frac12\left[\frac{x^5}{5}\right]_{0}^{1} =\frac{1}{10}. \end{aligned}

Again, the result is unchanged by the substitution.

A Common Mistake: Forgetting the Area Factor

A substitution changes more than the symbols in the integrand. It also changes the area element. For example, in the area calculation, a tempting but incorrect step is to replace $D$ by the unit square and integrate $1$ without including the Jacobian:

\text{Incorrect:}\qquad \iint_D 1\,dA \overset{\text{wrong}}{=} \int_0^1\int_0^1 1\,dv\,du =1.

This answer is wrong because the map does not preserve area: a small parameter rectangle near $u=0$ is compressed more strongly than one near $u=1$. The determinant calculates that local scaling. The corrected calculation is

\iint_D 1\,dA = \int_0^1\int_0^1 u^2\,dv\,du =\frac13.

A second check is geometric: the region lies inside the unit square, so its area cannot be $1$. The Jacobian both corrects the calculation and reflects the narrowing of the region near $x=0$.

A Reliable Construction Checklist

Read the boundary inequalities.
Here, $y$ lies between $0$ and the positive boundary height $x^2$, with $x$ between $0$ and $1$.

Normalize the position between the boundaries.
Use $v=y/x^2$ where $x>0$, so the two $y$-boundaries become $v=0$ and $v=1$.

Write the forward map and bounds.
With $u=x$, use $x=u,\;y=u^2v$, where $0\le u,v\le1$.

Compute and inspect the Jacobian.
The determinant is $u^2$; it is nonzero for $u>0$ and vanishes on the collapsed edge $u=0$.

Transform every part of the integral.
Substitute into the integrand and multiply by the absolute Jacobian before integrating over the rectangle.

When This Design Is Useful

The method applies whenever a region is bounded vertically by a lower curve $y=g(x)$ and an upper curve $y=h(x)$, with $h(x)>g(x)$ over the relevant $x$-interval. For points where the two curves differ, the normalized coordinate

v=\frac{y-g(x)}{h(x)-g(x)}

ranges from $0$ on the lower boundary to $1$ on the upper boundary. Solving gives $y=g(x)+v\bigl(h(x)-g(x)\bigr)$. The particular region here has $g(x)=0$ and $h(x)=x^2$, producing $y=u^2v$ after setting $u=x$.

This general observation is a way to design a candidate substitution, not a guarantee that every resulting map is usable everywhere. The boundary gap must be positive where the inverse ratio is used, and the Jacobian and one-to-one behavior must still be checked. In this example, the gap vanishes at $x=0$, which is why the inverse fails there and the edge collapses. Explicitly identifying such a set is part of a correct change-of-variables setup.

Check Your Understanding

Use the map $x=u,\;y=u^2v$ and its domain carefully in each question.

Why does $v=y/x^2$ range from $0$ to $1$ when $x>0$ and $(x,y)\in D$?
What point does the entire edge $u=0$ map to, and why does this prevent a one-to-one inverse on the closed square?
Compute the Jacobian determinant of $F(u,v)=(u,u^2v)$, and state where it vanishes.
Set up, but do not evaluate, the transformed integral for $\iint_D x\,dA$.
Explain why integrating $1$ over the unit square without a Jacobian gives the wrong area of $D$.