Product-And-Ratio Substitutions for Positive Regions

Why Use Product-and-Ratio Coordinates?

Some regions in the positive $xy$-plane are described more naturally by the product $xy$ and the ratio $x/y$ than by $x$ and $y$ separately. Their boundary curves may look complicated in Cartesian coordinates, while conditions on $xy$ and $x/y$ become simple bounds in new coordinates.

For positive $x$ and $y$, set $x=uv$ and $y=u/v$. Then the product depends only on $u$, and the ratio depends only on $v$. This separation can simplify both the region and the integrand. The cost is a Jacobian factor, which must be included when changing the variables of integration.

Key idea: In the positive quadrant, $u$ records the square root of the product $xy$, while $v$ records the square root of the ratio $x/y$. Product boundaries become $u$-bounds, and ratio boundaries become $v$-bounds.

Definition and Inverse Coordinates

Consider the map from positive $u,v$ to positive $x,y$ given by $x=uv$ and $y=u/v$. Multiplying and dividing these equations gives $xy=u^2$ and $x/y=v^2$. Since all four variables are positive, the inverse is uniquely determined by taking positive square roots.

u=\sqrt{xy}, \qquad v=\sqrt{\frac{x}{y}}.

Definition: The product-and-ratio substitution on the positive quadrant is the change of variables $$ x=uv,\qquad y=\frac{u}{v},\qquad u>0,\quad v>0. $$ Its inverse is $u=\sqrt{xy}$, $v=\sqrt{x/y}$. It is one-to-one from the positive $uv$-quadrant onto the positive $xy$-quadrant.

The positivity condition matters. It ensures that the positive square roots give the inverse and that no sign choices or multiple branches are needed. For instance, if $x$ or $y$ were allowed to be negative, these formulas would not describe the whole region without additional cases.

Computing the Jacobian

The Jacobian determinant measures the local area-scaling factor of the map. Differentiate $x=uv$ and $y=u/v$ with respect to $u$ and $v$:

\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} v & u\\[4pt] 1/v & -u/v^2 \end{vmatrix} = -\frac{u}{v}-\frac{u}{v} = -\frac{2u}{v}.

Because $u$ and $v$ are positive, this determinant is never zero in the domain. Its sign records orientation, not the sign of area. A change-of-variables formula for an ordinary scalar area integral uses the absolute value of the determinant:

\left|\frac{\partial(x,y)}{\partial(u,v)}\right|=\frac{2u}{v}, \qquad dx\,dy=\frac{2u}{v}\,du\,dv.

Theorem: Let $D$ be a region in the positive $xy$-quadrant, and let $R$ be its image in the positive $uv$-quadrant under $x=uv,\ y=u/v$. If $f$ is integrable on $D$, then $$ \iint_D f(x,y)\,dx\,dy = \iint_R f\!\left(uv,\frac{u}{v}\right)\frac{2u}{v}\,du\,dv. $$ The absolute Jacobian factor is required because the map reverses orientation but area is nonnegative.

Why the Bounds Separate

The substitution has been designed so that its two coordinates isolate the product and ratio:

xy=(uv)\left(\frac{u}{v}\right)=u^2, \qquad \frac{x}{y}=\frac{uv}{u/v}=v^2.

Thus, for positive constants $a<b$, the condition $a\le xy\le b$ is equivalent to $\sqrt a\le u\le\sqrt b$. Likewise, $c\le x/y\le d$, for positive $c<d$, is equivalent to $\sqrt c\le v\le\sqrt d$. When the product and ratio bounds are independent, their transformed region is a rectangle. If one condition relates the product and ratio, the transformed bounds can instead depend on one another.

This translation must preserve every inequality. For example, $x/y\le xy$ becomes $v^2\le u^2$, and because $u,v>0$, this is equivalent to $v\le u$. The positivity is what permits taking square roots without reversing or branching the inequality.

Worked Examples

Worked Example: Area of a Product-and-Ratio Rectangle

Find the area of the positive region $D$ defined by $1\le xy\le4$ and $1\le x/y\le9$.

Use $xy=u^2$ and $x/y=v^2$. The product inequalities give $1\le u^2\le4$, so $1\le u\le2$. The ratio inequalities give $1\le v^2\le9$, so $1\le v\le3$. Therefore the transformed region is the rectangle $1\le u\le2,\ 1\le v\le3$.

Area is the integral of $1$. Substitute the absolute Jacobian:

\begin{aligned} \iint_D 1\,dx\,dy &=\int_1^2\int_1^3 \frac{2u}{v}\,dv\,du\\ &=\int_1^2 2u\,[\ln v]_1^3\,du\\ &=\int_1^2 2u\ln 3\,du\\ &=\ln 3\,[u^2]_1^2\\ &=3\ln 3. \end{aligned}

Check the bounds by mapping back: $xy=u^2$ ranges from $1$ to $4$, and $x/y=v^2$ ranges from $1$ to $9$, exactly as required. The final antiderivative is consistent because $\frac{d}{du}(u^2)=2u$; the area is positive, as it must be.

Worked Example: A Triangular Transformed Region

Evaluate $\iint_D (x/y)\,dx\,dy$, where $D$ is defined by $1\le xy\le4$ and $1\le x/y\le xy$.

The product condition gives $1\le u\le2$. The ratio condition gives $1\le v^2\le u^2$. Since $u,v>0$, the transformed bounds are $1\le v\le u$. Thus the region is not a rectangle: its upper $v$-bound depends on $u$.

The integrand $x/y$ becomes $v^2$. Multiplying by the absolute Jacobian gives the transformed integrand $v^2(2u/v)=2uv$:

\begin{aligned} \iint_D \frac{x}{y}\,dx\,dy &=\int_1^2\int_1^u 2uv\,dv\,du\\ &=\int_1^2 2u\left[\frac{v^2}{2}\right]_1^u\,du\\ &=\int_1^2 u(u^2-1)\,du\\ &=\int_1^2 (u^3-u)\,du\\ &=\left[\frac{u^4}{4}-\frac{u^2}{2}\right]_1^2\\ &=(4-2)-\left(\frac14-\frac12\right)\\ &=\frac94. \end{aligned}

Check the bounds: $1\le v\le u$ implies $1\le x/y=v^2\le u^2=xy$, so the ratio condition is recovered. For the evaluation, differentiating $u^4/4-u^2/2$ gives $u^3-u$, the outer integrand. The result is positive, consistent with the positive integrand on $D$.

Worked Example: A Ratio Integrand with Dependent Bounds

Evaluate $\iint_D (y/x)\,dx\,dy$, where $D$ is given by $1\le xy\le4$ and $xy\le x/y\le4xy$.

The product condition gives $1\le u\le2$. The ratio conditions become $u^2\le v^2\le4u^2$. Positivity gives $u\le v\le2u$. Also, $y/x=1/(x/y)=1/v^2$. Including the Jacobian, the transformed integrand is $2u/v^3$.

\begin{aligned} \iint_D \frac{y}{x}\,dx\,dy &=\int_1^2\int_u^{2u}\frac{2u}{v^3}\,dv\,du\\ &=\int_1^2 2u\left[-\frac{1}{2v^2}\right]_{u}^{2u}\,du\\ &=\int_1^2 \left(\frac{1}{u}-\frac{1}{4u}\right)\,du\\ &=\frac34[\ln u]_1^2\\ &=\frac34\ln2. \end{aligned}

Check the inner step: evaluating $-u/v^2$ at $2u$ and $u$ gives $-1/(4u)+1/u=3/(4u)$, as used. Differentiating $\ln u$ gives $1/u$, so the outer evaluation is correct. The bounds $u\le v\le2u$ recover $xy\le x/y\le4xy$.

A Derivation of the Change-of-Variables Factor

The Jacobian factor is not an extra adjustment chosen to make an answer work. It follows from how the map changes a small area. Near a point $(u,v)$, a small rectangle with side lengths $\Delta u$ and $\Delta v$ maps approximately to a parallelogram. Its two side vectors are approximately the two columns of the derivative matrix, scaled by $\Delta u$ and $\Delta v$.

Form the derivative matrix.
The columns are $(x_u,y_u)=(v,1/v)$ and $(x_v,y_v)=(u,-u/v^2)$.

Compute the parallelogram area.
The area of the parallelogram spanned by two planar vectors is the absolute value of their determinant. Here that determinant is $-2u/v$, so the area scale is $2u/v$.

Replace area elements.
A small $uv$-rectangle of area $\Delta u\,\Delta v$ therefore corresponds locally to $xy$-area approximately $(2u/v)\Delta u\,\Delta v$. Summing these local contributions gives $dx\,dy=(2u/v)\,du\,dv$ in the change-of-variables formula.

This argument explains why the determinant appears and why its absolute value is used: a parallelogram has positive area even when its spanning vectors have negative orientation. The change-of-variables theorem makes this local area scaling valid for the integral over the whole region.

Common Pitfall: Keeping the Signed Determinant

A frequent error is to substitute the determinant $-2u/v$ directly for the area factor. For the area region in the first example, that mistake would give

\int_1^2\int_1^3 -\frac{2u}{v}\,dv\,du=-3\ln3,

which cannot be the area of a region. The map reverses orientation, so its determinant is negative, but area integrals use the absolute value. The correction is to use $2u/v$, giving $3\ln3>0$. Another essential check is to translate inequalities only after noting $u,v>0$; that condition justifies replacing $v^2\le u^2$ by $v\le u$.

When This Substitution Is Useful

Look for boundaries or integrands involving $xy$, $x/y$, or simple powers of these expressions. In the new coordinates, $(xy)^m(x/y)^n$ becomes $u^{2m}v^{2n}$, when the powers are defined on the positive quadrant. The Jacobian then contributes the additional factor $2u/v$. This may turn a complicated-looking integrand into a product of a function of $u$ and a function of $v$, or make the limits easier to describe.

The substitution is not automatically advantageous. If the region is already simple in $x,y$ and the transformed bounds or integrand become more difficult, there may be no benefit. Before integrating, check that the map is one-to-one on the chosen region, translate every boundary, and include the absolute Jacobian. For this map, the entire positive quadrant is a valid one-to-one region, so those checks are especially direct.

Check Your Understanding

Use the inverse coordinates and the absolute Jacobian when answering these questions.

If $xy$ ranges from $4$ to $25$, what are the corresponding bounds for $u$?
If $x/y$ ranges from $1$ to $16$, what are the corresponding bounds for $v$?
What is the absolute value of the Jacobian determinant, and why is it used rather than the signed determinant?
Translate $1\le xy\le9$ and $xy\le x/y\le4xy$ into bounds for $u$ and $v$.
For $1\le xy\le4$ and $1\le x/y\le xy$, explain why the transformed region is not rectangular.