Sum-and-Difference Substitutions for Symmetric Integrands

Why use sums and differences?

Some integrands are awkward when written in \(x\) and \(y\), but simple when expressed through \(x+y\) and \(x-y\). This happens especially when the same expressions also describe the boundaries of the region. The substitution \(u=x+y\), \(v=x-y\) makes those expressions into coordinate variables, so the transformed integrand and region may both become easier to handle.

The substitution is also well suited to integrands symmetric in \(x\) and \(y\). Swapping \(x\) and \(y\) leaves \(u\) unchanged and changes \(v\) to \(-v\). Consequently, a symmetric integrand written in the new variables is unchanged when \(v\) changes sign. This symmetry can make the transformed integral easier to organize, although the region and the Jacobian must still be handled correctly.

The substitution and its Jacobian

Definition: Use the linear change of variables \(u=x+y\), \(v=x-y\). Solving these equations for \(x\) and \(y\) gives the inverse map \(x=(u+v)/2\), \(y=(u-v)/2\). The Jacobian determinant of the inverse map is \(-1/2\), so the area element transforms according to \(dx\,dy=\frac12\,du\,dv\).

To compute that determinant, differentiate the inverse formulas. The resulting matrix has rows corresponding to \(x\) and \(y\), and columns corresponding to \(u\) and \(v\):

\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \frac12 & \frac12\\ \frac12 & -\frac12 \end{vmatrix} = -\frac14-\frac14 = -\frac12.

For a double integral, the area-scaling factor is the absolute value of this determinant. Thus \(dx\,dy=\frac12\,du\,dv\). The negative sign records orientation; it does not make area negative.

This map is invertible everywhere: each pair \((u,v)\) determines exactly one pair \((x,y)\), and conversely. Therefore, for a region on which the integrand is integrable, the change-of-variables formula is

\iint_D f(x,y)\,dx\,dy = \iint_{D'} f\!\left(\frac{u+v}{2},\frac{u-v}{2}\right) \frac12\,du\,dv,

where \(D'\) is the image of the original region \(D\) under \(u=x+y,\ v=x-y\). In practice, it is often simplest to rewrite the integrand directly using \(x+y=u\) and \(x-y=v\), then determine \(D'\) from the original boundary conditions.

Useful algebra in the new coordinates

The substitution makes expressions involving sums and differences immediate. For example, the square terms can be rewritten by adding or subtracting \(u^2\) and \(v^2\):

x^2+y^2=\frac{u^2+v^2}{2}, \qquad xy=\frac{u^2-v^2}{4}, \qquad (x+y)^2=u^2.

For instance, \(u^2+v^2=(x+y)^2+(x-y)^2=2x^2+2y^2\), which gives the first identity. Likewise, \(u^2-v^2=(x+y)^2-(x-y)^2=4xy\), giving the second. These identities can turn a symmetric polynomial in \(x,y\) into a particularly simple expression in \(u,v\).

Squares become diamonds

A square aligned with the \(x\)- and \(y\)-axes generally does not remain a rectangle in the \(u,v\)-plane. The square \(-1\leq x\leq1,\ -1\leq y\leq1\), for example, becomes the diamond \(|u|+|v|\leq2\).

To see why, the inverse formulas show that the square’s conditions are \(|u+v|\leq2\) and \(|u-v|\leq2\). For real \(u,v\), the larger of \(|u+v|\) and \(|u-v|\) equals \(|u|+|v|\). If \(u\) and \(v\) have the same sign, \(|u+v|=|u|+|v|\); if they have opposite signs, \(|u-v|=|u|+|v|\). The other absolute value is no larger. Hence both square conditions hold exactly when \(|u|+|v|\leq2\).

Worked Example: Integrate over a square by using its diamond image

Evaluate \(\iint_D (x+y)^2\,dx\,dy\), where \(D\) is the square \(-1\leq x\leq1,\ -1\leq y\leq1\).

Set \(u=x+y,\ v=x-y\). The integrand becomes \(u^2\), the area factor is \(1/2\), and the square becomes \(|u|+|v|\leq2\). For each fixed \(u\) between \(-2\) and \(2\), the allowed \(v\)-values run from \(-(2-|u|)\) to \(2-|u|\). Thus

\begin{aligned} \iint_D (x+y)^2\,dx\,dy &=\frac12\int_{-2}^{2}\int_{-(2-|u|)}^{2-|u|}u^2\,dv\,du\\ &=\frac12\int_{-2}^{2}2u^2(2-|u|)\,du\\ &=\int_{-2}^{2}u^2(2-|u|)\,du\\ &=2\int_0^2 u^2(2-u)\,du\\ &=2\left[\frac{2u^3}{3}-\frac{u^4}{4}\right]_0^2\\ &=2\left(\frac{16}{3}-4\right) =\frac83. \end{aligned}

Check by integrating in the original square. Expand \((x+y)^2=x^2+2xy+y^2\). The integral of \(2xy\) vanishes because the domain is symmetric in each variable, while each square term contributes \((\int_{-1}^1 x^2\,dx)(\int_{-1}^1 1\,dy)=\frac23\cdot2=\frac43\). The total is \(\frac43+\frac43=\frac83\), agreeing with the transformed calculation.

Diamond-shaped regions can become rectangles

The transformation also works in the opposite direction. A region described by separate bounds on \(x+y\) and \(x-y\) becomes a rectangle in the \(u,v\)-plane. In the original \(x,y\)-plane, those same bounds generally describe a tilted parallelogram. This is useful when the integrand also has a simple expression in \(u\) and \(v\).

Worked Example: Transform a parallelogram described by two linear expressions

Evaluate \(\iint_D xy\,dx\,dy\), where \(D\) is given by \(|x+y|\leq1\) and \(|x-y|\leq2\).

In the new coordinates the region is the rectangle \(-1\leq u\leq1,\ -2\leq v\leq2\). Since \(xy=(u^2-v^2)/4\) and \(dx\,dy=\frac12\,du\,dv\),

\begin{aligned} \iint_D xy\,dx\,dy &=\frac18\int_{-1}^{1}\int_{-2}^{2}(u^2-v^2)\,dv\,du\\ &=\frac18\left( 4\int_{-1}^{1}u^2\,du - 2\int_{-2}^{2}v^2\,dv \right)\\ &=\frac18\left(4\cdot\frac23-2\cdot\frac{16}{3}\right)\\ &=\frac18\left(\frac83-\frac{32}{3}\right) =-1. \end{aligned}

For a check, the \(u^2\) contribution before multiplying by \(1/8\) is \(4(2/3)=8/3\). The \(v^2\) contribution is \(2(16/3)=32/3\). Their difference is \(-8\), and \(-8/8=-1\). The negative result is consistent with the transformed expression: across this rectangle, the \(v^2\) term has the larger integral.

When the integrand and region use the same coordinates

A particularly efficient setup occurs when both the integrand and the boundaries are already expressed using \(x+y\) and \(x-y\). Replacing those expressions by \(u\) and \(v\) then gives a rectangular integral without further boundary calculations. The inverse map remains important: it confirms that each point of the new rectangle corresponds to exactly one point of the original region.

Worked Example: Evaluate on a region bounded by sums and differences

Let \(D\) be the region \(0\leq x+y\leq2,\ 0\leq x-y\leq1\). Evaluate \(\iint_D e^{x+y}(x-y)\,dx\,dy\).

Set \(u=x+y,\ v=x-y\). The region becomes \(0\leq u\leq2,\ 0\leq v\leq1\), and the integrand becomes \(e^u v\). Including the area factor gives

\begin{aligned} \iint_D e^{x+y}(x-y)\,dx\,dy &=\frac12\int_0^2\int_0^1 e^u v\,dv\,du\\ &=\frac12\left(\int_0^2 e^u\,du\right) \left(\int_0^1 v\,dv\right)\\ &=\frac12(e^2-1)\left(\frac12\right)\\ &=\frac{e^2-1}{4}. \end{aligned}

To check the transformed bounds, the inverse map sends the rectangle’s corners \((u,v)=(0,0),(2,0),(0,1),(2,1)\) to \((x,y)=(0,0),(1,1),(1/2,-1/2),(3/2,1/2)\), respectively. These are precisely the four vertices of the parallelogram described by the stated inequalities. The integrals used above also check directly: \(\int_0^2 e^u\,du=e^2-1\) and \(\int_0^1 v\,dv=1/2\).

Why the method works—and a common error

The substitution is effective because it aligns the coordinates with the expressions that occur in the problem. The change-of-variables theorem accounts for how the map scales area; the inverse determinant shows that a small area in the \(u,v\)-plane corresponds to half as much area in the \(x,y\)-plane. That factor is present no matter how simple the transformed integrand looks.

A frequent mistake is to transform the integrand and bounds correctly but omit the factor \(1/2\). For the last example, this error would give \((e^2-1)/2\) instead of \((e^2-1)/4\). The correction follows directly from the inverse Jacobian:

\left|\frac{\partial(x,y)}{\partial(u,v)}\right| = \left|\begin{vmatrix}\frac12&\frac12\\[2pt]\frac12&-\frac12\end{vmatrix}\right| =\frac12.

Always distinguish the transformed integrand from the transformed area element. The integrand is rewritten by substituting expressions such as \(x+y=u\); the area element must separately be multiplied by the absolute Jacobian determinant. Also check that the transformed region matches the original inequalities. A correct integrand and Jacobian cannot compensate for incorrect bounds.

Key takeaway: For \(u=x+y,\ v=x-y\), use \(x=(u+v)/2,\ y=(u-v)/2\) and \(dx\,dy=\frac12\,du\,dv\). Transform every boundary condition as well as the integrand, and use absolute values when expressing the image of a square as a diamond.

Check Your Understanding

Use the inverse map and the transformed region carefully in each question.

Find \(x\) and \(y\) in terms of \(u\) and \(v\), and compute the absolute value of the inverse Jacobian determinant.
Rewrite \(x^2+y^2\) and \(xy\) using \(u=x+y\) and \(v=x-y\).
What region in the \(u,v\)-plane corresponds to the square \(-2\leq x\leq2,\ -2\leq y\leq2\)?
What rectangle in the \(u,v\)-plane corresponds to \(|x+y|\leq3,\ |x-y|\leq1\)?
Why does the transformed area element contain a factor of \(1/2\), even if the transformed integrand contains no fractions?