How to Change Variables and Bounds in a Double Integral

A change of variables can make a double integral easier to evaluate, but it changes more than the formula being integrated. You must also describe the original region in the new coordinates and account for how areas scale under the transformation.

The general rule is: substitute into the integrand, use bounds for the transformed region, and multiply by the absolute value of the appropriate Jacobian determinant. The example below shows how to find all three pieces.

The change-of-variables formula

Suppose you introduce new coordinates $u$ and $v$ through $u=g(x,y)$ and $v=h(x,y)$. Assume the transformation is one-to-one and continuously differentiable on an open set containing the region, has a nonzero Jacobian determinant there, and has a continuously differentiable inverse on its image. Write that inverse as $x=x(u,v)$ and $y=y(u,v)$. For an integrable function $f$, the change-of-variables formula is

$$\iint_R f(x,y)\,dx\,dy = \iint_S f\bigl(x(u,v),y(u,v)\bigr) \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv,$$

where $S$ is the image of the original region $R$ in the $uv$-plane. The Jacobian determinant is

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

Its absolute value converts a small area element $du\,dv$ into the corresponding area $dx\,dy$. You can instead calculate $\partial(u,v)/\partial(x,y)$ and use the reciprocal of its absolute value, provided that determinant is nonzero. The absolute value matters because area is positive, regardless of orientation.

How to find the new bounds

Use this sequence:

1. Solve for the old coordinates. Find $x$ and $y$ as functions of $u$ and $v$. This also reveals whether the transformation is one-to-one on the region.
2. Rewrite the region. Substitute the inverse formulas into the original inequalities or boundary equations. The resulting conditions describe $S$.
3. Choose an order of integration. Arrange the conditions as limits for the inner and outer variables. If no single pair of nested limits describes the whole region conveniently, divide it into simpler pieces.
4. Transform the integrand and area element. Substitute for $x$ and $y$, then multiply by the absolute Jacobian.
5. Integrate over the new bounds. Check that the bounds cover the mapped region exactly once.

It is usually safer to transform the region’s inequalities than to guess the new limits from a sketch alone. A sketch is still useful for checking that the limits have the right shape.

Worked example: a triangular region

Evaluate

$$\iint_R (x+y)\,dx\,dy, \qquad R=\{(x,y):x\geq 0,\ y\geq 0,\ x+y\leq 1\}.$$

Choose $u=x+y$ and $v=x-y$. Adding and subtracting these equations gives the inverse transformation

$$x=\frac{u+v}{2},\qquad y=\frac{u-v}{2}.$$

Now transform the region. The condition $x+y\leq 1$ becomes $u\leq 1$. Since $x,y\geq 0$, the inverse formulas give $u+v\geq 0$ and $u-v\geq 0$, or $-u\leq v\leq u$. Also, $u=x+y\geq 0$. Therefore the transformed region is

$$0\leq u\leq 1,\qquad -u\leq v\leq u.$$

The Jacobian determinant of the inverse transformation is

$$\frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{vmatrix} =-\frac{1}{2},$$

so $dx\,dy=\frac12\,du\,dv$. The integrand becomes $x+y=u$. Thus

$$\begin{aligned} \iint_R (x+y)\,dx\,dy &=\int_0^1\int_{-u}^{u} u\cdot\frac12\,dv\,du\\ &=\int_0^1 u^2\,du\\ &=\frac13. \end{aligned}$$

The inner integral ranges from $-u$ to $u$ because each fixed value of $u$ cuts the transformed triangle in a segment of that length. As a check, integrating directly in $x,y$ also gives $1/3$.

Common mistakes

• Changing the integrand but keeping the old bounds. The old bounds describe points in the $xy$-plane; the new bounds must describe the same region in the $uv$-plane.
• Forgetting the Jacobian. Substitution in the function alone does not account for the change in area.
• Using the wrong determinant. If you calculate $\partial(x,y)/\partial(u,v)$, use its absolute value directly. If you calculate the forward determinant $\partial(u,v)/\partial(x,y)$, use the absolute value of its reciprocal.
• Treating a curved or dependent region as a rectangle. New bounds may depend on the outer variable. For example, the worked example has $v$ limits that depend on $u$.
• Counting part of the region more than once. A transformation that is not one-to-one may require splitting the original region into pieces before applying the formula.

Changing variables versus reversing the order

Reversing the order of integration changes which variable is integrated first, but it does not change coordinates: the variables remain $x$ and $y$, and no Jacobian is needed. A change of variables introduces different coordinates, such as $u$ and $v$, so you must transform both the region and the area element. Sometimes both techniques can be used together, but they are separate steps.