A Reliable Setup for General Substitution

Why a Reliable Setup Matters

A change of variables can simplify an integral by replacing a complicated region or integrand with a more manageable one. But a correct substitution changes more than the symbols in the integrand. It also changes the region of integration and introduces a volume-scaling factor, the absolute value of the Jacobian determinant.

A dependable setup keeps these tasks in a fixed order: define the map and its parameter domain, describe the image region, compute the Jacobian, and then transform the integrand and integration limits. Keeping the steps separate makes it easier to detect missing factors or mismatched bounds.

Definition: Let $D$ be a region in parameter coordinates $\mathbf{u}$, and let $T:D\to R$ be a coordinate map into a region $R$. In two variables, write $T(u,v)=(x(u,v),y(u,v))$; in three variables, write $T(u,v,w)=(x(u,v,w),y(u,v,w),z(u,v,w))$. The Jacobian determinant is the determinant of the matrix of first partial derivatives of the map.

The domain $D$ describes the parameter values. The image $R=T(D)$ is the region in the original coordinates. Bounds in the integral must describe one of these regions consistently: either integrate over $R$ in $x,y,z$, or over $D$ in $u,v,w$.

The Change-of-Variables Formula

For a two-variable map, its Jacobian determinant is

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

For a three-variable map, the Jacobian determinant is the determinant of the corresponding $3\times3$ matrix. The notation $dA$ denotes area and $dV$ denotes volume.

Theorem: Suppose $T:D\to R$ is continuously differentiable and one-to-one on the interior of $D$, with nonzero Jacobian determinant there. Assume the boundaries do not affect the integral and the integrand is integrable. Then $$ \iint_R f(x,y)\,dA = \iint_D f(x(u,v),y(u,v)) \left|\frac{\partial(x,y)}{\partial(u,v)}\right|\,du\,dv. $$ In three variables, the corresponding formula is $$ \iiint_R f(x,y,z)\,dV = \iiint_D f(x(u,v,w),y(u,v,w),z(u,v,w)) \left|\frac{\partial(x,y,z)}{\partial(u,v,w)}\right|\,du\,dv\,dw. $$

The absolute value is essential: the determinant can be negative if the map reverses orientation, but area and volume are nonnegative. If a map covers a region more than once, the one-to-one hypothesis fails; the contributions from different parameter points must not be treated as though the map were injective. The earlier material on multiplicity addresses that issue.

A Practical Setup Procedure

Before integrating, make each part of the substitution explicit. First specify the map from parameter coordinates to original coordinates and give the proposed parameter domain. Next use the map to determine the image region, checking that the stated bounds describe exactly that image. Then compute the Jacobian determinant, substitute the coordinate formulas into the integrand, and multiply by the absolute Jacobian. Finally, write the integral over the parameter domain using bounds that match the chosen order of integration.

Define the map and parameter domain.
Write $T$ explicitly and state the region $D$ on which it is being used.

Determine the image region.
Translate the restrictions on the parameters into bounds or inequalities in the original coordinates, and check that the map covers the intended region.

Compute the Jacobian.
Form the derivative matrix, evaluate its determinant, and use its absolute value in the integral.

Transform the integrand and bounds.
Replace each original coordinate using the map, include the Jacobian factor, and integrate over $D$.

Why the Jacobian Appears

Near a point where the map is differentiable, a small change in parameter coordinates is approximated by the derivative matrix acting on that change. In two dimensions, that matrix sends a small rectangle to an approximately parallelogram-shaped piece. The area of the parallelogram is the original small area multiplied by the absolute determinant of the matrix. In three dimensions, the same reasoning applies to a small box and its approximately parallelepiped-shaped image.

Thus the Jacobian accounts for how much area or volume each small parameter region contributes in the original coordinates. The substitution theorem makes this local scaling rule precise for integrals. This explains why changing the integrand and bounds alone is not enough: the area or volume element must also be scaled.

Worked Examples

Worked Example: A Curved Boundary in a Double Integral

Let $D$ be the unit square $0\leq u\leq1,\;0\leq v\leq1$, and define $x=u,\;y=(1+u)v$. Set up and evaluate the integral of $xy$ over the image $R=T(D)$.

For $0\leq u\leq1$, $x=u$ ranges from $0$ to $1$. Since $y=(1+u)v$ and $v$ ranges from $0$ to $1$, the $y$-bounds are $0\leq y\leq1+u$. Replacing $u$ by $x$, the image is $0\leq x\leq1,\;0\leq y\leq1+x$. The inverse formulas are $u=x,\;v=y/(1+x)$, so each point of this region comes from exactly one point of the square.

The derivative matrix and determinant are

\frac{\partial(x,y)}{\partial(u,v)} = \det \begin{pmatrix} 1 & 0\\ v & 1+u \end{pmatrix} =1+u.

This determinant is positive on $D$. The transformed integrand is $xy=u(1+u)v$, so including the Jacobian gives $u(1+u)^2v$. Therefore

\begin{aligned} \iint_R xy\,dA &=\int_0^1\int_0^1 u(1+u)^2v\,dv\,du\\ &=\frac12\int_0^1 (u+2u^2+u^3)\,du\\ &=\frac12\left(\frac12+\frac23+\frac14\right)\\ &=\frac12\left(\frac{6+8+3}{12}\right) =\frac{17}{24}. \end{aligned}

Check directly over $R$:

\begin{aligned} \iint_R xy\,dA &=\int_0^1\int_0^{1+x}xy\,dy\,dx\\ &=\frac12\int_0^1 x(1+x)^2\,dx\\ &=\frac12\left(\frac12+\frac23+\frac14\right) =\frac{17}{24}. \end{aligned}

The two calculations agree, including the nonconstant Jacobian factor.

Worked Example: Transforming Bounds for a Shifted Region

Let $0\leq u\leq1,\;0\leq v\leq1$, with $x=u,\;y=v+u^2$. Evaluate the integral of $x+y$ over the image region.

From $x=u$, we have $0\leq x\leq1$. For each fixed $u$, $v$ ranges from $0$ to $1$, so $u^2\leq y\leq u^2+1$. Replacing $u$ by $x$, the image is $0\leq x\leq1,\;x^2\leq y\leq x^2+1$. The map is one-to-one because its inverse is $u=x,\;v=y-x^2$.

The Jacobian determinant is

\frac{\partial(x,y)}{\partial(u,v)} = \det \begin{pmatrix} 1 & 0\\ 2u & 1 \end{pmatrix} =1.

Substitute $x=u$ and $y=v+u^2$. The integral becomes

\begin{aligned} \iint_R (x+y)\,dA &=\int_0^1\int_0^1 (u+v+u^2)\,dv\,du\\ &=\int_0^1\left(u+\frac12+u^2\right)\,du\\ &=\frac12+\frac12+\frac13 =\frac43. \end{aligned}

To check, integrate directly over the image:

\begin{aligned} \iint_R(x+y)\,dA &=\int_0^1\int_{x^2}^{x^2+1}(x+y)\,dy\,dx\\ &=\int_0^1\left(x+x^2+\frac12\right)\,dx\\ &=\frac12+\frac13+\frac12 =\frac43. \end{aligned}

Worked Example: A Triple-Integral Setup

Let $D$ be the unit cube $0\leq u,v,w\leq1$, and define $x=u,\;y=v,\;z=(1+u+v)w$. Find the volume of its image $R$.

The first two coordinates range over $0\leq x\leq1,\;0\leq y\leq1$. For each $u,v$, the parameter $w$ ranges from $0$ to $1$; therefore $z$ ranges from $0$ to $1+u+v$. In original coordinates the image is $0\leq x\leq1,\;0\leq y\leq1,\;0\leq z\leq1+x+y$. The inverse is $u=x,\;v=y,\;w=z/(1+x+y)$, which is defined throughout the image.

The Jacobian determinant is

\frac{\partial(x,y,z)}{\partial(u,v,w)} = \det \begin{pmatrix} 1&0&0\\ 0&1&0\\ w&w&1+u+v \end{pmatrix} =1+u+v.

It is positive throughout the cube. For volume, the integrand is $1$, so

\begin{aligned} \iiint_R 1\,dV &=\int_0^1\int_0^1\int_0^1(1+u+v)\,dw\,dv\,du\\ &=\int_0^1\int_0^1(1+u+v)\,dv\,du\\ &=\int_0^1\left(\frac32+u\right)\,du\\ &=\frac32+\frac12=2. \end{aligned}

Check in the original region by integrating first with respect to $z$:

\iiint_R1\,dV =\int_0^1\int_0^1\int_0^{1+x+y}1\,dz\,dy\,dx =\int_0^1\int_0^1(1+x+y)\,dy\,dx =2.

Common Pitfall: Omitting the Jacobian

In the first example, substituting $x=u,\;y=(1+u)v$ into $xy$ gives $u(1+u)v$. It would be wrong to stop there and integrate that expression over the unit square. That calculation gives

\int_0^1\int_0^1u(1+u)v\,dv\,du =\frac12\left(\frac12+\frac13\right) =\frac{5}{12},

which does not equal the correct value $17/24$. The error is the missing area-scaling factor $1+u$. The corrected integrand is $u(1+u)v(1+u)=u(1+u)^2v$, and its integral is $17/24$, as verified above.

Other setup errors have the same underlying cause: treating the substitution as a relabeling rather than a transformation. Bounds must describe the actual parameter domain, each original coordinate must be replaced in the integrand, and the absolute Jacobian must multiply that new integrand.

Final Checks Before Integrating

Before evaluating a general substitution, check that the map covers the intended region and is one-to-one on the chosen domain, or account properly for any multiplicity. Confirm that the Jacobian is computed from the stated map, not its inverse, and use its absolute value. Finally, check that the transformed bounds and transformed integrand use the same parameter coordinates. When the original integral is manageable, a direct evaluation provides a useful independent check.

Check Your Understanding

For each question, identify the map, region, Jacobian, or condition that governs the setup.

What are the four main steps in setting up a change of variables?
Why does the integral use the absolute value of the Jacobian determinant?
For $x=u,\;y=v+u^2$ on the unit square, what inequalities describe the image in $x,y$ coordinates?
What factor must multiply the transformed integrand in a double integral?
Why must injectivity, or a careful treatment of multiplicity, be considered before applying the formula?