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Jacobians and General Substitutions · Tutorial 684 of 999

Linear Changes of Variables in the Plane

Use an invertible linear map to rewrite planar integrals, transform region bounds, and account correctly for area scaling.

Advanced 10 min read

What You'll Learn

  • Represent planar linear substitutions with a matrix and identify when the map is invertible.
  • Derive the constant area-scaling factor from the determinant.
  • Transform region bounds using the inverse linear map.
  • Set up and evaluate integrals after shear, rotation, and mixed-coordinate substitutions.
  • Distinguish orientation from area scaling by using the absolute determinant.
  • Diagnose the error caused by omitting the determinant factor.

Why Linear Changes of Variables Are Useful

A planar region may have awkward boundaries in the coordinates \(x,y\) but simple bounds in different coordinates. An invertible linear change of variables can turn tilted parallelograms into rectangles, or express a rotated region using ordinary rectangular bounds. The integral becomes easier to set up, provided the region, integrand, and area factor are all transformed consistently.

The essential feature of a linear map is that it sends the origin to the origin and scales every direction in a fixed way. Its Jacobian determinant is constant, so the area correction does not vary from point to point. The determinant's sign records orientation, while its absolute value gives the factor by which area is scaled.

Definition: An invertible linear change of variables in the plane has the form \(\mathbf{x}=\mathbf{A}\mathbf{u}\), where \(\mathbf{x}=(x,y)^{\top}\), \(\mathbf{u}=(u,v)^{\top}\), and \(\mathbf{A}\) is a \(2\times2\) matrix with \(\det\mathbf{A}\ne0\). The map is invertible, with \(\mathbf{u}=\mathbf{A}^{-1}\mathbf{x}\).

The Integral Transformation Formula

Write the matrix and its determinant as

$$ \mathbf{A} = \begin{pmatrix} a&b\\ c&d \end{pmatrix}, \qquad \det\mathbf{A}=ad-bc. $$

The coordinate equations are \(x=au+bv\) and \(y=cu+dv\). The Jacobian matrix of this map is \(\mathbf{A}\) itself, so the Jacobian determinant is the constant \(ad-bc\). If \(G\) is a region in the \(uv\)-plane and \(D=\mathbf{A}(G)\) is its image in the \(xy\)-plane, then for an integrable function \(f\),

$$ \iint_D f(x,y)\,dA = \iint_G f(au+bv,cu+dv)\,|ad-bc|\,du\,dv. $$

Here \(dA\) denotes area in the original \(xy\)-plane. This formula applies when the map is invertible and the regions and integrand satisfy the usual hypotheses of the change-of-variables theorem; for example, it is valid for a continuous integrand over a bounded region with a piecewise-smooth boundary. The transformed region \(G\) must correspond to \(D\) under the stated map.

Key point: Transform both the integrand and the region, and multiply by \(|\det\mathbf{A}|\). The determinant may be negative, but an area factor cannot be negative.

Why the Determinant Scales Area

The columns of \(\mathbf{A}\), \((a,c)^{\top}\) and \((b,d)^{\top}\), are the images of the unit coordinate directions. A rectangle with side lengths \(\Delta u\) and \(\Delta v\) has those side directions after mapping, so its image is a parallelogram with adjacent side vectors \((a,c)^{\top}\Delta u\) and \((b,d)^{\top}\Delta v\).

The area of a parallelogram spanned by \((p_1,p_2)^{\top}\) and \((q_1,q_2)^{\top}\) is \(|p_1q_2-p_2q_1|\). Applying this formula to the image sides gives

$$ \left| (a\Delta u)(d\Delta v)-(c\Delta u)(b\Delta v)\right| = |ad-bc|\,\Delta u\,\Delta v. $$

Thus every small rectangle's area is multiplied by the same factor \(|\det\mathbf{A}|\). To see why this yields the integral formula, partition \(G\) into small rectangles. Each rectangle maps to a parallelogram whose area is its original area multiplied by \(|\det\mathbf{A}|\). On each small piece, the function's value is \(f(au+bv,cu+dv)\) at the corresponding mapped point. Therefore the Riemann sums for the integral over \(D\) are the sums over \(G\) with each area contribution multiplied by \(|\det\mathbf{A}|\). Taking the limit gives the displayed transformation formula.

If \(\det\mathbf{A}<0\), the map reverses orientation: the order of the two coordinate directions is reversed. The parallelogram's area remains positive, which is why the formula uses the absolute value.

Worked Example: A Shear Turns a Tilted Region into a Rectangle

Worked Example: A Shear Turns a Tilted Region into a Rectangle

Evaluate \(\iint_D (x-y)\,dA\), where \(D\) is bounded by \(0\le y\le1\), \(y\le x\), and \(x\le y+2\).

Use the shear \(x=u+v,\ y=v\). Its matrix and determinant are

$$ \mathbf{A} = \begin{pmatrix} 1&1\\ 0&1 \end{pmatrix}, \qquad \det\mathbf{A}=1\cdot1-1\cdot0=1. $$

The inverse equations are \(v=y\) and \(u=x-y\). The bounds \(0\le y\le1\) become \(0\le v\le1\). The bounds \(y\le x\le y+2\), after subtracting \(y\), become \(0\le u\le2\). Thus \(G\) is the rectangle \(0\le u\le2,\ 0\le v\le1\).

The integrand becomes \(x-y=(u+v)-v=u\), and the area factor is \(|\det\mathbf{A}|=1\). Therefore

$$ \begin{aligned} \iint_D (x-y)\,dA &=\int_0^1\int_0^2 u\,du\,dv\\ &=\int_0^1\left[\frac{u^2}{2}\right]_0^2\,dv\\ &=\int_0^1 2\,dv\\ &=[2v]_0^1\\ &=2. \end{aligned} $$

As a check in the original coordinates, for each \(y\) from \(0\) to \(1\), \(x-y\) ranges from \(0\) to \(2\). Hence the inner integral is \(\int_y^{y+2}(x-y)\,dx=[(x-y)^2/2]_y^{y+2}=2\), and the outer integral is \(\int_0^1 2\,dy=2\), as required.

Worked Example: A Rotation Preserves Area

Worked Example: A Rotation Preserves Area

Evaluate \(\iint_D (x^2+y^2)\,dA\), where \(D\) is described by \(0\le(x+y)/\sqrt{2}\le2\) and \(0\le(y-x)/\sqrt{2}\le1\).

Set \(u=(x+y)/\sqrt{2}\) and \(v=(y-x)/\sqrt{2}\). Solving for \(x,y\) gives \(x=(u-v)/\sqrt{2}\) and \(y=(u+v)/\sqrt{2}\), so the forward map is

$$ \mathbf{A} = \frac{1}{\sqrt{2}} \begin{pmatrix} 1&-1\\ 1&1 \end{pmatrix}, \qquad \det\mathbf{A} = \frac{1}{2}\bigl(1\cdot1-(-1)\cdot1\bigr) =1. $$

The given inequalities are exactly the rectangular bounds \(0\le u\le2,\ 0\le v\le1\). Also,

$$ x^2+y^2 = \frac{(u-v)^2+(u+v)^2}{2} = \frac{2u^2+2v^2}{2} = u^2+v^2. $$

Consequently,

$$ \begin{aligned} \iint_D (x^2+y^2)\,dA &=\int_0^1\int_0^2 (u^2+v^2)\,du\,dv\\ &=\int_0^1\left[\frac{u^3}{3}+v^2u\right]_0^2\,dv\\ &=\int_0^1\left(\frac{8}{3}+2v^2\right)\,dv\\ &=\left[\frac{8v}{3}+\frac{2v^3}{3}\right]_0^1\\ &=\frac{8}{3}+\frac{2}{3}= \frac{10}{3}. \end{aligned} $$

The determinant equals \(1\), consistent with a rotation preserving area. As a check on the transformed integrand, substituting the inverse expressions into \(x^2+y^2\) gives \(u^2+v^2\), exactly the function integrated above. The bounds are the original inequalities rewritten as \(0\le u\le2\) and \(0\le v\le1\), so the transformed rectangle covers precisely \(D\).

Worked Example: Mixed Coordinates and a Nonunit Area Factor

Worked Example: Mixed Coordinates and a Nonunit Area Factor

Find the area of the region \(D\) defined by \(0\le(2x-y)/3\le1\) and \(-1\le(-x+2y)/3\le1\).

Choose \(u=(2x-y)/3\) and \(v=(-x+2y)/3\). The inverse relations are \(x=2u+v\) and \(y=u+2v\), giving

$$ \mathbf{A} = \begin{pmatrix} 2&1\\ 1&2 \end{pmatrix}, \qquad \det\mathbf{A}=2\cdot2-1\cdot1=3. $$

The inequalities defining \(D\) say exactly \(0\le u\le1,\ -1\le v\le1\). Since the function for area is \(f(x,y)=1\), its transformed value is also \(1\). The determinant factor is \(|3|=3\), so

$$ \begin{aligned} \operatorname{Area}(D) &=\int_{-1}^1\int_0^1 1\cdot3\,du\,dv\\ &=\int_{-1}^1 [3u]_0^1\,dv\\ &=\int_{-1}^1 3\,dv\\ &=[3v]_{-1}^1\\ &=3-(-3)=6. \end{aligned} $$

For an independent geometric check, the rectangle in \(uv\)-coordinates has area \(1\cdot2=2\). Its image is a parallelogram scaled in area by \(3\), so its area is \(3\cdot2=6\). This agrees with the integral.

A Reliable Setup

For a linear substitution, the most dependable order is to identify the map and its inverse, translate the region, transform the integrand, and then include the constant area factor. The following checklist keeps those tasks distinct:

1
Specify the map.
Write \(x=au+bv\) and \(y=cu+dv\), or state the matrix \(\mathbf{A}\) clearly.
2
Check invertibility.
Compute \(ad-bc\). If it is zero, this is not an invertible linear change of variables in the plane.
3
Rewrite the bounds.
Use the inverse relations when the original region is specified in \(x,y\); determine the corresponding bounds in \(u,v\).
4
Transform and integrate.
Substitute into \(f\), multiply by \(|ad-bc|\), and integrate over the transformed region.

Common Pitfall: Dropping the Determinant

In the mixed-coordinate example, the \(uv\)-rectangle has area \(2\), but its image has area \(6\). If the determinant factor is omitted, the incorrect calculation is

$$ \int_{-1}^1\int_0^1 1\,du\,dv = \int_{-1}^1 1\,dv = 2. $$

This answer is too small by a factor of \(3\), because the map expands area by \(|\det\mathbf{A}|=3\). The corrected calculation includes that factor and gives \(3\cdot2=6\). A related error is to use \(\det\mathbf{A}\) without absolute value. That would produce a negative integral of \(1\) if the map reversed orientation, even though the area of the region is positive. For scalar area integrals, use the absolute determinant.

When This Method Matters

Linear substitutions are particularly effective when the boundaries of a region are lines through the origin. Linear combinations of \(x\) and \(y\) can then become the new coordinates, turning slanted inequalities into simple bounds. They are also useful when an integrand has a simple expression in coordinates aligned with a region's symmetry or orientation.

Every choice still has to be checked. The matrix must be invertible, the transformed bounds must describe exactly the original region, and the determinant factor must be included. A neat-looking integrand does not compensate for incorrect bounds, and correct bounds do not compensate for a missing area factor.

Check Your Understanding

Use the determinant and transformed bounds carefully when answering these questions.

  1. For \(x=u+2v,\ y=3u+v\), compute the determinant and state the area-scaling factor.
  2. Under \(x=u+v,\ y=v\), what are the \(u,v\) bounds if \(0\le y\le2\) and \(y\le x\le y+3\)?
  3. For \(x=(u-v)/\sqrt{2},\ y=(u+v)/\sqrt{2}\), express \(x^2+y^2\) in terms of \(u,v\).
  4. Why does a negative determinant not make the area factor negative?
  5. If an invertible linear map has determinant \(4\), what is the area of the image of a region of area \(5\)?

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