Case Study 2 — A Skewed Panel and the Linear Change of Variables

Field: Engineering and physics (structural mechanics) Calculus used: Jacobian determinant of a linear map (Section 33.3), inverse-Jacobian shortcut (Section 33.8), and the 2D change-of-variables formula (Section 33.4)

A structural engineer is analyzing a flat composite panel cut into the shape of a parallelogram — a common geometry where two stiffening members cross at an angle, leaving a skewed bay of material between them. She needs the panel's total strain energy, which for her loading model reduces to integrating a stress-density function over the parallelogram's area. The integrand is simple. The region is the problem: its four sides are slanted lines, none of them parallel to the coordinate axes, and setting up the integral in $x$ and $y$ would force her into a tangle of piecewise limits and a region split into triangles. This is precisely the situation the chapter promised to dissolve (Section 33.1): when the region is awkward but bounded by a family of straight lines, you change variables so the boundary lines become the coordinate grid.

The Region and the Integral

The parallelogram $R$ is bounded by the four lines

$$x + y = 1, \qquad x + y = 3, \qquad 2x - y = 0, \qquad 2x - y = 2.$$

Her loading produces a stress density proportional to $(x+y)^2$, so the quantity she wants is

$$E = \iint_R (x+y)^2\,dA.$$

In Cartesian coordinates this is genuinely unpleasant. Solving the four lines pairwise for the corners, the vertices of $R$ are $\left(\tfrac13,\tfrac23\right)$, $\left(1,2\right)$, $\left(\tfrac53,\tfrac43\right)$, and $\left(1,0\right)$ — no two sides axis-aligned, so any iterated $x$–$y$ setup requires breaking the region into pieces and tracking which slanted line bounds each piece. The engineer instead reads the structure straight off the boundary equations.

Designing the Transformation

The four boundary lines come in two parallel families: $x+y = \text{const}$ and $2x - y = \text{const}$. That observation is the change of variables. Define

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

In these coordinates the region is no longer skewed — it is the rectangle $1 \le u \le 3$, $0 \le v \le 2$, because each boundary line is now simply "$u = $ constant" or "$v = $ constant." The miserable parallelogram has become the friendliest region in the plane, exactly as in Worked Example 1 of Section 33.4.

To apply the change-of-variables formula she needs the area element, which means the Jacobian. Here the chapter's reciprocal shortcut (Section 33.8) saves work: rather than first invert the system to get $x$ and $y$ as functions of $u$ and $v$ and differentiate those, she differentiates the easy direction directly. The map $(x,y)\mapsto(u,v)$ has Jacobian

$$\det\frac{\partial(u,v)}{\partial(x,y)} = \det\begin{pmatrix} \partial u/\partial x & \partial u/\partial y \\ \partial v/\partial x & \partial v/\partial y \end{pmatrix} = \det\begin{pmatrix} 1 & 1 \\ 2 & -1 \end{pmatrix} = (1)(-1) - (1)(2) = -3.$$

By the inverse rule, the Jacobian of the transformation she actually integrates with — the one sending $(u,v)$ back to $(x,y)$ — is the reciprocal:

$$\det\frac{\partial(x,y)}{\partial(u,v)} = \frac{1}{-3} = -\frac13, \qquad \text{so} \qquad \lvert\det J\rvert = \frac13.$$

This is the heart of the method and the most common place students slip (Section 33.4's pitfall): the area element belongs to the variables being integrated. Since she will integrate over $u$ and $v$, the factor that multiplies $du\,dv$ is $\lvert\det\partial(x,y)/\partial(u,v)\rvert = \tfrac13$, not the $3$ she would get by carelessly grabbing the determinant of the easy direction. Geometrically the panel $R$ is one-third the area of the $(u,v)$-rectangle, so each unit of $du\,dv$ corresponds to only one-third as much physical area — the Jacobian is bookkeeping for that shrinkage.

Confirming the Inverse Directly

A careful engineer never trusts a reciprocal she has not checked, so she inverts the system explicitly. Adding the two equations, $u + v = 3x$, gives $x = (u+v)/3$; then $y = u - x = (2u - v)/3$. Differentiating,

$$J = \begin{pmatrix} \partial x/\partial u & \partial x/\partial v \\ \partial y/\partial u & \partial y/\partial v \end{pmatrix} = \begin{pmatrix} 1/3 & 1/3 \\ 2/3 & -1/3 \end{pmatrix}, \qquad \det J = \left(\tfrac13\right)\!\left(-\tfrac13\right) - \left(\tfrac13\right)\!\left(\tfrac23\right) = -\tfrac19 - \tfrac29 = -\tfrac13.$$

The two routes agree: $\lvert\det J\rvert = \tfrac13$. The reciprocal shortcut was correct, and now she trusts it.

Evaluating the Integral

With the region rectangular and the Jacobian in hand, the integral collapses. The integrand $(x+y)^2$ is simply $u^2$ in the new variables, so

$$E = \iint_R (x+y)^2\,dA = \int_{1}^{3}\!\int_{0}^{2} u^2 \cdot \underbrace{\frac13}_{\lvert\det J\rvert}\,dv\,du.$$

The integrand has no $v$-dependence, so the inner integral just contributes the length of the $v$-interval, a factor of $2$:

$$E = \frac13\int_1^3 u^2\,(2)\,du = \frac23\int_1^3 u^2\,du = \frac23\cdot\left[\frac{u^3}{3}\right]_1^3 = \frac23\cdot\frac{27 - 1}{3} = \frac23\cdot\frac{26}{3} = \frac{52}{9}.$$

The total strain energy is $E = \tfrac{52}{9} \approx 5.78$ in the engineer's working units. A computation that would have demanded splitting the panel into triangles and grinding through slanted limits became three lines of arithmetic, all because the boundary lines told her which coordinates to use.

A Sanity Check via Area

She checks the result's plausibility with a quantity she can verify another way: the panel's area. Setting the integrand to $1$,

$$\text{area}(R) = \iint_R dA = \int_1^3\!\int_0^2 \frac13\,dv\,du = \frac13\cdot 2\cdot 2 = \frac43.$$

This matches the geometric relation $\lvert R\rvert = \lvert\det J\rvert\cdot\lvert S\rvert$ from Section 33.4: the $(u,v)$-rectangle $S$ has area $2\times 2 = 4$, and $\tfrac13\cdot 4 = \tfrac43$. The Jacobian converts the area of the model rectangle into the area of the physical panel exactly. She can also bound the energy by hand: on $R$, $x+y$ ranges from $1$ to $3$, so $(x+y)^2$ ranges from $1$ to $9$, and the energy must lie between $1\cdot\tfrac43 = \tfrac43$ and $9\cdot\tfrac43 = 12$. Her answer $\tfrac{52}{9}\approx 5.78$ sits comfortably in that window — a reassuring, independent confirmation.

# Verify the strain-energy integral over the skewed panel (Section 33.4).
import sympy as sp

u, v = sp.symbols('u v', positive=True)
detJ = sp.Rational(1, 3)                      # |det J| from the linear map
E = sp.integrate(sp.integrate(u**2 * detJ, (v, 0, 2)), (u, 1, 3))
print("E =", E)                               # exact value
# Output: E = 52/9

Why the Linear Case Is the Template

The engineer notes for her report that this was the easy case in a precise sense: a linear transformation has a constant Jacobian, so the factor $\lvert\det J\rvert$ slides outside the integral as a pure number (Section 33.5). Every parallelogram, every skew bay, every sheared coordinate frame in structural and continuum mechanics reduces this way — and the same idea, with a non-constant Jacobian, handles curved regions through polar, cylindrical, or spherical coordinates (Section 33.7). The lesson generalizes directly into physics: a linear change of variables that removes a cross term from a quadratic energy (diagonalizing the strain tensor, or rotating to principal axes) is this exact computation, with the Jacobian guaranteeing the integral's value survives the change of frame. When the determinant of such a rotation is $1$, the transformation preserves area outright — the same unit-Jacobian fact that underlies Liouville's theorem in mechanics (Section 33.10). The humble parallelogram and the deep conservation law are two faces of one Jacobian.

Discussion Questions

  1. The engineer used the reciprocal shortcut (Section 33.8) to avoid inverting the system. Restate, in your own words, why $\det\partial(x,y)/\partial(u,v)$ is the reciprocal of $\det\partial(u,v)/\partial(x,y)$, and what could go wrong if you used the wrong one here.
  2. The Jacobian came out as $\tfrac13$, less than $1$. Interpret this geometrically: what does it say about the size of the physical panel relative to the model rectangle?
  3. Why was the integral separable (the $v$-integral just a constant factor) once she changed variables, and why was it not separable in $x$ and $y$?
  4. Suppose the stress density were $(2x - y)^2$ instead of $(x+y)^2$. How would the calculation change, and which variable would now carry the dependence?
  5. Connect this case to the economics decorrelation problem (Exercise F2): both use a linear change of variables to simplify a quadratic. What is the shared mathematical structure, and what role does the Jacobian play in each?

Annotated Reading

  • Stewart, J. (2021). Calculus: Early Transcendentals (9th ed.), Cengage, §15.10 (Change of Variables in Multiple Integrals). Worked parallelogram-to-rectangle examples nearly identical to this case; the standard undergraduate treatment.
  • Strang, G., & Herman, E. Calculus, Volume 3 (OpenStax), §5.7. Free, with the linear and polar change-of-variables formula stated and applied to skewed regions.
  • Boresi, A. P., & Schmidt, R. J. (2003). Advanced Mechanics of Materials (6th ed.), Wiley. Shows where strain-energy integrals over panel geometries arise in practice, and why principal-axis (linear) transformations are the working engineer's first move.
  • Marsden, J. E., & Tromba, A. J. (2011). Vector Calculus (6th ed.), W. H. Freeman, §6.2. A clean derivation of the change-of-variables theorem with the Jacobian as the area-scaling factor, including the inverse-Jacobian shortcut used above.