5.2 Change of Variables and the Jacobian Determinant

5.2 Change of Variables and the Jacobian Determinant

Transforming coordinate systems is the multivariable analogue of $u$-substitution in single-variable calculus. On the GRE Subject Test in Mathematics, the Change of Variables Theorem enables candidates to transform irregular polygonal, elliptical, and hyperbolic domains into standard rectangular regions, replacing unwieldy boundary curves with constant limits of integration.

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The Change of Variables Theorem

Let $T: S \to R$ be a $C^1$ transformation (a continuously differentiable mapping) that is injective (one-to-one) on the interior of an open bounded set $S \subset \mathbb{R}^2$, mapping $S$ onto $R \subset \mathbb{R}^2$. Suppose the Jacobian determinant is non-zero almost everywhere on $S$. For any integrable function $f$ on $R$: $\iint_R f(x, y) \, dx \, dy = \iint_S f(x(u, v), y(u, v)) \left| \frac{\partial(x, y)}{\partial(u, v)} \right| \, du \, dv.$

Geometric Derivation of the Jacobian

Consider an infinitesimal rectangle in the $uv$-plane with vertices $(u, v)$, $(u + du, v)$, and $(u, v + dv)$. Under the mapping $T(u, v) = \langle x(u, v), y(u, v) \rangle$, the displacement vectors along the coordinate directions map to tangent vectors in the $xy$-plane: $\mathbf{T}_u \, du = \left\langle \frac{\partial x}{\partial u}, \frac{\partial y}{\partial u} \right\rangle du, \quad \mathbf{T}_v \, dv = \left\langle \frac{\partial x}{\partial v}, \frac{\partial y}{\partial v} \right\rangle dv.$ These two vectors span an infinitesimal parallelogram in the $xy$-plane. The area of this parallelogram is given by the norm of their cross product: $dA_{xy} = \|(\mathbf{T}_u \, du) \times (\mathbf{T}_v \, dv)\| = \left| \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u} \right| du \, dv.$

Requirement of the Absolute Value

Because integration of scalar functions measures unsigned area (positive measure), the absolute value $|J|$ is essential. A mapping with negative determinant reverses orientation (such as reflection), but the area element must remain strictly non-negative.

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The Jacobian Determinant and the Inverse Jacobian Rule

Definition in Two Dimensions

The Jacobian matrix of the transformation $(x, y) = T(u, v)$ is the matrix of first partial derivatives, and the Jacobian determinant is: $J(u, v) = \frac{\partial(x, y)}{\partial(u, v)} = \det \begin{pmatrix} \frac{\partial x}{\partial u} & \frac{\partial x}{\partial v} \\ \frac{\partial y}{\partial u} & \frac{\partial y}{\partial v} \end{pmatrix} = \frac{\partial x}{\partial u} \frac{\partial y}{\partial v} - \frac{\partial x}{\partial v} \frac{\partial y}{\partial u}.$

Definition in Three Dimensions

For a transformation $(x, y, z) = T(u, v, w)$: $\frac{\partial(x, y, z)}{\partial(u, v, w)} = \det \begin{pmatrix} x_u & x_v & x_w \\ y_u & y_v & y_w \\ z_u & z_v & z_w \end{pmatrix}.$

The Inverse Jacobian Rule

By the chain rule and the Inverse Function Theorem, if $T$ is invertible with inverse $T^{-1}(x, y) = (u(x, y), v(x, y))$, the Jacobian matrices satisfy $D(T^{-1})(T(u, v)) \cdot DT(u, v) = I$. Taking determinants yields the indispensable Inverse Jacobian Rule: $\frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{\frac{\partial(u, v)}{\partial(x, y)}}.$

GRE Pro-Tip: In many exam problems, the boundaries are presented as $u = g(x, y)$ and $v = h(x, y)$ (e.g., $u = xy, ; v = y/x^2$). Solving algebraically for $x(u, v)$ and $y(u, v)$ is time-consuming and error-prone. Instead, calculate $\frac{\partial(u, v)}{\partial(x, y)}$ directly, take its reciprocal, and express the result in terms of $u$ and $v$.

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Linear Transformations and Area Scaling

Let $T: \mathbb{R}^2 \to \mathbb{R}^2$ be a linear transformation defined by $\mathbf{x} = A \mathbf{u}$, where $A = \begin{pmatrix} a & b \ c & d \end{pmatrix}$. The partial derivatives are constants ($x_u = a, x_v = b, y_u = c, y_v = d$), so the Jacobian is constant: $J = \det A = ad - bc.$ For any rectifiable region $S$ in the $uv$-plane, the area of its image $R = T(S)$ satisfies: $\operatorname{Area}(T(S)) = |\det A| \cdot \operatorname{Area}(S).$

Elliptical Scaling

To evaluate integrals over the elliptical region $R = {(x, y) : \frac{x^2}{a^2} + \frac{y^2}{b^2} \le 1}$:

1. Apply the transformation $x = a u, ; y = b v$. The region $S$ is the unit disk $u^2 + v^2 \le 1$.
2. The Jacobian is $\frac{\partial(x, y)}{\partial(u, v)} = \det \begin{pmatrix} a & 0 \ 0 & b \end{pmatrix} = ab$.
3. The area of the ellipse is $\operatorname{Area}(R) = ab \iint_{u^2+v^2 \le 1} du , dv = \pi ab$.
4. In modified polar coordinates ($x = a r \cos \theta, ; y = b r \sin \theta$), the Jacobian is $J = abr$, and the bounds become $0 \le r \le 1, ; 0 \le \theta \le 2\pi$.

Ellipsoids in Three Dimensions

Similarly, for the solid ellipsoid $\frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} \le 1$, the linear scaling $x = au, y = bv, z = cw$ has Jacobian $J = abc$, mapping the ellipsoid to the unit ball. The volume is immediately $abc \left( \frac{4}{3}\pi \right) = \frac{4}{3}\pi abc$.

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Nonlinear Substitutions: Hyperbolic and Sheared Regions

When region boundaries are level curves of multivariable functions, define $u$ and $v$ as those functions.

Example: Hyperbolic Quadrilaterals

Consider a region $R$ in the first quadrant bounded by $xy = 1, xy = 4, y = 2x, y = 5x$:

• Define $u = xy \implies 1 \le u \le 4$.
• Define $v = y/x \implies 2 \le v \le 5$.
• Compute the inverse Jacobian: $\frac{\partial(u, v)}{\partial(x, y)} = \det \begin{pmatrix} y & x \\ -y/x^2 & 1/x \end{pmatrix} = \frac{y}{x} - \left(-\frac{y}{x}\right) = \frac{2y}{x} = 2v.$
• Therefore, $\left| \frac{\partial(x, y)}{\partial(u, v)} \right| = \frac{1}{2v}$. The double integral simplifies to $\int_1^4 du \int_2^5 \frac{f(u, v)}{2v} , dv$.

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Canonical Transformations and Jacobians

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Step-by-Step Worked Problems

Problem 1: Parallelogram Domain Transformation

Evaluate $\iint_R (2x - y) , dA$, where $R$ is the parallelogram bounded by $2x - y = 1$, $2x - y = 4$, $x + y = 2$, and $x + y = 5$.

Solution:

1. Choose natural coordinates aligned with boundaries: $u = 2x - y$ and $v = x + y$.
2. The transformed region $S$ in the $uv$-plane is the rectangle $[1, 4] \times [2, 5]$.
3. Compute the Jacobian of the mapping $(x, y) \mapsto (u, v)$: $\frac{\partial(u, v)}{\partial(x, y)} = \det \begin{pmatrix} 2 & -1 \\ 1 & 1 \end{pmatrix} = 2(1) - (-1)(1) = 3.$
4. By the Inverse Jacobian Rule: $\left| \frac{\partial(x, y)}{\partial(u, v)} \right| = \frac{1}{3}.$
5. The integrand is $2x - y = u$. Apply the Change of Variables Theorem: $\iint_R (2x - y) \, dA = \int_1^4 du \int_2^5 u \cdot \frac{1}{3} \, dv = \frac{1}{3} \left( \int_1^4 u \, du \right) \left( \int_2^5 1 \, dv \right).$
6. Evaluate the simple single integrals: $\int_1^4 u \, du = \left[ \frac{u^2}{2} \right]_1^4 = \frac{16 - 1}{2} = \frac{15}{2}, \quad \int_2^5 dv = 3.$
7. The final result is $\frac{1}{3} \cdot \frac{15}{2} \cdot 3 = \frac{15}{2}$.

Problem 2: Elliptical Region Integration

Compute the area of the region $R$ bounded by the ellipse $4x^2 + 9y^2 = 36$.

Solution:

1. Standardize the ellipse: $\frac{x^2}{9} + \frac{y^2}{4} = 1$, so $a = 3$ and $b = 2$.
2. Define $x = 3u$ and $y = 2v$, mapping $R$ to the unit disk $S: u^2 + v^2 \le 1$.
3. The Jacobian determinant is $J = ab = 3 \times 2 = 6$.
4. $\operatorname{Area}(R) = \iint_R 1 , dA_{xy} = \iint_S 6 , du , dv = 6 \operatorname{Area}(S) = 6(\pi \cdot 1^2) = 6\pi$.

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GRE Exam Traps & Pitfalls

Trap 1: Inverting the Jacobian Direction When substituting $u(x, y)$ and $v(x, y)$, remember that $dx , dy = |\frac{\partial(x,y)}{\partial(u,v)}| , du , dv = \frac{1}{|\frac{\partial(u,v)}{\partial(x,y)}|} , du , dv$. Multiplying by $\frac{\partial(u,v)}{\partial(x,y)}$ instead of its reciprocal is a classic trap.

Trap 2: Forgetting the Absolute Value on Negative Jacobians Determinants of reflection or shear matrices can be negative. For example, if $u = y$ and $v = x$, $J = -1$. Areas and integrals of non-negative functions must remain non-negative; always take $|J|$.

Trap 3: Non-Injective Transformations The change of variables formula requires $T$ to be one-to-one on the interior of $S$. Applying transformations that fold or wrap domains (such as squaring coordinates over intervals containing both positive and negative values) invalidates the theorem.