Difficulty: Intermediate–Advanced | Prerequisites: Double integrals, polar coordinates, cylindrical and spherical coordinates (§15.3–15.8)
Change of variables lets you transform a messy integral over a complicated region into a cleaner one by switching to new coordinates (u, v). The trade-off: you multiply the integrand by the absolute value of the Jacobian determinant, which measures how the substitution stretches or compresses area. The technique generalises polar, cylindrical, and spherical substitutions into a single framework.
Change of variables (substitution in multiple integrals)
A technique for evaluating ∫∫_R f(x, y) dA by introducing new coordinates u = u(x, y), v = v(x, y) that simplify either the integrand, the region of integration, or both. Think of it as the 2D version of u-substitution from single-variable calculus.
Jacobian determinant (∂(x, y)/∂(u, v))
The determinant of the 2×2 matrix of partial derivatives [∂x/∂u, ∂x/∂v; ∂y/∂u, ∂y/∂v]. It measures the local area scaling factor when you change coordinates. In simple terms, it tells you how much a tiny patch of area in the (u, v) plane gets stretched or compressed when mapped back to (x, y).
Inverse Jacobian
If you define the transformation as u = u(x, y) and v = v(x, y) (mapping from old to new), the Jacobian you need is ∂(x, y)/∂(u, v). A useful shortcut: ∂(x, y)/∂(u, v) = 1 / [∂(u, v)/∂(x, y)]. So you can compute whichever direction is easier and take the reciprocal.
Linear change of variables
A substitution where u and v are linear functions of x and y (e.g., u = 2x + y, v = x − y). The Jacobian is constant, and the transformation maps parallelograms to rectangles. Useful when the region of integration is a parallelogram or when the integrand factors neatly under a linear substitution.
Transformation of the region
When you change variables, the region R in the xy-plane maps to a region S in the uv-plane. The boundary curves of R become the boundary curves of S under the substitution. Choosing u and v so that S is a rectangle is the ideal outcome.
The general formula for a transformation T: (u, v) → (x, y):
∫∫_R f(x, y) dA = ∫∫_S f(x(u, v), y(u, v)) |∂(x, y)/∂(u, v)| du dv
where S is the image of R under the inverse transformation. The absolute value of the Jacobian replaces the simple "r" in polar coordinates or "ρ² sin φ" in spherical. Those are specific instances of this general formula.
The homework asks for ∫∫_R (3y² − 4xy − 4x²) dA over the quadrilateral with vertices (0, 2), (3, 0), (5, 4), (2, 6).
The key insight: notice that 3y² − 4xy − 4x² factors as (3y + 2x)(y − 2x). This suggests setting u = y − 2x and v = 3y + 2x (or something proportional).
Find the equations of the four edges of the quadrilateral. Two pairs of opposite sides will be level curves of u and v.
The vertices (0, 2), (3, 0), (5, 4), (2, 6) lie on the lines: y − 2x = 2, y − 2x = −6, y + (2/3)x = 2, y + (2/3)x = 6 (verify by substituting each vertex).
With suitable u and v, the quadrilateral maps to a rectangle in the uv-plane, and the integrand becomes simply u · v.
Compute the Jacobian, set up the integral over the rectangle, and evaluate.
The "subtle" method the homework hints at is exactly this factoring approach, compared to the "tedious" method of integrating directly with complicated limits.
The region E is in the first quadrant, above y = x/3, below y = 3x, between the curves xy = 3 and xy = 27.
The curves xy = 3 and xy = 27 are hyperbolas. Setting u = xy makes them u = 3 and u = 27 (constant bounds).
The lines y = x/3 and y = 3x give y/x = 1/3 and y/x = 3. Setting v = y/x makes them v = 1/3 and v = 3 (constant bounds).
The region E, which has curved boundaries in xy-space, maps to the rectangle [3, 27] × [1/3, 3] in uv-space.
To evaluate ∫∫_E (x²/y² + x²y²) dA:
Express x²/y² and x²y² in terms of u and v. Since u = xy and v = y/x, we get x² = u/v and y² = uv. So x²/y² = 1/v² and x²y² = u².
Compute the Jacobian: ∂(u, v)/∂(x, y) = det[y, x; −y/x², 1/x] = y/x + y/x = 2y/x = 2v. So ∂(x, y)/∂(u, v) = 1/(2v).
The integral becomes ∫₃²⁷ ∫_(1/3)³ (1/v² + u²) · (1/(2v)) dv du.
Why the hint works: the substitution was chosen so that the boundary curves become constant in the new coordinates. The curves xy = c and y/x = c are the natural "coordinate lines" for this region. This is the guiding principle: choose u and v so that the boundary of your region maps to lines of constant u or constant v.
Look for these clues:
The region is bounded by curves of the form g(x, y) = c₁ and g(x, y) = c₂. Set u = g(x, y).
The integrand factors into functions that look like natural coordinate functions.
The region is a parallelogram (use a linear transformation to map it to a rectangle).
The integrand contains xy, x/y, x² + y², or x² − y² (these suggest specific standard substitutions).
∫∫_R f(x, y) dA = ∫∫_S f(x(u, v), y(u, v)) |∂(x, y)/∂(u, v)| du dv
For T: (u, v) → (x, y):
Jacobian matrix = [∂x/∂u, ∂x/∂v; ∂y/∂u, ∂y/∂v]
∂(x, y)/∂(u, v) = (∂x/∂u)(∂y/∂v) − (∂x/∂v)(∂y/∂u)
∂(x, y)/∂(u, v) = 1 / [∂(u, v)/∂(x, y)]
This means you can compute the Jacobian in whichever direction is more convenient and invert.
Polar coordinates (r, θ): x = r cos θ, y = r sin θ. Jacobian = r.
Cylindrical (r, θ, z): Jacobian = r.
Spherical (ρ, θ, φ): Jacobian = ρ² sin φ.
u = xy, v = y/x: ∂(u, v)/∂(x, y) = 2y/x = 2v, so |∂(x, y)/∂(u, v)| = 1/(2v).
Linear transformation u = ax + by, v = cx + dy: Jacobian = 1/(ad − bc), a constant.
Forgetting the absolute value on the Jacobian. The formula uses |∂(x, y)/∂(u, v)|, not ∂(x, y)/∂(u, v). If your Jacobian comes out negative, you take its absolute value. The sign reflects orientation, not area.
Computing the Jacobian in the wrong direction. If you define u and v as functions of x and y, you have ∂(u, v)/∂(x, y). The formula needs ∂(x, y)/∂(u, v), which is the reciprocal. Students often compute the forward Jacobian and forget to invert.
Not transforming the region. Changing the integrand without also changing the limits of integration is a common error. You must re-express the boundary of R in terms of u and v to get the new region S.
Assuming every substitution simplifies things. A poor choice of u and v can make the integral worse. The test of a good substitution: does the region become simpler (ideally a rectangle), and does the integrand become manageable after including the Jacobian?
⚠️ "Evaluate the integral using a suitable change of variables" is a standard exam format. Practise identifying when the integrand factors and when the region's boundary curves suggest natural coordinates.
⚠️ The Jacobian computation is nearly always tested. Be fluent with 2×2 determinants and with the inverse-Jacobian shortcut.
⚠️ Parallelogram regions with a linear change of variables are a favourite exam problem. The vertices tell you the edge directions, which determine the substitution.
⚠️ Regions bounded by hyperbolas xy = c and lines y/x = c (or y = mx) appear frequently. The substitution u = xy, v = y/x (or variants) is the standard approach.
True or false: If ∂(u, v)/∂(x, y) = 5, then |∂(x, y)/∂(u, v)| = 5. (False: it equals 1/5.)
Fill in the blank: The Jacobian for polar coordinates (x = r cos θ, y = r sin θ) is ___. (r.)
True or false: The change of variables formula requires you to take the absolute value of the Jacobian determinant. (True.)
Fill in the blank: If a region is bounded by xy = 3, xy = 27, y/x = 1/3, and y/x = 3, a natural substitution is u = ___ and v = ___. (u = xy, v = y/x.)
True or false: A linear change of variables always produces a constant Jacobian. (True.)
Q: The quadrilateral with vertices (0, 2), (3, 0), (5, 4), (2, 6) has two pairs of parallel sides. Find equations for all four sides and identify a linear substitution that maps the quadrilateral to a rectangle.
A: The sides lie on the lines y − 2x = 2, y − 2x = −6, 3y + 2x = 6, and 3y + 2x = 22 (verify each pair of adjacent vertices). Set u = y − 2x and v = 3y + 2x. Then the quadrilateral maps to the rectangle −6 ≤ u ≤ 2, 6 ≤ v ≤ 22.
Q: For the substitution u = xy, v = y/x, compute ∂(u, v)/∂(x, y) and hence find |∂(x, y)/∂(u, v)|.
A: ∂u/∂x = y, ∂u/∂y = x, ∂v/∂x = −y/x², ∂v/∂y = 1/x. The determinant is y/x − (−y/x) = 2y/x = 2v. So |∂(x, y)/∂(u, v)| = 1/(2v) (valid in the first quadrant where v > 0).
Q: Evaluate ∫∫_E (x²/y² + x²y²) dA where E is bounded by xy = 3, xy = 27, y/x = 1/3, and y/x = 3, using u = xy and v = y/x.
A: In terms of u and v: x² = u/v and y² = uv, so x²/y² = (u/v)/(uv) = 1/v² and x²y² = (u/v)(uv) = u². The integral becomes ∫₃²⁷ ∫_(1/3)³ (1/v² + u²)(1/(2v)) dv du. Evaluate the inner integral with respect to v, then the outer with respect to u.
Q: Explain in one or two sentences why the substitution u = xy, v = y/x was a reasonable guess for the region bounded by xy = 3, xy = 27, y = x/3, y = 3x.
A: The four boundary curves are level curves of the functions xy and y/x. Setting u equal to one and v equal to the other maps each boundary curve to a line of constant u or constant v, turning the curved region into a rectangle.
Q: For a linear transformation u = ax + by, v = cx + dy, what conditions on a, b, c, d ensure the transformation is invertible?
A: The transformation is invertible if and only if ad − bc ≠ 0. This is the Jacobian ∂(u, v)/∂(x, y), and it must be non-zero for the mapping to be one-to-one.
Change of variables is the unifying idea behind polar, cylindrical, and spherical integration from §15.3-15.8. Each of those is a specific instance of this general framework, with a specific Jacobian. This section makes the general theory explicit.
The technique connects to linear algebra: a linear change of variables is a matrix transformation, and the Jacobian determinant equals the determinant of that matrix. Students who have taken or are taking linear algebra will recognise the area-scaling interpretation of the determinant.
§16.1 (line integrals of scalar functions) follows this material and uses parametric curves, which are themselves a kind of coordinate change from a single parameter t to (x, y) or (x, y, z).
change of variables, Jacobian, Jacobian determinant, substitution in double integrals, u-substitution 2D, linear transformation, coordinate change, MATH 215, Calculus III, University of Michigan, §15.9, §16.1, multivariable calculus, parallelogram to rectangle, inverse Jacobian, hyperbolic region, xy = constant, y/x = constant, area element, dA transformation, Stewart multivariable calculus