Difficulty: Intermediate to Advanced | Prerequisites: Double and triple integrals, partial derivatives, determinants of 2×2 and 3×3 matrices.
This material deals with two related skills: rearranging the order of integration in a triple integral (Fubini's theorem in practice), and transforming an integral into a new coordinate system using a substitution and its Jacobian. Both come up constantly in applications where the natural description of a region does not match the most convenient integration order or coordinate system. You need to be comfortable sketching 3D regions from inequalities and computing determinants.
Fubini's theorem says you can integrate in any order, provided you adjust the bounds correctly. Change of variables lets you swap to coordinates that simplify the region or the integrand, at the cost of multiplying by the absolute value of the Jacobian determinant.
Fubini's theorem
If f is continuous on a bounded region, the iterated integral gives the same value regardless of the order of integration, provided the bounds are adjusted to describe the same region. In simple terms, you can integrate x first then y, or y first then x, and get the same answer.
Order of integration
The sequence in which you evaluate the nested integrals. For a triple integral, there are six possible orders: dz dy dx, dz dx dy, dy dz dx, dy dx dz, dx dz dy, dx dy dz. Some orders produce simpler bounds than others.
Jacobian determinant
The determinant of the matrix of partial derivatives when changing variables. For a transformation (x, y) = T(u, v), the Jacobian is J = |∂x/∂u · ∂y/∂v - ∂x/∂v · ∂y/∂u|. It tells you how much area (or volume) is stretched or compressed by the transformation.
Change of variables
Replacing (x, y) with new coordinates (u, v) in an integral, so that either the region becomes simpler (e.g. a rectangle) or the integrand becomes simpler, or both. The trade-off is that you must multiply by |J| and rewrite the bounds.
Level curves
Curves along which a function takes a constant value (e.g. u = 3 or v = 1/3). When the boundary of the original region D consists of level curves of u and v, the transformed region in (u, v)-space is a rectangle, which is the ideal situation for a change of variables.
The problem
Given the integral ∫₀³ ∫₀^(9-x²) ∫₀^(9-y) f(x, y, z) dz dy dx, sketch the region and rewrite it in three of the five other possible orders.
Reading off the original bounds
From the integral:
0 ≤ x ≤ 3
0 ≤ y ≤ 9 - x²
0 ≤ z ≤ 9 - y
The region is bounded by the parabolic cylinder y = 9 - x² (or x² + y = 9), the plane z = 9 - y (or y + z = 9), and the coordinate planes.
Strategy for reordering
To rewrite in a different order, project the 3D region onto each pair of coordinate planes and determine the shadow. Then express the bounds for the "free" variable in terms of the other two.
For example, projecting onto the yz-plane:
z ranges from 0 to 9
For each z, y ranges from 0 to 9 - z (from the plane y + z = 9)
For each (y, z), x ranges from 0 to √(9 - y) (from the parabola x² = 9 - y)
This gives the order dx dy dz: ∫₀⁹ ∫₀^(9-z) ∫₀^(√(9-y)) f(x, y, z) dx dy dz.
Other orders shown in the solutions
Order dx dz dy: ∫₀⁹ ∫₀^(9-y) ∫₀^(√(9-y)) f(x, y, z) dx dz dy.
Order dz dx dy: ∫₀⁹ ∫₀^(√(9-y)) ∫₀^(9-y) f(x, y, z) dz dx dy.
Each reordering requires re-examining which variable's bounds depend on which others. The 3D sketch and the projected 2D shadows are essential tools.
Practical tip
Draw the projections onto all three coordinate planes. Label each shadow with its boundary curves. Then read off the bounds for each order directly from the sketches.
The setup
Given the substitution y² = uv and x² = u/v, the goal is to express (x, y) in terms of (u, v), compute the Jacobian, and evaluate an integral over a region D whose boundaries are level curves of u and v.
Solving: y = √(uv) and x = √(u/v).
Computing the Jacobian
The Jacobian matrix is:
J = | ∂x/∂u ∂x/∂v | | ∂y/∂u ∂y/∂v |
With x = u^(1/2) v^(-1/2) and y = u^(1/2) v^(1/2):
∂x/∂u = (1/2) u^(-1/2) v^(-1/2)
∂x/∂v = -(1/2) u^(1/2) v^(-3/2)
∂y/∂u = (1/2) u^(-1/2) v^(1/2)
∂y/∂v = (1/2) u^(1/2) v^(-1/2)
The determinant: (1/4) v^(-1) + (1/4) v^(-1) = (1/2) v^(-1).
So |J| = 1/(2v).
Why the boundaries become a rectangle
The original region D is bounded by curves of the form u = constant (u = 3, u = 27) and v = constant (v = 3, v = 1/3). In (u, v)-space, this is the rectangle [3, 27] × [1/3, 3], which is far simpler to integrate over than the original curved region.
Evaluating the integral
The integrand (y² + x²) = uv + u/v = u(v + 1/v). With the Jacobian:
∬D (y² + x²) dA = ∫{1/3}^3 ∫_3^{27} u(v + 1/v) · (1/(2v)) du dv
This simplifies to (1/2) ∫_{1/3}^3 ∫_3^{27} (u + u/v²) du dv, which after integration gives 100/2 + 6552 ln 3. (The exact arithmetic is involved, but the method is what matters.)
Jacobian for a 2D change of variables
J = ∂(x, y)/∂(u, v) = ∂x/∂u · ∂y/∂v - ∂x/∂v · ∂y/∂u
Change of variables formula (2D)
∬D f(x, y) dA = ∬{D'} f(x(u,v), y(u,v)) |J| du dv
where D' is the transformed region in (u, v)-space.
Jacobian for a 3D change of variables
The determinant of the 3×3 matrix of partial derivatives ∂(x, y, z)/∂(u, v, w).
Standard Jacobians
Polar/cylindrical: |J| = r. Spherical: |J| = ρ² sin φ. These are special cases of the general formula.
Changing integration order is routine in probability (computing expectations of joint distributions), physics (evaluating integrals over oddly shaped charge distributions), and engineering (thermal analysis of complex geometries). The ability to spot the easiest order can turn an hours-long computation into a five-minute one.
Change of variables with a Jacobian is the mathematical backbone of coordinate transformations in computer graphics, fluid dynamics, and general relativity. Every time you see a simulation in polar, cylindrical, or spherical coordinates, the Jacobian is doing the bookkeeping behind the scenes.
Forgetting to include |J| when changing variables. The Jacobian accounts for how the transformation stretches area. Omitting it is like converting miles to kilometres but not updating the speed.
Confusing the Jacobian of the forward transformation with the inverse. If you define u, v in terms of x, y, you need the Jacobian ∂(x, y)/∂(u, v), not ∂(u, v)/∂(x, y). (They are reciprocals of each other.)
When reordering a triple integral, students often get the correct region but write bounds that describe a different region. Always verify by checking corner points.
Assuming every integral can be rewritten with constant bounds. Only rectangular regions have constant bounds in every order. Non-rectangular regions always have at least some variable-dependent bounds.
⚠️ Rewriting a triple integral in a different order is a near-certain exam question. Practise by starting from the bounds, sketching the region, then reading off new bounds from the sketch.
⚠️ The Jacobian computation is tested directly: expect to compute a 2×2 or 3×3 determinant and apply it. Errors in partial derivatives cascade through the entire problem.
⚠️ Recognising when the boundary curves are level curves of a substitution (so the region becomes a rectangle) is a powerful exam technique.
True or false: Fubini's theorem says you can always swap the order of integration without changing the bounds. (False. You can swap the order, but you must adjust the bounds to describe the same region.)
Fill in the blank: the Jacobian of polar coordinates (x = r cos θ, y = r sin θ) is ____. (r)
True or false: if |J| = 3, then each unit of area in (u, v)-space corresponds to 3 units of area in (x, y)-space. (True.)
Fill in the blank: a triple integral has ____ possible orders of integration. (6)
True or false: when the boundary of D consists of level curves u = c₁, u = c₂, v = c₃, v = c₄, the region in (u, v)-space is a rectangle. (True.)
Q: The integral ∫₀¹ ∫₀^x ∫₀^y f(x,y,z) dz dy dx is given. Rewrite it in the order dx dy dz.
A: From the bounds: 0 ≤ z ≤ y ≤ x ≤ 1. In dx dy dz order: z from 0 to 1, y from z to 1, x from y to 1. The integral is ∫₀¹ ∫_z¹ ∫_y¹ f(x,y,z) dx dy dz.
Q: A transformation is given by u = xy, v = x/y. Find the Jacobian ∂(x,y)/∂(u,v).
A: From u = xy and v = x/y, solve for x and y: x = √(uv), y = √(u/v). Compute the four partial derivatives and take the determinant. The result is |J| = 1/(2v).
Q: Why is a change of variables useful even when the integrand is already simple (e.g. f = 1)?
A: Because the region of integration may have complicated boundaries in (x, y) but simple (rectangular) boundaries in (u, v). Volume and area calculations often benefit from a coordinate change that simplifies the region, even when the integrand is trivial.
Q: Given ∫₀² ∫₀^(4-x²) ∫₀^(4-y) f dz dy dx, what are the bounds on z when you project onto the xz-plane?
A: z ranges from 0 to 4 overall. For a given (x, z), the constraint is y ≤ 4 - z (from the plane) and y ≤ 4 - x² (from the parabola). The relevant bound depends on which is smaller.
Fubini's theorem is the reason you can evaluate double and triple integrals as iterated integrals at all. Without it, you would need to work with Riemann sums directly. The theorem underpins everything in this course from here on.
Change of variables generalises to the substitutions used in surface integrals and line integrals later in the course. The Jacobian concept reappears in the Divergence theorem (where it becomes the determinant of the metric tensor in more advanced treatments).
The specific substitution y² = uv, x² = u/v is an example of a broader technique: whenever the boundary curves of a region are level curves of some natural pair of functions, those functions are your new coordinates.
Fubini's theorem, order of integration, iterated integral reorder, triple integral six orders, Jacobian determinant, change of variables, coordinate transformation, level curves, rectangular region in uv-space, parabolic cylinder, MATH 215, Calculus III, multivariable calculus, integration bounds, projection onto coordinate planes