Change of Variables & the Jacobian

Study Sheet

Change of Variables & the Jacobian

How substitutions stretch area and volume

The Jacobian Determinant

Tip
Substitution in several variables

One-variable substitution carries a factor dxdu\tfrac{dx}{du}; in several variables the factor is a DETERMINANT: for a map (u,v)↦(x(u,v),y(u,v))(u, v) \mapsto (x(u,v), y(u,v)), dx dy=∣det⁡∂(x,y)∂(u,v)∣ du dvdx\,dy = \left|\det\dfrac{\partial(x, y)}{\partial(u, v)}\right|\,du\,dv, where the Jacobian matrix collects all four partials.

The reason is geometric: the map sends a tiny du×dvdu \times dv rectangle to a tiny parallelogram spanned by the column vectors (xu,yu)(x_u, y_u) and (xv,yv)(x_v, y_v), and a parallelogram's area is exactly the absolute determinant. Every coordinate system you have used is a special case: polar's factor rr, cylindrical's rr, spherical's ρ2sin⁡ϕ\rho^2\sin\phi — all Jacobians, all computable in two minutes from the definition rather than memorized.

Side note
Which direction is the determinant?

The classic trap: if you define the NEW variables from the old (u=x+yu = x + y, v=x−yv = x - y), you naturally compute det⁡∂(u,v)∂(x,y)\det\tfrac{\partial(u,v)}{\partial(x,y)} — but the integral needs the INVERSE factor ∣det⁡∂(x,y)∂(u,v)∣\left|\det\tfrac{\partial(x,y)}{\partial(u,v)}\right|, its reciprocal. Rule of thumb: the Jacobian you multiply by always has the OLD variables on top. When in doubt, test on a unit square.

Try it
Try it: the area of an ellipse in one line

Find the area of x2a2+y2b2≤1\tfrac{x^2}{a^2} + \tfrac{y^2}{b^2} \le 1. Work: the map x=aux = au, y=bvy = bv turns it into the unit disk u2+v2≤1u^2 + v^2 \le 1; the Jacobian is det⁡(a00b)=ab\det\begin{pmatrix}a & 0\\ 0 & b\end{pmatrix} = ab; area =ab×π=πab= ab \times \pi = \pi ab. Setting a=b=ra = b = r recovers πr2\pi r^2 — the circle formula was the special case all along.

Proofs & Why It Matters

Tip
Proof: why the determinant appears

Near a point, the substitution map is approximated by its derivative — the linear map with the Jacobian matrix JJ (this is the tangent-plane idea applied to a MAP rather than a function).

A linear map multiplies all areas by ∣det⁡J∣|\det J|: the unit square spanned by e1,e2\mathbf e_1, \mathbf e_2 lands on the parallelogram spanned by the columns of JJ, whose area is ∣det⁡J∣|\det J|, and general regions are limits of little squares. Summing f×(image area)f \times (\text{image area}) over a fine grid gives Riemann sums of ∬f ∣det⁡J∣ du dv\iint f\,|\det J|\,du\,dv. ■\blacksquare SIGNIFICANCE: this is the bridge between linear algebra ("determinant = volume factor") and calculus ("substitution rule") — and in probability it is exactly how densities transform, the change-of-variables formula behind every normalizing flow in machine learning.

Tip
Proof: polar coordinates, this time with no hand-waving

x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta: J=(cos⁡θ−rsin⁡θsin⁡θrcos⁡θ)J = \begin{pmatrix}\cos\theta & -r\sin\theta\\ \sin\theta & r\cos\theta\end{pmatrix}, det⁡J=rcos⁡2θ+rsin⁡2θ=r\det J = r\cos^2\theta + r\sin^2\theta = r.

The "curvy rectangle" argument from the double-integrals topic was the geometric shadow of this two-line computation — and now spherical's ρ2sin⁡ϕ\rho^2\sin\phi is just a bigger determinant, not a new idea to memorize. ■\blacksquare

Going Deeper: Worked Problems

Example
Worked: an integral that only the right substitution can do

Compute ∬R(x+y)2 dA\displaystyle\iint_R (x + y)^2\,dA over the square RR with vertices (1,0),(0,1),(−1,0),(0,−1)(1,0), (0,1), (-1,0), (0,-1).

Step 1 — the region is a diamond, awkward in x,yx,y; substitute u=x+yu = x + y, v=x−yv = x - y, which maps it to the square ∣u∣≤1|u| \le 1, ∣v∣≤1|v| \le 1.

Step 2 — Jacobian: ∂(u,v)∂(x,y)=det⁡(111−1)=−2\dfrac{\partial(u,v)}{\partial(x,y)} = \det\begin{pmatrix}1&1\\1&-1\end{pmatrix} = -2, so dx dy=12 du dvdx\,dy = \tfrac12\,du\,dv.

Step 3 — the integrand is u2u^2: ∫−11 ⁣∫−11u2⋅12 du dv=12⋅23⋅2=23\displaystyle\int_{-1}^1\!\int_{-1}^1 u^2\cdot\tfrac12\,du\,dv = \tfrac12\cdot\tfrac23\cdot2 = \tfrac23.

Step 4 — check plausibility: the diamond has area 22 and (x+y)2(x+y)^2 averages less than 11 on it, so an answer below 22 is right-sized; the exact value 23\tfrac23 took three lines because the substitution matched BOTH region and integrand.

Example
Worked: deriving the cylindrical volume element

Show that dV=r dr dθ dzdV = r\,dr\,d\theta\,dz for x=rcos⁡θx = r\cos\theta, y=rsin⁡θy = r\sin\theta, z=zz = z.

Step 1 — the Jacobian matrix of (r,θ,z)↦(x,y,z)(r, \theta, z) \mapsto (x, y, z) is (cos⁡θ−rsin⁡θ0sin⁡θrcos⁡θ0001)\begin{pmatrix}\cos\theta & -r\sin\theta & 0\\ \sin\theta & r\cos\theta & 0\\ 0 & 0 & 1\end{pmatrix}.

Step 2 — expand along the last row: the determinant is 1⋅(rcos⁡2θ+rsin⁡2θ)=r1\cdot(r\cos^2\theta + r\sin^2\theta) = r.

Step 3 — so dV=∣r∣ dr dθ dz=r dr dθ dzdV = |r|\,dr\,d\theta\,dz = r\,dr\,d\theta\,dz since r≥0r \ge 0.

Step 4 — the geometric reading: the zz-direction is untouched, so the 3D element is just the polar area element r dr dθr\,dr\,d\theta times dzdz — the same determinant computed in the Jacobian topic, with one extra dimension riding along for free.