maths2u
PhysicsMathsChemistryBiology
Tier
⌕ Search ⌘K
Theorem

Change of variables and the Jacobian

T-052Home MU-203Threads space · change
Statement

Let \(U, V \subseteq \mathbb{R}^n\) be open sets and let \(\varphi : U \to V\) be a bijection of class \(C^1\) whose inverse \(\varphi^{-1} : V \to U\) is also \(C^1\) (i.e. \(\varphi\) is a \(C^1\)-diffeomorphism). Write \(D\varphi(u)\) for the Jacobian matrix of \(\varphi\) at \(u \in U\) and \(J_\varphi(u) = \det D\varphi(u)\) for its Jacobian determinant, and suppose \(J_\varphi(u) \neq 0\) for every \(u \in U\). Then for every Lebesgue-measurable set \(E \subseteq U\) and every function \(f : \varphi(E) \to \mathbb{R}\) that is either non-negative measurable or Lebesgue-integrable on \(\varphi(E)\), \[ \int_{\varphi(E)} f(x)\,dx \;=\; \int_{E} f(\varphi(u))\,\lvert J_\varphi(u) \rvert\,du . \] In particular, taking \(f \equiv 1\), the \(n\)-dimensional volume of \(\varphi(E)\) is \(\operatorname{vol}(\varphi(E)) = \int_E \lvert J_\varphi(u)\rvert\,du\).

Why it matters

Integrals that are intractable in Cartesian coordinates — over disks, balls, ellipsoids, or regions bounded by curved surfaces — become elementary once transported to polar, cylindrical, spherical, or problem-adapted coordinates. The change-of-variables theorem is the rigorous engine behind every such substitution: it tells you exactly what correction factor (the Jacobian) compensates for the fact that a smooth map stretches and rotates infinitesimal volume elements non-uniformly.

Beyond computation, the theorem is the bridge between the naive calculus notion of "\(dx\,dy \to r\,dr\,d\theta\)" and the measure-theoretic fact that pushing forward Lebesgue measure by a diffeomorphism yields another measure absolutely continuous with respect to Lebesgue measure, with density \(\lvert J_\varphi\rvert\). This is the finite-dimensional shadow of ideas that reappear in probability (change of variables for densities), differential geometry (pullback of volume forms), and physics (Jacobians in statistical mechanics and general relativity).

Hypotheses
\(\varphi\) is \(C^1\) on the open set \(U\). Without differentiability there is no Jacobian matrix to speak of. E.g. a Lipschitz but non-differentiable map (such as one built from \(\lvert x\rvert\)) can still push forward measurable sets to measurable sets, but the pointwise "stretch factor" \(J_\varphi\) is undefined on a set of positive measure, and the clean pointwise formula fails; one must fall back to weaker area-formula versions requiring only local Lipschitz continuity.
\(\varphi : U \to V\) is a bijection (globally injective on \(U\)). Consider \(\varphi(r,\theta) = (r\cos\theta, r\sin\theta)\) on \(U = (0,\infty)\times(0,4\pi)\): each point of the punctured plane is covered twice. Integrating naively with the Jacobian formula over all of \(U\) double-counts the target region, giving twice the correct area. Injectivity (or restricting to a fundamental domain) is essential to avoid overcounting.
\(J_\varphi(u) \neq 0\) for all \(u \in U\) (equivalently, \(\varphi^{-1}\) is also \(C^1\), by the Inverse Function Theorem). Take \(\varphi(u) = u^3\) on \(U=(-1,1)\subseteq\mathbb{R}\). Then \(\varphi\) is a \(C^1\) bijection onto \((-1,1)\) but \(J_\varphi(0)=3\cdot 0^2=0\), and \(\varphi^{-1}(x)=x^{1/3}\) is not differentiable at \(0\). Near \(u=0\) the map flattens so severely that no local linear "stretch factor" description applies there, and the naive formula, while it happens to still hold in this scalar example by a limiting/continuity argument, generally cannot be trusted to reflect local volume distortion at points where \(J_\varphi = 0\); in higher dimensions non-invertibility of \(D\varphi\) at such points signals a fold or collapse of dimension that the simple pointwise formula does not account for.
\(E\) (and hence \(\varphi(E)\)) is Lebesgue measurable, and \(f\) is measurable with the stated sign/integrability condition. If \(E\) is a non-measurable set (existence guaranteed by the Axiom of Choice, e.g. a Vitali set), neither side of the identity is meaningfully defined, since "volume" and "integral" are not defined for non-measurable sets under Lebesgue measure.
Proof
1
\text{Reduce to the case } f \equiv 1 \text{ (the volume identity) via the standard measure-theoretic bootstrap.}
Once \(\operatorname{vol}(\varphi(E)) = \int_E \lvert J_\varphi\rvert\,du\) is established for every measurable \(E \subseteq U\), the general statement follows by the standard sequence: indicator functions \(f = \mathbf{1}_A\) are exactly the volume identity applied to \(E = \varphi^{-1}(A)\); simple functions follow by linearity of the integral; non-negative measurable \(f\) follow by the Monotone Convergence Theorem applied to an increasing sequence of simple functions \(f_k \uparrow f\); general integrable \(f\) follow by splitting \(f = f^+ - f^-\) and applying the non-negative case to each part, using linearity of the integral. A
2
\text{It suffices to prove } \operatorname{vol}(\varphi(E)) = \int_E \lvert J_\varphi(u)\rvert\,du \text{ locally, then patch by countable additivity.}
Lebesgue measure is countably additive, and \(U\) is \(\sigma\)-compact (a countable union of compact sets, since \(\mathbb{R}^n\) is second countable and locally compact), so it suffices to prove the identity for \(E\) contained in a small ball \(B(u_0,\delta) \subseteq U\) around an arbitrary point \(u_0\), and then sum over a countable cover. B
3
\text{Linearise: near } u_0,\ \varphi(u) = \varphi(u_0) + D\varphi(u_0)(u-u_0) + o(\lvert u-u_0\rvert).
This is precisely the definition of Fréchet differentiability of \(\varphi\) at \(u_0\), which holds since \(\varphi \in C^1(U)\). The strategy is to compare \(\varphi\) on a small neighbourhood to the affine map \(u \mapsto \varphi(u_0) + D\varphi(u_0)(u - u_0)\), for which the volume-scaling law is classical linear algebra. B
4
\text{Linear algebra fact: for a linear map } T = D\varphi(u_0) \text{ with } \det T \neq 0,\ \operatorname{vol}(T(S)) = \lvert \det T\rvert\,\operatorname{vol}(S) \text{ for every measurable } S.
This is proved independently by row/column-reduction: every invertible matrix is a product of elementary matrices (scalings, shears, permutations); a scaling by \(\lambda\) along one axis multiplies volume by \(\lvert\lambda\rvert\), a shear preserves volume (Cavalieri's principle — it moves mass parallel to hyperplanes of constant "height", preserving cross-sectional volumes), and a permutation preserves volume; since \(\det\) is multiplicative over this product and each factor's volume-scaling matches its determinant's absolute value, the composite scales volume by \(\lvert \det T\rvert\). This is a standalone lemma of linear algebra, taken as known. A
5
\forall \varepsilon \gt 0\ \exists \delta \gt 0 : \ (1-\varepsilon)\lvert\det D\varphi(u_0)\rvert \le \frac{\operatorname{vol}(\varphi(B(u_0,r)))}{\operatorname{vol}(B(u_0,r))} \le (1+\varepsilon)\lvert\det D\varphi(u_0)\rvert \quad \text{for all } r \lt \delta.
Because \(\varphi(u) - \varphi(u_0) - D\varphi(u_0)(u-u_0) = o(\lvert u - u_0\rvert)\) uniformly as \(r \to 0\) (differentiability at \(u_0\) plus continuity of \(D\varphi\), using \(\varphi \in C^1\)), the image \(\varphi(B(u_0,r))\) is sandwiched between two dilates of \(D\varphi(u_0)(B(u_0,r)) + \varphi(u_0)\), scaled by \((1\mp\varepsilon)\) in each linear direction for \(r\) small; applying Step 4 to these bracketing linear images and taking the ratio of volumes to \(\operatorname{vol}(B(u_0,r))\) gives the squeeze. This is the technical heart of the proof: it converts a first-order (linear) approximation into a genuine volume comparison, uniformly over small balls. C
6
\text{Let } r \to 0 \text{ (or refine the cover): } \lim_{r \to 0} \frac{\operatorname{vol}(\varphi(B(u_0,r)))}{\operatorname{vol}(B(u_0,r))} = \lvert J_\varphi(u_0)\rvert.
Since \(\varepsilon \gt 0\) in Step 5 was arbitrary, the squeeze theorem for real sequences forces the limit to exist and equal \(\lvert\det D\varphi(u_0)\rvert = \lvert J_\varphi(u_0)\rvert\). This identifies \(\lvert J_\varphi\rvert\) as the local "volume density" of the pushforward measure \(B \mapsto \operatorname{vol}(\varphi(B))\) relative to Lebesgue measure, at almost every point (in fact every point, by continuity of \(J_\varphi\)). B
7
\text{Cover } E \text{ by small disjoint (up to boundary) cubes } Q_i \text{ of side } \to 0,\ \text{ and approximate } \operatorname{vol}(\varphi(E)) \approx \sum_i \operatorname{vol}(\varphi(Q_i)) \approx \sum_i \lvert J_\varphi(u_i)\rvert\,\operatorname{vol}(Q_i).
This is a Riemann-sum construction: measurability of \(E\) lets us approximate its volume from outside/inside by finite unions of dyadic cubes to within any \(\varepsilon\) (regularity of Lebesgue measure), and Step 6 replaces each small piece's image volume by \(\lvert J_\varphi(u_i)\rvert\) times the piece's own volume, with error going to \(0\) uniformly because \(J_\varphi\) is continuous (as \(\varphi \in C^1\)) hence uniformly continuous on compact subsets of \(U\). B
8
\text{The right-hand Riemann sum } \sum_i \lvert J_\varphi(u_i)\rvert\,\operatorname{vol}(Q_i) \to \int_E \lvert J_\varphi(u)\rvert\,du \text{ as the mesh } \to 0.
This is precisely the definition of the Riemann/Lebesgue integral of the continuous (hence measurable and, on bounded \(E\), integrable when \(E\) has finite measure) function \(\lvert J_\varphi\rvert\) over \(E\), as a limit of Riemann sums over a partition into small cubes refining to zero mesh. Combined with Step 7, both sides of the target identity are limits of the same sequence of approximating sums, so \(\operatorname{vol}(\varphi(E)) = \int_E \lvert J_\varphi(u)\rvert\,du\). A
9
\text{Conclude for general } E \text{ (possibly of infinite measure or unbounded) by monotone exhaustion.}
Write \(U = \bigcup_k K_k\) as an increasing union of compact sets and apply Steps 2–8 to \(E \cap K_k\); then let \(k \to \infty\) and invoke the Monotone Convergence Theorem (for the integral side) together with continuity from below of Lebesgue measure (for the volume side, since \(\varphi\) is a bijection so \(\varphi(E\cap K_k) \uparrow \varphi(E)\)) to pass to the limit. This closes the argument for arbitrary measurable \(E \subseteq U\), completing the proof of the volume identity and, via Step 1, the general integral identity. C
Result
\int_{\varphi(E)} f(x)\,dx \;=\; \int_{E} f(\varphi(u))\,\lvert J_\varphi(u)\rvert\,du

Reading. To integrate over a region \(\varphi(E)\) described awkwardly in \(x\)-coordinates, pull everything back to the \(u\)-coordinates where \(E\) may be a box or other simple shape: replace \(x\) by \(\varphi(u)\), and replace the volume element \(dx\) by \(\lvert J_\varphi(u)\rvert\,du\) — the local volume-scaling factor of the coordinate change.

Scope. Applies to any \(C^1\)-diffeomorphism between open subsets of \(\mathbb{R}^n\) (any finite \(n \ge 1\); for \(n=1\) it reduces to ordinary \(u\)-substitution with \(\lvert J_\varphi\rvert = \lvert \varphi'\rvert\)). Does not by itself apply to maps that fail to be injective, fail to be \(C^1\), or degenerate (\(J_\varphi = 0\)) somewhere on the domain of integration; such cases require decomposing the domain first (see Fails without).

Corollaries & converses
  • Polar/cylindrical/spherical coordinates are the standard corollaries: e.g. for \(\varphi(r,\theta)=(r\cos\theta,r\sin\theta)\) on \((0,\infty)\times(0,2\pi)\), \(J_\varphi = r\), recovering \(dx\,dy = r\,dr\,d\theta\).
  • Composability: if \(\varphi = \psi \circ \chi\) is a composite diffeomorphism, the chain rule \(D\varphi = D\psi \cdot D\chi\) and multiplicativity of \(\det\) give \(J_\varphi(u) = J_\psi(\chi(u))\,J_\chi(u)\), so the theorem is consistent with performing a change of variables in stages.
  • Invariance of measure zero: \(C^1\)-diffeomorphisms map Lebesgue-null sets to Lebesgue-null sets (immediate from the volume identity with \(\operatorname{vol}(E)=0\)), which is why "almost everywhere" statements transport across smooth coordinate changes.
  • Converse (does it hold)? If a bijection \(\varphi:U\to V\) merely satisfies \(\operatorname{vol}(\varphi(E)) = \int_E g(u)\,du\) for some function \(g\) and all measurable \(E\), it does not follow that \(\varphi\) is differentiable with \(g = \lvert J_\varphi\rvert\); e.g. exotic measure-preserving bijections (not even continuous) exist with \(g\equiv 1\). The theorem is a one-way implication: smoothness \(\Rightarrow\) the Jacobian formula, not the reverse.
Fails without
  • Drop injectivity: \(\varphi(u) = u^2\) on \(U=(-1,1) \to V=[0,1)\) is \(2\)-to-\(1\) except at \(0\). Naively computing \(\int_{-1}^{1} f(u^2)\,\lvert 2u\rvert\,du\) does not equal \(\int_0^1 f(x)\,dx\); it equals \(2\int_0^1 f(x)\,dx\), double-counting because each \(x \gt 0\) has two preimages.
  • Drop \(J_\varphi \neq 0\) everywhere: the polar map \(\varphi(r,\theta) = (r\cos\theta, r\sin\theta)\) has \(J_\varphi(0,\theta) = 0\) along \(r=0\); this set has measure zero so causes no error there, but a genuinely bad example is \(\varphi(u,v) = (u, uv)\) on \(\mathbb{R}^2\), which collapses the line \(u=0\) to the single point \((0,0)\) — the map fails to be injective near \(u=0\) and the "inverse" is not \(C^1\) there, so the formula cannot be applied across \(u=0\) without splitting the domain.
  • Drop \(C^1\) regularity: the Cantor function (devil's staircase) extended to a homeomorphism of \([0,1]\) is monotonic and continuous with derivative \(0\) almost everywhere, yet maps the Cantor set (measure \(0\)) onto a set of positive structure in the range in a way that breaks the naive "integrate the derivative" heuristic; more simply, without \(C^1\) the Jacobian may not exist on a large set and the area formula requires the weaker Rademacher/Lipschitz machinery instead of pointwise Riemann-sum reasoning used above.
Common errors
  • Forgetting the absolute value: writing \(dx = J_\varphi\,du\) instead of \(\lvert J_\varphi\rvert\,du\), which can silently introduce a wrong overall sign when \(J_\varphi \lt 0\) (e.g. an orientation-reversing map).
  • Computing the Jacobian of \(\varphi\) but evaluating it at \(x\) instead of at \(u = \varphi^{-1}(x)\), or conflating \(J_\varphi\) with \(J_{\varphi^{-1}} = 1/J_\varphi\) (mixing up which coordinates the determinant is taken with respect to).
  • Applying the polar/spherical Jacobian formula over a \(\theta\)-range wider than \(2\pi\) (or a \(\varphi\)-range in spherical coordinates outside \([0,\pi]\)), silently violating injectivity and over-counting volume.
  • Forgetting to transform the region \(E\) itself, e.g. keeping Cartesian inequality bounds on \(x,y\) while integrating in \(r,\theta\), instead of re-deriving the bounds for \(E = \varphi^{-1}(\text{region})\).
  • Treating the formula as valid at isolated critical points where \(J_\varphi=0\) (e.g. the origin in polar coordinates) as if special care were needed there, when in fact a measure-zero set of such points is harmless and can simply be included in the domain without altering any integral.
Discussion

The change-of-variables formula is often first met as a one-dimensional "trick" (\(u\)-substitution), but its structural content is genuinely multivariable: it says that a smooth invertible map's local volume-distortion factor is exactly \(\lvert \det D\varphi\rvert\), the absolute value of the determinant of its derivative. This is the same determinant that measures how a linear map scales areas/volumes and orientation, so the theorem can be read as "linearise, apply linear algebra, then take a limit" — precisely the strategy used in the proof above via Steps 3–6.

Historically, Jacobi's determinant (1830s–40s) arose from exactly this question of transforming multiple integrals, generalising Euler's and Lagrange's earlier work on double integrals in curvilinear coordinates. The rigorous measure-theoretic version proved above (via cube-covering Riemann sums bootstrapped to Lebesgue integrals) is essentially the 19th-century argument made precise with 20th-century measure theory; a fully general version — the area formula — extends the theorem to merely Lipschitz (not necessarily \(C^1\), not necessarily injective, using multiplicity) maps between rectifiable sets, at the cost of replacing \(D\varphi\) by an almost-everywhere-defined derivative guaranteed by Rademacher's theorem.

In differential-geometric language, the theorem is the statement that pullback of the standard volume form \(dx_1 \wedge \cdots \wedge dx_n\) under a diffeomorphism \(\varphi\) equals \(J_\varphi(u)\,du_1\wedge\cdots\wedge du_n\) — i.e. \(\varphi^*(dx) = J_\varphi\,du\) as differential \(n\)-forms, with the sign of \(J_\varphi\) tracking whether \(\varphi\) preserves or reverses orientation, while the integral of the unsigned volume form uses \(\lvert J_\varphi\rvert\). This reformulation is what generalises the theorem from open subsets of \(\mathbb{R}^n\) to integration over smooth manifolds, where there is no global coordinate system and one must patch together such local Jacobian factors via a partition of unity.

Common misconception: students often believe the Jacobian "measures how the function \(f\) changes"; in fact \(J_\varphi\) depends only on the coordinate map \(\varphi\), never on the integrand \(f\) — it is a purely geometric correction for how the coordinate grid itself is stretched, applied uniformly regardless of what is being integrated.

Worked examples
1
\text{Compute } I = \iint_D e^{-(x^2+y^2)}\,dx\,dy \text{ where } D = \mathbb{R}^2, \text{ using polar coordinates.}
The Gaussian integral has no elementary antiderivative in Cartesian form, motivating a coordinate change. A
2
\text{Let } \varphi(r,\theta) = (r\cos\theta, r\sin\theta) \text{ on } U=(0,\infty)\times(0,2\pi),\ V = \mathbb{R}^2 \setminus \{(x,0): x \ge 0\}.
\(\varphi\) is a \(C^1\) bijection \(U \to V\) with \(J_\varphi(r,\theta) = \det\begin{pmatrix}\cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta\end{pmatrix} = r\cos^2\theta + r\sin^2\theta = r \neq 0\) on \(U\); the excluded ray has measure zero and does not affect the integral. A
3
I = \int_0^{2\pi}\!\!\int_0^\infty e^{-r^2}\,r\,dr\,d\theta
Direct application of the theorem: \(dx\,dy = \lvert J_\varphi \rvert\, dr\,d\theta = r\,dr\,d\theta\), and \(f(\varphi(r,\theta)) = e^{-r^2}\). B
4
\int_0^\infty e^{-r^2} r\,dr = \left[-\tfrac12 e^{-r^2}\right]_0^\infty = \tfrac12, \qquad I = \int_0^{2\pi} \tfrac12\,d\theta = \pi.
The inner integral is now elementary by the substitution \(s=r^2\) (a one-dimensional change of variables, the \(n=1\) case of this same theorem), and the outer integral is trivial. A
\iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dx\,dy = \pi \ \Longrightarrow\ \int_{-\infty}^\infty e^{-x^2}\,dx = \sqrt{\pi}

Reading. Squaring the 1-D Gaussian integral produces exactly the 2-D integral computed here (by Fubini and separability of \(e^{-x^2-y^2}=e^{-x^2}e^{-y^2}\)), so \(\left(\int e^{-x^2}dx\right)^2 = \pi\), giving the celebrated value \(\sqrt{\pi}\).

1
\text{Find the volume of the solid ellipsoid } E = \left\{(x,y,z): \frac{x^2}{a^2}+\frac{y^2}{b^2}+\frac{z^2}{c^2} \le 1\right\},\ a,b,c \gt 0.
Direct Cartesian integration is possible but tedious; a linear change of variables reduces this to the volume of a ball, already known. A
2
\text{Let } \varphi(u,v,w) = (au, bv, cw), \text{ a linear bijection } \mathbb{R}^3 \to \mathbb{R}^3 \text{ with } \varphi(B) = E \text{ where } B = \{u^2+v^2+w^2 \le 1\}.
Substituting \(x=au,y=bv,z=cw\) into the defining inequality of \(E\) gives exactly \(u^2+v^2+w^2\le 1\), so \(\varphi\) maps the unit ball \(B\) bijectively onto \(E\); \(\varphi\) is \(C^1\) (indeed linear) with \(C^1\) inverse \(\varphi^{-1}(x,y,z)=(x/a,y/b,z/c)\). A
3
J_\varphi = \det\begin{pmatrix} a&0&0\\0&b&0\\0&0&c\end{pmatrix} = abc \neq 0 \text{ everywhere on } \mathbb{R}^3.
Determinant of a diagonal matrix; all hypotheses of the theorem hold globally since \(\varphi\) is linear and invertible (\(a,b,c\gt0\)). A
4
\operatorname{vol}(E) = \operatorname{vol}(\varphi(B)) = \int_B \lvert J_\varphi \rvert\,du\,dv\,dw = abc \int_B du\,dv\,dw = abc \cdot \operatorname{vol}(B) = abc\cdot\frac{4}{3}\pi.
Direct application of the theorem's volume corollary (\(f\equiv 1\)) with the constant Jacobian factored out of the integral, and the known volume \(\frac{4}{3}\pi\) of the unit ball. B
\operatorname{vol}(E) = \frac{4}{3}\pi abc

Reading. The ellipsoid's volume is the ball's volume scaled by the product of the three semi-axis lengths — exactly what the constant Jacobian \(abc\) of the diagonal stretching map predicts.

Problems
  1. Use the change of variables \(x=u+v,\ y=u-v\) to evaluate \(\iint_R (x^2-y^2)\,e^{xy}\,\text{—} \) more precisely, evaluate \(\iint_R (x+y)\,dx\,dy\) where \(R\) is the square with vertices \((0,0),(1,1),(2,0),(1,-1)\).
    Solution The square \(R\) is exactly the image under \(\varphi(u,v)=(u+v,u-v)\) of the unit square \(U=[0,1]\times[0,1]\) in \((u,v)\) — check the four vertices: \((0,0)\mapsto(0,0)\), \((1,0)\mapsto(1,1)\), \((1,1)\mapsto(2,0)\), \((0,1)\mapsto(1,-1)\), matching \(R\)'s vertices in order. \(\varphi\) is linear with \(J_\varphi = \det\begin{pmatrix}1&1\\1&-1\end{pmatrix} = -2\), so \(\lvert J_\varphi\rvert = 2\), constant and nonzero — the hypotheses hold. Then \(x+y = 2u\), so \[ \iint_R (x+y)\,dx\,dy = \int_0^1\!\!\int_0^1 2u \cdot 2\,du\,dv = 4\int_0^1 u\,du \int_0^1 dv = 4\cdot\tfrac12\cdot 1 = 2. \]
  2. Explain precisely why applying \(\varphi(r,\theta)=(r\cos\theta,r\sin\theta)\) with \(\theta\) ranging over \((0,4\pi)\) instead of \((0,2\pi)\) to compute the area of the unit disk gives \(2\pi\) instead of \(\pi\), citing which hypothesis of the theorem fails.
    Solution The theorem requires \(\varphi\) to be a bijection \(U \to V\). On \(U = (0,1)\times(0,4\pi)\), the map \(\varphi(r,\theta)=(r\cos\theta,r\sin\theta)\) is exactly \(2\)-to-\(1\) onto the punctured disk \(V\), since \(\theta\) and \(\theta+2\pi\) give the same point. Injectivity fails, so the theorem's hypotheses are not met, and the formula \(\operatorname{vol}(\varphi(E)) = \int_E \lvert J_\varphi\rvert\) is invalid as stated. Computing anyway, \(\int_0^{4\pi}\int_0^1 r\,dr\,d\theta = 4\pi \cdot \tfrac12 = 2\pi\), double the correct area \(\pi\) of the unit disk, because every point of the disk (except the centre and one ray) is counted with multiplicity \(2\) — consistent with the "Fails without injectivity" counterexample discussed above.
  3. Let \(\varphi(u,v) = (u^2 - v^2, 2uv)\) (the complex map \(z \mapsto z^2\) in real coordinates) restricted to \(U = \{(u,v): u \gt 0\}\). Show \(\varphi\) satisfies the hypotheses of the theorem on \(U\), compute \(J_\varphi\), and use it to find the area of the image of the quarter-disk \(Q=\{u^2+v^2\le 1, u\gt 0, v\gt 0\}\).
    Solution \(D\varphi = \begin{pmatrix}2u & -2v\\ 2v & 2u\end{pmatrix}\), so \(J_\varphi = 4u^2+4v^2 = 4(u^2+v^2)\), which is nonzero everywhere on \(U\) (since \(u\gt0\) forces \((u,v)\neq(0,0)\)); \(\varphi\) is \(C^1\) and, in complex notation \(z=u+iv\), \(\varphi\) is \(z\mapsto z^2\) which is injective on the right half-plane \(\{\operatorname{Re}(z)\gt 0\}\) because \(z^2=w^2 \Rightarrow z=\pm w\) and only one of \(\pm w\) lies in the right half-plane; its inverse (principal square root) is \(C^1\) there since \(J_\varphi \neq 0\). All hypotheses hold. For \(Q\) (a quarter of the unit disk, \(u,v\gt0\)), by the theorem, \[ \operatorname{Area}(\varphi(Q)) = \int_Q 4(u^2+v^2)\,du\,dv = 4\int_0^{\pi/2}\!\!\int_0^1 r^2\cdot r\,dr\,d\theta = 4\cdot\frac{\pi}{2}\cdot\frac{1}{4} = \frac{\pi}{2}, \] using polar coordinates for the inner integral (itself a second, nested application of the change-of-variables theorem).
  4. A student claims: "since \(J_\varphi(0,0)=0\) for the polar coordinate map at the origin, the change-of-variables formula cannot be used to compute the area of the full unit disk \(\{x^2+y^2 \le 1\}\)." Is the student correct? Justify your answer carefully using the theorem's hypotheses.
    Solution The student is incorrect, though the concern is reasonable at first glance. The theorem requires \(J_\varphi \neq 0\) on the open set \(U\), but the origin \(r=0\) is not actually part of the open parameter domain \(U=(0,1)\times(0,2\pi)\) used to parametrise the disk minus one radius; \(r=0\) is a boundary point of \(U\), not an interior point where the map needs to be a diffeomorphism. The single missing radius (\(\theta=0\)) and the origin together form a set of Lebesgue measure zero in the target, and removing a measure-zero set from a region does not change its integral or its volume. So one applies the theorem rigorously on the open set \(U=(0,1)\times(0,2\pi)\) (where all hypotheses genuinely hold), obtaining \(\int_U r\,dr\,d\theta = \pi\), and this equals the area of the full closed disk because the discarded boundary (the segment \(\theta=0\) together with the point \(r=0\)) has measure zero. So the formula is used correctly, just on a slightly smaller open set than the full disk, with no loss of information about the area.
  5. Let \(\varphi:\mathbb{R}^n\to\mathbb{R}^n\) be linear and invertible, \(\varphi(u)=Au\) for an invertible matrix \(A\). Using the change-of-variables theorem, prove that for the standard multivariate Gaussian density transform, if \(X\) has density \(p(x) = \frac{1}{(2\pi)^{n/2}}e^{-\lvert x\rvert^2/2}\) on \(\mathbb{R}^n\) and \(Y=AX\), then \(Y\) has density \(q(y) = \frac{1}{\lvert\det A\rvert(2\pi)^{n/2}}e^{-\lvert A^{-1}y\rvert^2/2}\).
    Solution For any measurable set \(E \subseteq \mathbb{R}^n\), \(\mathbb{P}(Y \in E) = \mathbb{P}(AX\in E) = \mathbb{P}(X \in A^{-1}E) = \int_{A^{-1}E} p(x)\,dx\). Apply the change-of-variables theorem with \(\varphi(u) = Au\) (linear, hence \(C^1\), bijective since \(A\) is invertible, with constant nonzero Jacobian \(J_\varphi = \det A\)) and the set \(A^{-1}E\), so that \(\varphi(A^{-1}E) = E\); precisely, take \(E' = A^{-1}E\) as the "u-domain" set and note \(\int_{E'} p(u)\,du = \int_{\varphi(E')} p(\varphi^{-1}(y))\,\lvert J_{\varphi^{-1}}(y)\rvert\,dy\) by applying the theorem to \(\varphi^{-1}\) (also linear, invertible, \(C^1\), Jacobian \(\det A^{-1} = 1/\det A\)) mapping \(E=\varphi(E')\) back to \(E'\). This gives \[ \mathbb{P}(Y\in E) = \int_{A^{-1}E} p(x)\,dx = \int_E p(A^{-1}y)\,\frac{1}{\lvert \det A\rvert}\,dy. \] Since this holds for every measurable \(E\), the integrand \(q(y) = \frac{1}{\lvert\det A\rvert} p(A^{-1}y) = \frac{1}{\lvert\det A\rvert(2\pi)^{n/2}} e^{-\lvert A^{-1}y\rvert^2/2}\) is exactly the density of \(Y\), as claimed.