Integral depth: coordinate coverage, slicing and improper failures

Learning goals and theorem boundary

Choose variables by region geometry, distinguish signed orientation from scalar area scaling, and verify global coverage instead of trusting a local Jacobian alone. We also compare continuous finite-region integration with an improper example where two iterated values differ.

The sufficient proper-integral theorem used here is the version proved in the Jordan/Riemann chapters. Let $W,W'\subset\mathbb R^2$ be open, let $T:W\to W'$ be a $C^1$ diffeomorphism on these open neighborhoods, and let $K\subset W$ be compact and Jordan measurable. For $f$ continuous on $T(K)$,

$$\iint_{T(K)}f(x)\,dA=\iint_K f(T(u))|\det DT(u)|\,du.$$

Bounded boundary values do not change these proper Riemann-region integrals. The neighborhood diffeomorphism hypothesis is stronger than merely requiring an interior bijection and a nonzero Jacobian on the closure. For example, polar coordinates on $(1,2)\times(-\pi,\pi)$ are an interior diffeomorphism but identify the two closed seam edges; this is not a single neighborhood diffeomorphism around that closed rectangle. Such a seam must be treated by an explicit truncation and omission estimate. More general multiplicity and singular-boundary versions require additional results. The linear argument below explains local scale; the complete nonlinear proof under the stated compact-neighborhood hypotheses is provided by the two Jordan/Riemann chapters.

1. Local scale and orientation are different data

For an invertible linear map $x=A u$ in the plane, a unit square becomes the parallelogram spanned by the matrix columns. Its oriented area is $\det A$ and its ordinary area is $|\det A|$, from the two-dimensional cross-product/determinant formula. General rectangles scale by the same factor through bilinearity. An affine translation does not change those areas.

Thus for scalar area integration the factor is $|\det A|$. For example $T(u,v)=(u,-v)$ has determinant minus one but maps a unit square to a region of area one, not minus one. Oriented line and flux integrals retain directional information in a different way; replacing every determinant or cross product by its absolute magnitude would be equally wrong there.

For a nonlinear map, differentiability provides the linear approximation at each point, and the Jacobian describes its local scale. That fact alone says nothing about how many parameter points represent the same image point. Local invertibility must be combined with a domain choice giving the required global one-to-one coverage.

2. Polar variables need an angular range and an origin check

The map $T(r,\theta)=(r\cos\theta,r\sin\theta)$ has determinant $r$. For $r>0$ and an angular interval of length $2\pi$ with one endpoint omitted, it describes the punctured disk once, apart from a chosen radial seam in an open-interior formulation. Letting the angle range over length $4\pi$ describes almost every point twice, even though the Jacobian is still nonzero at every positive radius.

For a bounded continuous integrand on a disk, the coordinate singularity at $r=0$ can be handled by excluding a small radius $\varepsilon$. The omitted contribution has magnitude at most $M\pi\varepsilon^2$ if $|f|\le M$, tending to zero. Piecewise smooth boundary curves and a radial seam have zero area in the standard Riemann-region theory, so they do not alter scalar area integrals. These are hypotheses and limiting facts, not a claim that an arbitrary unbounded singular integrand can be ignored at the origin.

For an annulus $1\le x^2+y^2\le4$, a full once-only polar description is $1\le r\le2$, $0\le\theta<2\pi$. Its area is $\int_0^{2\pi}\int_1^2r\,dr\,d\theta=3\pi$. Extending the same angle integral to $4\pi$ gives $6\pi$, correctly measuring two parameter traversals rather than the annulus's ordinary area.

3. Match an ellipse before integrating its polynomial

Let $a,b>0$ and $D=\{x^2/a^2+y^2/b^2\le1\}$. Set $x=ar\cos\theta,y=br\sin\theta$ with $0\le r\le1$ and one full angular turn. The Jacobian magnitude is $ab r$. Therefore

$$\iint_D(x^2+y^2)\,dA =ab\int_0^{2\pi}\int_0^1r^3(a^2\cos^2\theta+b^2\sin^2\theta)\,dr\,d\theta =\frac{\pi ab(a^2+b^2)}4.$$

The transformed region is simple, the coefficients stay positive and the integrand becomes separable. Those are reasons to choose these variables; the goal is not to introduce a Jacobian for its own sake. A negative scale parameter would require an absolute determinant and a revised interpretation of the semiaxes, so positivity is stated rather than assumed silently.

4. Choose order by drawing the slice inequalities

Consider $D=\{0\le y\le x\le1\}$. A horizontal slice fixes $y$ and has $y\le x\le1$; a vertical slice fixes $x$ and has $0\le y\le x$. For continuous $g$,

$$\int_0^1\int_y^1g(x)\,dx\,dy =\int_0^1\int_0^xg(x)\,dy\,dx =\int_0^1xg(x)\,dx.$$

For $g(x)=e^{x^2}$, the result is $(e-1)/2$. The first order asks for a non-elementary antiderivative, while the second inserts the missing factor $x$ naturally through slice length. The equality is justified because the integrand is continuous on this compact ordinary region, not merely because the same symbols occur in two integrals.

A slice can also have multiple components. In the annulus above, for $|x|\le1$ the allowed $y$ values are

$$[-\sqrt{4-x^2},-\sqrt{1-x^2}]\ \cup\ [\sqrt{1-x^2},\sqrt{4-x^2}],$$

whereas for $1<|x|\le2$ they form the entire interval $[-\sqrt{4-x^2},\sqrt{4-x^2}]$. Integrating the entire outer interval for every $x$ would fill in the hole. A double integral over a disconnected slice is the sum over its components, and changes in slice structure require splitting the outer interval or choosing better coordinates.

5. Two finite iterated improper values can disagree

On $(0,1]^2$, let

$$f(x,y)=\frac{x^2-y^2}{(x^2+y^2)^2}.$$

It is smooth at every included point but unbounded near the missing origin. For each fixed $y>0$,

$$\int_0^1f(x,y)\,dx=\left[-\frac{x}{x^2+y^2}\right]_0^1=-\frac1{1+y^2},$$

so the subsequent $y$ integral is $-\pi/4$. In the reverse order, for each fixed $x>0$,

$$\int_0^1f(x,y)\,dy=\left[\frac{y}{x^2+y^2}\right]_0^1=\frac1{1+x^2},$$

giving $+\pi/4$. Each displayed one-variable integral is legitimate, but the results differ. Thus one cannot assert a Fubini interchange merely from the existence of both iterated improper values.

The obstruction is not hidden: in polar coordinates $f=\cos(2\theta)/r^2$. On the truncated sector $\varepsilon\le r\le1$, $\pi/16\le\theta\le\pi/8$, inside the square, its absolute area integral is bounded below by a positive constant times $\int_\varepsilon^1dr/r$, diverging as $\varepsilon\downarrow0$. Change of variables is valid on each nonsingular truncated sector; the positive lower bound then proves failure of absolute integrability near the origin. Continuity only on the punctured interior does not meet the compact-continuity hypothesis used in the earlier triangle example.

Graded coordinate and diagnosis tasks

1. Affine map. Let $T(u,v)=(2u+v,u+3v)$ map $[0,1]^2$ onto a parallelogram $D$. Compute $\iint_D(x-y)\,dA$ and show the coverage and determinant checks.
2. Orientation. Explain why using a signed determinant for scalar area under $(u,v)\mapsto(u,-v)$ gives an impossible result, and why the correction is not a universal instruction to erase orientation from flux integrals.
3. Multiplicity. Evaluate the polar parameter integral of one over $1\le r\le2$, $0\le\theta\le4\pi$, and identify exactly why it is not the annulus's area.
4. Order selection. Prove the formula $\int_0^1\int_y^1g(x)dxdy=\int_0^1xg(x)dx$ for continuous $g$ by describing the same region in both orders. Why does choosing $g=e^{x^2}$ make the change useful?
5. Improper audit. In the rational counterexample, identify the failed hypotheses behind the attempted order interchange. Why does removing one point from the written domain not make its singular contribution harmless?
6. Synthesis. Use $u=x+y,v=x-y$ to evaluate $\iint_{|x|+|y|\le1}(x^2+y^2)\,dA$, proving the transformed region and absolute Jacobian rather than guessing them.

Full solutions

1. The determinant is $2\cdot3-1\cdot1=5\ne0$, so the linear map is a bijection onto its parallelogram image. The transformed integrand is $(2u+v)-(u+3v)=u-2v$. Thus the integral is $5\int_0^1\int_0^1(u-2v)dudv=5(1/2-1)=-5/2$. Positivity of the determinant makes its magnitude five; no multiple coverage occurs.
2. The image has positive area one, while multiplying the constant-one integrand by determinant minus one would give negative one. Ordinary area uses absolute scale. Flux instead uses a chosen normal and directional cross product; replacing that vector by its magnitude would lose the direction of flow. The two integral types encode different data.
3. The parameter integral is $(4\pi)[(2^2-1^2)/2]=6\pi$. Every ordinary annulus point has two angular representatives in that range, apart from endpoint redundancies, so this counts the region twice. Restricting to one full turn restores one-to-one interior coverage and yields $3\pi$.
4. The inequalities describe $0\le y\le x\le1$. Fixing $x$ gives an inner interval of $y$ length $x$, and compact continuity permits the order change. The resulting one-variable integral has factor $x$. With $g=e^{x^2}$, substitute $w=x^2$ to obtain $(e-1)/2$, avoiding an elementary-antiderivative request for $e^{x^2}$ alone.
5. The function is not continuous and bounded on the closure, and its absolute integral diverges near the omitted origin. Neither standard compact Riemann/Fubini hypotheses nor absolute improper integrability holds. Its behavior on neighborhoods of a missing point can accumulate unbounded integral magnitude even though the point itself has zero area. Omitting a point does not remove every surrounding neighborhood.
6. The identity $|x|+|y|=\max(|x+y|,|x-y|)$ follows by considering whether $x,y$ have the same or opposite signs. Hence the transformed region is $[-1,1]^2$. The inverse is $x=(u+v)/2,y=(u-v)/2$, with determinant $-1/2$, so the area factor is $1/2$. Also $x^2+y^2=(u^2+v^2)/2$. The integral is $\frac14\int_{-1}^1\int_{-1}^1(u^2+v^2)dudv=2/3$. The negative inverse determinant changes orientation but not scalar area.

Summary and terms

A Jacobian describes local linear scale; coverage multiplicity is a separate global issue. A slice may have several intervals, and its structure determines integration bounds. An absolute-integrability hypothesis prevents cancellation-dependent order failures in improper settings. Coordinate choice is justified by geometry, invertibility, coverage and integrability together, not by a nonzero determinant alone.