Module 41: Jacobians and general changes of variables

8 exercises, each with a complete solution. 1 figure.

On this page

A. Prerequisites and observable goals
B. The central idea: coordinate changes distort small areas
C. The change-of-variables theorem
D. Why a determinant measures local area
E. A complete substitution checklist
F. Worked examples
Computing bridge: Changing a probability representation
G. Independent practice
H. Complete solutions: Module 41
I. Retention and readiness

A. Prerequisites and observable goals#

Before

Double and triple integrals, partial derivatives, vector area, and coordinate geometry.

After

Compute a Jacobian, transform both a region and an integrand, use absolute volume scaling, and recognize when a coordinate map double-counts a region.

B. The central idea: coordinate changes distort small areas#

A substitution changes not only the names inside the function but also the size of the tiny pieces being added. A map $T(u,v)=(x(u,v),y(u,v))$ transforms a small input rectangle into an approximately parallelogram-shaped output cell. The Jacobian determinant measures its signed area scale; its absolute value measures geometric area scale. This generalizes the extra factors seen in one-dimensional substitution and polar coordinates.

C. The change-of-variables theorem#

For a sufficiently smooth one-to-one coordinate map between suitable regions, with nonzero Jacobian in the interiors and appropriate integrability,

\iint_D f(x,y)\dd A =\iint_E f(x(u,v),y(u,v))\left|\frac{\partial(x,y)}{\partial(u,v)}\right|\dd u\dd v,

where

\frac{\partial(x,y)}{\partial(u,v)}= \begin{vmatrix}x_u&x_v\\y_u&y_v\end{vmatrix}=x_uy_v-x_vy_u.

Standard coordinate maps may have exceptional boundary points or seams, such as the polar origin; these can be handled by splitting or limiting arguments when they do not alter the integral. In three dimensions the corresponding factor is the absolute determinant of the $3\times3$ derivative matrix. [O3Jac]

The determinant of the forward map is the reciprocal of that of the inverse map at corresponding regular points. Do not accidentally use $\partial(u,v)/\partial(x,y)$ as though it were $\partial(x,y)/\partial(u,v)$ . Absolute values belong to area and volume integrals. Oriented line and surface integrals later carry signs differently.

D. Why a determinant measures local area#

The images of tiny coordinate sides are approximately $T_u\Delta u$ and $T_v\Delta v$ . Their parallelogram area is the absolute planar determinant of those vectors, giving $|x_uy_v-x_vy_u|\Delta u\Delta v$ . In three dimensions, the scalar triple product of the three derivative columns supplies the volume scale.

For polar coordinates, the derivative matrix is

\begin{pmatrix}\cos\theta&-r\sin\theta\\\sin\theta&r\cos\theta\end{pmatrix},

whose determinant is $r(\cos^2\theta+\sin^2\theta)=r$ . This derives the polar factor rather than adding it from memory. Cylindrical coordinates retain this factor because the vertical coordinate is unchanged.

E. A complete substitution checklist#

Choose new coordinates that straighten the region or simplify the integrand. Solve for the original coordinates when needed. Transform every boundary and specify a one-to-one branch. Compute the forward Jacobian and its absolute value on that branch. Substitute the function and the area or volume factor together, evaluate, and check the resulting geometry. For a linear map, compare with a parallelogram area or transformed vertices; for nonlinear maps, check a few boundary points without mistaking those checks for a proof of global injectivity.

F. Worked examples#

A rotated and stretched rectangle. Let $u=x+y$ , $v=x-y$ , with $0\le u\le2$ , $0\le v\le1$ . Inverting gives $x=(u+v)/2$ , $y=(u-v)/2$ , whose determinant is $-1/2$ . The mapped region has area $\int_0^2\int_0^1(1/2)\dd v\dd u=1$ . The negative determinant reverses orientation, not area. Also $\iint_D(x+y)\dd A=\int_0^2\int_0^1u/2\dd v\dd u=1$ .

Exact image of a rectangular grid under

x=2u+v

y=u-v

. Equal input cells map to parallelograms with three times their area. The negative determinant records reversed orientation, while geometric area uses its magnitude.

An ellipse becomes a disk. For $x=3u,y=2v$ , the ellipse $x^2/9+y^2/4\le1$ becomes $u^2+v^2\le1$ . The constant area scale is $6$ . Further changing $u,v$ to polar coordinates produces the combined factor $6r$ , not just $6$ and not just $r$ .

Computing bridge: Changing a probability representation#

A density is probability per unit of coordinate volume. Under a one-to-one differentiable transformation, preserving probability requires the appropriate absolute Jacobian, just as preserving mass does. Forgetting it changes the amount assigned to a region, not merely its notation. A03 defines densities and conditional probabilities; this module supplies the geometric reason that a transformed density must account for stretching. Coordinate seams and multiple coverage still need separate treatment.

G. Independent practice#

F foundation C standard course work T transfer S optional stretch L runnable lab or project. Attempt the complete set before opening its solutions.

M41.01FSolution

For $x=2u+v,y=u-v$ on $0\le u\le1$ , $0\le v\le2$ , find the Jacobian and image area. Evaluate $\iint_D(x-y)\dd A$ .

M41.02CSolution

Use a change of variables to find the area of $x^2/9+y^2/4\le1$ .

M41.03TSolution

Evaluate $\iint_Dx^2\dd A$ over the ellipse in M41.02.

M41.04TSolution

The map $x=uv,y=v$ sends $1\le u\le2$ , $1\le v\le3$ to a region $D$ . Describe $D$ , find its area, and evaluate $\iint_Dy\dd A$ .

M41.05TSolution

In the first quadrant, $D$ is bounded by $1\le xy\le4$ and $1\le y/x\le9$ . Evaluate $\iint_D(y/x)\dd A$ using $u=xy,v=y/x$ .

M41.06CSolution

Compute the cylindrical-coordinate Jacobian from $(r,\theta,z)$ to $(x,y,z)$ directly as a determinant.

M41.07TSolution

Why does applying $x=u^2,y=v$ over $-1\le u\le1$ , $0\le v\le1$ double-count the unit square? Show this with an area integral and repair the parameter domain.

M41.08SSolution

Prove $\int_{-\infty}^{\infty}e^{-x^2}\dd x=\sqrt\pi$ by squaring finite symmetric integrals, comparing a square with inscribed and circumscribed disks, and using polar coordinates. Avoid assuming an elementary antiderivative.

Optional small hints

For M41.05, the positive inverse is $x=\sqrt{u/v},y=\sqrt{uv}$ . For M41.07, absolute values prevent cancellation but do not prevent duplicate coverage. For M41.08, a square $[-R,R]^2$ lies between disks of radii $R$ and $\sqrt2R$ .

H. Complete solutions: Module 41#

Compare the first step where your work differs. A corrected answer is not yet an independent success.

M41.01Show solution

A signed determinant becomes positive area scale

The determinant is $2(-1)-1(1)=\boxed{-3}$ . The input rectangle has area $2$ , so the image has area $\boxed6$ . Since $x-y=u+2v$ ,

\iint_D(x-y)\dd A=3\int_0^1\int_0^2(u+2v)\dd v\dd u =3\int_0^1(2u+4)\dd u=\boxed{15}.

The affine map is one-to-one because its determinant is nonzero.

Back to M41.01

M41.02Show solution

Stretch the unit disk by two axis factors

Set $x=3r\cos\theta,y=2r\sin\theta$ with $0\le r\le1$ , $0\le\theta\le2\pi$ . The area factor is $6r$ . Thus $A=\int_0^{2\pi}\int_0^16r\dd r\dd\theta=\boxed{6\pi}$ . This agrees with $\pi ab$ for semiaxes $3,2$ .

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M41.03Show solution

Transform the integrand as well as the region

Under the same map, $x^2=9r^2\cos^2\theta$ . Hence

I=54\int_0^{2\pi}\cos^2\theta\dd\theta\int_0^1r^3\dd r =54(\pi)(1/4)=\boxed{27\pi/2}.

The factor $6r$ is still needed even though the ellipse has already become a unit disk in the bounds.

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M41.04Show solution

A nonlinear map with a simple positive inverse

Since $y=v>0$ and $u=x/y$ , the image is $1\le y\le3$ , $y\le x\le2y$ . The Jacobian is $v$ , positive throughout. Area is $\int_1^3\int_1^2 v\dd u\dd v=\boxed4$ . For the other integral, $y=v$ , so $\iint_Dy\dd A=\int_1^3\int_1^2v^2\dd u\dd v=\boxed{26/3}$ . Horizontal slice width $y$ provides the same area and moment checks.

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M41.05Show solution

Invert the coordinates on the stated first-quadrant branch

The inverse formulas are $x=\sqrt{u/v}$ and $y=\sqrt{uv}$ , with $1\le u\le4$ , $1\le v\le9$ . Differentiating gives determinant $1/(2v)>0$ . The integrand $y/x$ becomes $v$ , so it cancels the denominator in the area factor:

I=\int_1^4\int_1^9v\frac1{2v}\dd v\dd u =\frac12(3)(8)=\boxed{12}.

The first-quadrant assumption makes the inverse single-valued and avoids confusing opposite-sign branches.

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M41.06Show solution

A block determinant retains the planar polar factor

The derivative matrix is

\begin{pmatrix} \cos\theta&-r\sin\theta&0\\ \sin\theta&r\cos\theta&0\\ 0&0&1 \end{pmatrix}.

Expanding along the last row gives $r\cos^2\theta+r\sin^2\theta=\boxed r$ . On $r\ge0$ , the absolute value is the same. The degeneracy at the axis is handled by the usual coordinate seam/limit treatment, not by claiming the map is regular there.

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M41.07Show solution

Multiplicity is a separate issue from determinant sign

The Jacobian is $2u$ , so an absolute-area calculation over the full proposed rectangle gives $\int_0^1\int_{-1}^1|2u|\dd u\dd v=2$ . Yet the image is the unit square of area one: almost every image point has two preimages $u=\pm\sqrt x$ . Restricting to $0\le u\le1$ makes the map one-to-one off its boundary and gives $\int_0^1\int_0^12u\dd u\dd v=\boxed1$ . Taking an absolute determinant does not repair double coverage by itself.

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M41.08Show solution

Bounded-domain calculations justify the infinite result

Let $I_R=\int_{-R}^Re^{-x^2}\dd x$ . By finite-domain Fubini, $I_R^2$ is the integral of $e^{-x^2-y^2}$ over $[-R,R]^2$ . Positivity and disk inclusion give

\iint_{r\le R}e^{-r^2}\dd A\le I_R^2\le\iint_{r\le\sqrt2R}e^{-r^2}\dd A.

Polar integration evaluates the disk of radius $A$ as $2\pi\int_0^A re^{-r^2}\dd r=\pi(1-e^{-A^2})$ . Thus $\pi(1-e^{-R^2})\le I_R^2\le\pi(1-e^{-2R^2})$ . Both bounds tend to $\pi$ , and $I_R\ge0$ , so $I_R\to\sqrt\pi$ . Since the integrand is even and nonnegative, each half-line integral is half this finite limit; the ordinary improper integral therefore exists and equals $\boxed{\sqrt\pi}$ . The argument did not presume an unjustified exchange over an unbounded domain.

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I. Retention and readiness#

Retrieve M40.04 and M39.05. Explain why a Jacobian has a direction (forward or inverse) and a sign, and why an ordinary area integral uses its magnitude. Advance when you can verify a coordinate map’s region coverage as well as its determinant.