Change of Variables and the Jacobian, multivariable and vector calculus lesson

Draw a grid of small squares on a rubber sheet and stretch the sheet. The squares become slanted, curved little quadrilaterals, some bigger and some smaller than before. A change of variables in a double integral does exactly this to the region, and the Jacobian is the number that says how much each tiny square grew or shrank. Polar, cylindrical and spherical coordinates were special cases. Here is the general rule behind all of them.

Key terms

Coordinate transformation

Definition A coordinate transformation of the plane is a map $T$ from a set in the $uv$ -plane to the $xy$ -plane,

T(u, v) = \big( x(u, v),\, y(u, v) \big).

We call $T$ a $C^1$ transformation on an open set $U$ when the component functions $x(u, v)$ and $y(u, v)$ have continuous first partial derivatives on $U$ . A transformation of space $T(u, v, w) = (x, y, z)$ is defined the same way with three component functions. When $T$ is one-to-one on a set, it has an inverse map there, which sends each image point back to the unique point it came from.

In plain words Each pair of new coordinates $(u, v)$ names a point $(x, y)$ of the ordinary plane. The formula tells you where every parameter point lands.

Example A region bounded by lines of the form $x + y = c$ and $x - y = c$ suggests the new variables $u = x + y$ and $v = x - y$ . Solving for $x$ and $y$ gives the transformation

T(u, v) = \left( \frac{u + v}{2},\, \frac{u - v}{2} \right).

Its inverse is the explicit map $T^{-1}(x, y) = (x + y,\, x - y)$ , so $T$ is one-to-one on the whole $uv$ -plane. Both components are linear, so $T$ is $C^1$ everywhere.

Image region

Definition If $S$ is a set in the $uv$ -plane, its image under $T$ is

T(S) = \{ T(u, v) \mid (u, v) \in S \}.

In a change of variables the region $R$ of the original integral must equal $T(S)$ for a region $S$ that is easier to describe. To find $S$ from $R$ , write each boundary curve of $R$ in the new variables, using the inverse map when one is available.

In plain words The image region is where the parameter region lands. You pick $S$ so that it lands exactly on the region you want to integrate over.

Example For the map above, take the square $S = [0, 2] \times [0, 2]$ . The corners go to

$(u, v)$	$T(u, v)$
$(0, 0)$	$(0, 0)$
$(2, 0)$	$(1, 1)$
$(2, 2)$	$(2, 0)$
$(0, 2)$	$(1, -1)$

Because $T$ is linear, the edges go to straight segments, so $R = T(S)$ is the tilted square with these four corners. Its sides lie on the lines $x + y = 0$ , $x + y = 2$ , $x - y = 0$ and $x - y = 2$ , which are exactly the lines $u = 0$ , $u = 2$ , $v = 0$ and $v = 2$ .

Derivative matrix

Definition Let $T(u, v) = (x(u, v), y(u, v))$ be $C^1$ on an open set. Its derivative matrix at a point $(u, v)$ is

DT(u, v) = \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix}.

The columns are the partial derivative vectors $T_u = \langle x_u, y_u \rangle$ and $T_v = \langle x_v, y_v \rangle$ . Near a point $(u_0, v_0)$ the map is approximated by its linearization

\begin{aligned} &T(u_0 + h, v_0 + k) \\[0.4em] &\approx T(u_0, v_0) + h\, T_u + k\, T_v. \end{aligned}

For a transformation of space the matrix is $3 \times 3$ , with columns $T_u$ , $T_v$ and $T_w$ .

In plain words Up close, any smooth transformation looks like a linear map. The two columns show where a tiny step in $u$ and a tiny step in $v$ get sent.

Example Polar coordinates are the transformation $T(r, \theta) = (r \cos \theta, r \sin \theta)$ , with

DT(r, \theta) = \begin{bmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{bmatrix}.

At $(r, \theta) = (2, \pi/2)$ this is the matrix with rows $(0, -2)$ and $(1, 0)$ . A step in $r$ moves the point straight up along the ray through $(0, 2)$ , and a step in $\theta$ moves it to the left, twice as fast because the circle there has radius 2.

Absolute Jacobian determinant

Definition The Jacobian of a $C^1$ transformation $T(u, v) = (x, y)$ is the determinant of its derivative matrix,

\frac{\partial(x, y)}{\partial(u, v)} = \det DT = x_u y_v - x_v y_u.

For a transformation of space it is the determinant of the $3 \times 3$ derivative matrix, written $\frac{\partial(x, y, z)}{\partial(u, v, w)}$ . A small rectangle $[u_0, u_0 + \Delta u] \times [v_0, v_0 + \Delta v]$ is sent to a curved region that is approximately the parallelogram spanned by $\Delta u\, T_u$ and $\Delta v\, T_v$ , whose area is

\left\lvert \frac{\partial(x, y)}{\partial(u, v)} \right\rvert \Delta u\, \Delta v.

So the absolute value of the Jacobian is the local scaling factor for area, and in space for volume. Its sign only records orientation. A negative Jacobian means the map flips the plane over, like a mirror, and area is never negative.

In plain words The Jacobian says how many times bigger a tiny patch becomes after the map. Throw away the sign, because a reflected patch has the same area as the original.

Example For $x = \frac{u + v}{2}$ and $y = \frac{u - v}{2}$ ,

\begin{aligned} \frac{\partial(x, y)}{\partial(u, v)} &= \frac{1}{2} \cdot \left( -\frac{1}{2} \right) - \frac{1}{2} \cdot \frac{1}{2} \\[0.4em] &= -\frac{1}{2}. \end{aligned}

The sign is negative because this map reverses orientation, and the area factor is $\frac{1}{2}$ . Check it on the square from before. The parameter square $S$ has area 4, and its image is a tilted square with diagonals of length 2, whose area is $\frac{2 \cdot 2}{2} = 2 = 4 \cdot \frac{1}{2}$ . For polar coordinates the same computation gives $\cos^2 \theta \cdot r + r \sin^2 \theta = r$ , which is the familiar factor in $dA = r \, dr \, d\theta$ .

Local invertibility

Definition The inverse function theorem says the following. Let $T$ be a $C^1$ transformation on an open set $U$ in the plane, and let $p$ be a point of $U$ with

\frac{\partial(x, y)}{\partial(u, v)}(p) \ne 0.

Then there is an open set $V$ containing $p$ such that $T$ is one-to-one on $V$ , the image $T(V)$ is open, and the inverse map from $T(V)$ back to $V$ is $C^1$ . At each point $(x, y) = T(u, v)$ with $(u, v)$ in $V$ , the Jacobian of the inverse is the reciprocal of the Jacobian of $T$ at $(u, v)$ ,

\frac{\partial(u, v)}{\partial(x, y)} = \frac{1}{\dfrac{\partial(x, y)}{\partial(u, v)}}.

The same holds in space. This is a local statement only. A nonzero Jacobian at every point does not make $T$ one-to-one on all of $U$ .

In plain words Where the Jacobian is not zero, the map can be undone near that point, and undoing it scales area by the reciprocal factor. Whether it can be undone everywhere at once is a separate question.

Example For $u = x + y$ and $v = x - y$ , the Jacobian in the other direction is

\frac{\partial(u, v)}{\partial(x, y)} = 1 \cdot (-1) - 1 \cdot 1 = -2,

the reciprocal of $-\frac{1}{2}$ , as the theorem promises. Now take polar coordinates on the open set where $r > 0$ and $\theta$ is any real number. The Jacobian $r$ is positive at every point, yet $(r, \theta)$ and $(r, \theta + 2\pi)$ both go to the same point. So the map is one-to-one near each point but not globally.

One-to-one coverage except boundary exceptions

Definition The change of variables theorem for double integrals reads as follows. Let $S$ be a bounded closed region in the $uv$ -plane whose boundary consists of finitely many piecewise smooth curves. Let $T$ be a $C^1$ transformation on an open set containing $S$ , and let $R = T(S)$ . Suppose that $T$ is one-to-one on the interior of $S$ and that its Jacobian is nonzero on the interior of $S$ . Then for every function $f$ continuous on $R$ ,

\begin{aligned} &\iint_R f(x, y) \, dA \\[0.4em] &= \iint_S f\big( T(u, v) \big) \left\lvert \frac{\partial(x, y)}{\partial(u, v)} \right\rvert du \, dv. \end{aligned}

The map may fail to be one-to-one, or the Jacobian may vanish, on the boundary of $S$ , because the boundary has zero area. What is not allowed is interior points of $S$ covering part of $R$ twice, or a region $T(S)$ that misses part of the region you meant to integrate over.

In plain words Every point of the region should be hit exactly once, apart from points on the edge, which carry no area. Overlaps count area twice and gaps leave area out.

Example Polar coordinates on $S = [0, 1] \times [0, 2\pi]$ give the unit disk. On the boundary of $S$ things go wrong. The whole edge $r = 0$ collapses to the origin, where the Jacobian $r$ is zero, and the edges $\theta = 0$ and $\theta = 2\pi$ land on the same segment. On the interior the map is one-to-one with $r > 0$ , so the theorem applies and

\text{Area} = \int_0^{2\pi} \int_0^1 r \, dr \, d\theta = \pi.

If instead you let $\theta$ run over $[0, 4\pi]$ , interior points cover the disk twice and the same integral returns $2\pi$ , because it counts the disk twice.

Double and triple integrals

Definition To change variables in a double integral, choose $T$ and $S$ with $T(S) = R$ , check the hypotheses above, replace $dA$ by $\left\lvert \frac{\partial(x, y)}{\partial(u, v)} \right\rvert du \, dv$ , and integrate over $S$ . For triple integrals the statement is the same with a bounded closed solid $S$ whose boundary consists of finitely many piecewise smooth surfaces, a map $T(u, v, w)$ that is $C^1$ on an open set containing $S$ and one-to-one with nonzero Jacobian on the interior of $S$ , a function $f$ continuous on $T(S)$ , and

dV = \left\lvert \frac{\partial(x, y, z)}{\partial(u, v, w)} \right\rvert du \, dv \, dw.

Cylindrical and spherical coordinates are the cases with factors $r$ and $\rho^2 \sin \phi$ .

In plain words Rewrite the region, rewrite the integrand, and multiply by the absolute Jacobian. Choose the new variables so that the new region is a box.

Example Let $R$ be the region in the first quadrant between the hyperbolas $xy = 1$ and $xy = 4$ and the lines $y = x$ and $y = 4x$ . Set $u = xy$ and $v = y/x$ , so the four boundary curves become $u = 1$ , $u = 4$ , $v = 1$ and $v = 4$ . Solving for $x$ and $y$ gives the explicit inverse

T(u, v) = \left( \sqrt{\frac{u}{v}},\, \sqrt{uv} \right).

This map is $C^1$ on the open quadrant $u, v > 0$ and sends it one-to-one onto the open first quadrant of the $xy$ -plane, with inverse $(x, y) \mapsto (xy, y/x)$ . The open quadrant contains the square $S = [1, 4] \times [1, 4]$ , and $T(S) = R$ . The Jacobian of the inverse is easier to compute,

\begin{aligned} \frac{\partial(u, v)}{\partial(x, y)} &= y \cdot \frac{1}{x} - x \cdot \left( -\frac{y}{x^2} \right) \\[0.4em] &= \frac{2y}{x} = 2v, \end{aligned}

so by the reciprocal rule

\frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{2v} > 0.

The area of $R$ is therefore

\int_1^4 \int_1^4 \frac{1}{2v} \, dv \, du = 3 \cdot \frac{\ln 4}{2} = 3 \ln 2.

In space, the solid ellipsoid $x^2 + \frac{y^2}{4} + \frac{z^2}{9} \le 1$ is the image of the unit ball under $x = u$ , $y = 2v$ , $z = 3w$ . That map is linear and one-to-one with Jacobian $1 \cdot 2 \cdot 3 = 6$ , so the volume is $6 \cdot \frac{4\pi}{3} = 8\pi$ .

Check yourself

Let $R$ be the tilted square with corners $(0, 0)$ , $(1, 1)$ , $(2, 0)$ and $(1, -1)$ . Use $u = x + y$ and $v = x - y$ to compute $\iint_R (x^2 - y^2) \, dA$ .

Show the answer

The integrand is $(x + y)(x - y) = uv$ , the region becomes $[0, 2] \times [0, 2]$ , and the absolute Jacobian is $\frac{1}{2}$ . So

\begin{aligned} &\iint_R (x^2 - y^2) \, dA \\[0.4em] &= \int_0^2 \int_0^2 \frac{uv}{2} \, du \, dv = \frac{2 \cdot 2}{2} = 2. \end{aligned}

The map $T(u, v) = (u^2 - v^2,\, 2uv)$ is one-to-one on the square $S = [0, 1] \times [0, 1]$ . Find its Jacobian and the area of $T(S)$ .

Show the answer

The Jacobian is $2u \cdot 2u - (-2v)(2v) = 4(u^2 + v^2)$ , which is positive except at the corner $(0, 0)$ . The area is

\begin{aligned} &\int_0^1 \int_0^1 4(u^2 + v^2) \, du \, dv \\[0.4em] &= 4 \left( \frac{1}{3} + \frac{1}{3} \right) = \frac{8}{3}. \end{aligned}

A student computes the area of the unit disk as $\int_0^{4\pi} \int_0^1 r \, dr \, d\theta$ . What goes wrong?

Show the answer

The answer $2\pi$ is twice the true area. With $\theta$ in $[0, 4\pi]$ the polar map is not one-to-one on the interior of the parameter region, since $\theta$ and $\theta + 2\pi$ give the same point, so every point of the open disk off one radius is counted twice.

Change of Variables and the Jacobian

Key terms

Coordinate transformation

Image region

Derivative matrix

Absolute Jacobian determinant

Local invertibility

One-to-one coverage except boundary exceptions

Double and triple integrals

Check yourself

How this lesson connects