Build coordinates from the boundaries, integrals from easy to advanced lesson

Integrate $e^{x+y}$ over the region $D$ where $0 \le x + y \le 1$ and $0 \le x - y \le 1$ . The four edges of $D$ are slanted lines, so vertical slices change their formula halfway across, yet the two expressions $x + y$ and $x - y$ that describe the edges also appear in the integrand. Making those expressions the coordinates turns $D$ into a unit square, but a unit square of the new coordinates does not have area $1$ in the plane. This page finds the factor that corrects every small area, the absolute value of a Jacobian determinant, and uses it to evaluate the integral.

Key terms

Parallelogram region

Definition Let $a$ , $b$ , $c$ and $d$ be real numbers with $ad - bc \ne 0$ , and let $p_1 < p_2$ and $q_1 < q_2$ . The set of points $(x, y)$ with

\begin{gathered} p_1 \le ax + by \le p_2, \\[0.4em] q_1 \le cx + dy \le q_2 \end{gathered}

is a parallelogram region. Its sides lie on the lines $ax + by = p_1$ , $ax + by = p_2$ , $cx + dy = q_1$ and $cx + dy = q_2$ , opposite sides are parallel, and each of its four corners is the point where a line of the first pair meets a line of the second pair. The condition $ad - bc \ne 0$ says that the two families of lines are not parallel to each other, so each such pair of lines meets in exactly one point. Every parallelogram in the plane can be described this way, using the lines through its sides.

In plain words Two pairs of parallel lines fence in a slanted four-sided region. Each pair bounds a band where one linear expression stays between two values, and the parallelogram is where the two bands overlap.

Example For $D$ the expressions are $x + y$ and $x - y$ , and $1 \cdot (-1) - 1 \cdot 1 = -2 \ne 0$ . Solving the four pairs of boundary equations gives the corners $(0, 0)$ , $(\tfrac12, \tfrac12)$ , $(1, 0)$ and $(\tfrac12, -\tfrac12)$ . The diagonals, from $(0, 0)$ to $(1, 0)$ and from $(\tfrac12, \tfrac12)$ to $(\tfrac12, -\tfrac12)$ , are perpendicular and both have length $1$ , so $D$ is a tilted square with area $\tfrac12$ . Slicing it by vertical lines needs two formulas. For $0 \le x \le \tfrac12$ a slice runs from $y = -x$ to $y = x$ , and for $\tfrac12 \le x \le 1$ it runs from $y = x - 1$ to $y = 1 - x$ . An iterated integral in $x$ and $y$ therefore splits into two pieces, which add up to the same value $\tfrac{e - 1}{2}$ found below.

Linear change of variables

Definition A linear change of variables in the plane introduces new coordinates $u = ax + by$ and $v = cx + dy$ with real constants satisfying $ad - bc \ne 0$ . Under this condition the map $(x, y) \mapsto (u, v)$ is one-to-one from the whole plane onto the whole plane, so every point has exactly one pair of new coordinates and every pair names exactly one point. If $ad - bc = 0$ , the map sends the whole plane onto a line or a single point, and different points share the same pair $(u, v)$ . Adding constants, as in $u = ax + by + k$ , only shifts the new coordinates and changes nothing that follows. For a parallelogram region, choosing $u$ and $v$ as the two boundary expressions makes the region the rectangle $p_1 \le u \le p_2$ , $q_1 \le v \le q_2$ .

In plain words Use the expressions that describe the edges as the coordinates. In the new coordinates each edge is a line where one coordinate is constant, so all the bounds of integration become constants.

Example For $D$ take $u = x + y$ and $v = x - y$ , with $ad - bc = -2$ . The inequalities become $0 \le u \le 1$ and $0 \le v \le 1$ , so $D$ corresponds to the unit square of the $uv$ -plane. The integrand $e^{x+y}$ becomes $e^u$ , with no $v$ in it. Here one choice simplifies both the region and the function, which is the best case.

Inverse map

Definition The change of variables formula integrates over the new region, so it needs the old coordinates as functions of the new ones, $x = x(u, v)$ and $y = y(u, v)$ . This map from the $uv$ -plane back to the $xy$ -plane is the inverse map. For $u = ax + by$ and $v = cx + dy$ with $\Delta = ad - bc \ne 0$ , solving the linear system gives

\begin{bmatrix} x \\ y \end{bmatrix} = \frac{1}{\Delta} \begin{bmatrix} d & -b \\ -c & a \end{bmatrix} \begin{bmatrix} u \\ v \end{bmatrix}.

The matrix of the inverse map is the inverse of the matrix of the forward map, so its determinant is $1/\Delta$ .

In plain words The forward map tells which point of the new region a point of the plane corresponds to. The inverse map goes back, from a point of the new region to the point of the plane it names, and that is the direction an integral over the new region needs.

Example For $u = x + y$ and $v = x - y$ , adding the equations gives $u + v = 2x$ and subtracting them gives $u - v = 2y$ , so

x = \frac{u + v}{2}, \quad y = \frac{u - v}{2}.

The point $(u, v) = (0.6, 0.2)$ of the square goes to $(x, y) = (0.4, 0.2)$ , and indeed $0.4 + 0.2 = 0.6$ and $0.4 - 0.2 = 0.2$ . The corners $(0, 0)$ , $(1, 0)$ , $(1, 1)$ and $(0, 1)$ of the square go to the corners $(0, 0)$ , $(\tfrac12, \tfrac12)$ , $(1, 0)$ and $(\tfrac12, -\tfrac12)$ of $D$ .

Jacobian determinant

Definition Let $x = x(u, v)$ and $y = y(u, v)$ have continuous first partial derivatives. The Jacobian determinant of this map at a point is

\begin{aligned} \frac{\partial(x, y)}{\partial(u, v)} &= \det \begin{bmatrix} x_u & x_v \\ y_u & y_v \end{bmatrix} \\[0.4em] &= x_u y_v - x_v y_u. \end{aligned}

The columns $(x_u, y_u)$ and $(x_v, y_v)$ are the velocities of the image point when $u$ alone or $v$ alone increases, so a small step $\Delta u$ moves the point by about $\Delta u\,(x_u, y_u)$ and a small step $\Delta v$ moves it by about $\Delta v\,(x_v, y_v)$ . For a linear map the partial derivatives are constant, these moves are exact, and the Jacobian is the determinant of the matrix of the map. When the map has an inverse with continuous partial derivatives, the Jacobians of the map and of its inverse at corresponding points multiply to $1$ , because by the chain rule their derivative matrices are inverse to each other.

In plain words The Jacobian packs the two edge directions of a tiny coordinate cell into one number. Its size measures how much the cell is stretched or shrunk, and its sign says whether the cell keeps or reverses its orientation.

Example For the inverse map $x = \tfrac{u + v}{2}$ , $y = \tfrac{u - v}{2}$ , a step $\Delta u$ moves the image point by $\Delta u\,(\tfrac12, \tfrac12)$ , up and to the right, and a step $\Delta v$ moves it by $\Delta v\,(\tfrac12, -\tfrac12)$ , down and to the right. These two vectors are the columns of the matrix below.

\begin{aligned} \frac{\partial(x, y)}{\partial(u, v)} &= \det \begin{bmatrix} \tfrac12 & \tfrac12 \\[0.2em] \tfrac12 & -\tfrac12 \end{bmatrix} \\[0.4em] &= -\tfrac14 - \tfrac14 = -\tfrac12. \end{aligned}

The forward map $u = x + y$ , $v = x - y$ has

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

and $(-2) \cdot (-\tfrac12) = 1$ , as the reciprocal rule says. The same determinant for polar coordinates $x = r\cos\theta$ , $y = r\sin\theta$ is $\cos\theta \cdot r\cos\theta + r\sin\theta \cdot \sin\theta = r$ , the factor in the polar area element.

Area scale factor

Definition Let $T$ be an invertible linear map from the $uv$ -plane to the $xy$ -plane with matrix $A$ . Every region $S$ that has an area satisfies

\begin{aligned} &\operatorname{area}\bigl(T(S)\bigr) \\[0.4em] &\quad = \lvert \det A \rvert \operatorname{area}(S), \end{aligned}

so $\lvert \det A \rvert$ is the area scale factor of $T$ , the same at every point. The inverse of $T$ has the reciprocal factor $1/\lvert \det A \rvert$ . For a map with continuous first partial derivatives the factor can change from point to point. A small rectangle with a corner at $(u, v)$ and sides $\Delta u$ and $\Delta v$ goes to a region whose area is $\bigl\lvert \frac{\partial(x, y)}{\partial(u, v)} \bigr\rvert\,\Delta u\,\Delta v$ , up to an error that is small compared with $\Delta u\,\Delta v$ as the rectangle shrinks.

In plain words A linear map stretches every region by the same factor, the absolute value of its determinant. Before using the factor, check which way the map goes. The map from the new coordinates to the plane and the map from the plane to the new coordinates have reciprocal factors.

Example The inverse map sends the unit square, with area $1$ , onto $D$ , with area $\tfrac12$ , so its scale factor is $\tfrac12 = \lvert -\tfrac12 \rvert$ . Cell by cell the same thing happens. The edges $\Delta u\,(\tfrac12, \tfrac12)$ and $\Delta v\,(\tfrac12, -\tfrac12)$ are perpendicular, with lengths $\tfrac{\sqrt2}{2}\Delta u$ and $\tfrac{\sqrt2}{2}\Delta v$ , so each small cell becomes a rectangle with area $\tfrac12\,\Delta u\,\Delta v$ . The forward map sends $D$ onto the unit square and doubles every area, with factor $2 = \lvert -2 \rvert$ . Using $2$ instead of $\tfrac12$ in the integral below would give $2(e - 1)$ , four times the correct value.

Absolute value of the Jacobian

Definition The change of variables theorem for double integrals says the following. Let $S$ be a closed bounded region of the $uv$ -plane whose boundary has zero area, for example a region with a piecewise smooth boundary. Let the map $x = x(u, v)$ , $y = y(u, v)$ have continuous first partial derivatives on an open set containing $S$ , let it be one-to-one with a nonzero Jacobian on the interior of $S$ , and let $D$ be the image of $S$ . Then for every function $f$ continuous on $D$

\begin{aligned} &\iint_D f\,dx\,dy \\[0.4em] &= \iint_S f \left\lvert \frac{\partial(x, y)}{\partial(u, v)} \right\rvert du\,dv, \end{aligned}

where on the right $f$ is evaluated at $\bigl(x(u, v), y(u, v)\bigr)$ . For a linear change of variables $u = ax + by$ , $v = cx + dy$ with $ad - bc \ne 0$ , the inverse map satisfies these conditions for every such $S$ , and its Jacobian is the constant $1/(ad - bc)$ , so $dx\,dy$ is replaced by $du\,dv / \lvert ad - bc \rvert$ . The absolute value is needed because a negative Jacobian only records that the map reverses orientation, like a mirror, while the areas of the cells are positive.

In plain words Rewrite three things together. The region becomes the new region, the integrand is written in the new coordinates, and each small area $dx\,dy$ is replaced by the area of the image of a small $du\,dv$ cell. The factor comes from the inverse map, the one that expresses $x$ and $y$ through $u$ and $v$ , and it is used without its sign.

Example Return to the integral

I = \iint_D e^{x+y}\,dx\,dy,

where $D$ is the parallelogram with corners $(0, 0)$ , $(\tfrac12, \tfrac12)$ , $(1, 0)$ and $(\tfrac12, -\tfrac12)$ . The forward map $u = x + y$ , $v = x - y$ is invertible with determinant $-2$ , its inverse $x = \tfrac{u + v}{2}$ , $y = \tfrac{u - v}{2}$ has determinant $-\tfrac12$ , and so $dx\,dy = \tfrac12\,du\,dv$ . The inverse map sends the unit square onto $D$ , and the integrand $e^{x+y} = e^u$ is continuous. Therefore

\begin{aligned} I &= \frac12 \int_0^1 \int_0^1 e^u\,dv\,du \\[0.4em] &= \frac12 \int_0^1 e^u\,du = \frac{e - 1}{2}. \end{aligned}

The inner integral is over $v$ , in which $e^u$ is constant, and the $v$ interval has length $1$ . A check in the plane agrees. On $D$ the value of $x + y$ runs from $0$ to $1$ , so $1 \le e^{x+y} \le e$ , and the integral must lie between $\tfrac12 \cdot 1$ and $\tfrac12 \cdot e$ . Indeed $\tfrac{e - 1}{2} \approx 0.859$ lies between $0.5$ and about $1.359$ . The negative sign of $-\tfrac12$ is visible in the corners. Going counterclockwise around the square through $(0, 0)$ , $(1, 0)$ , $(1, 1)$ and $(0, 1)$ , the image points $(0, 0)$ , $(\tfrac12, \tfrac12)$ , $(1, 0)$ and $(\tfrac12, -\tfrac12)$ go around $D$ clockwise.

Check yourself

The region $R$ is given by $0 \le x + y \le 2$ and $0 \le x - y \le 1$ . Evaluate the integral of $(x - y)\,e^{x+y}$ over $R$ .

Show the answer

With $u = x + y$ and $v = x - y$ the region becomes the rectangle $0 \le u \le 2$ , $0 \le v \le 1$ , and the integrand becomes $v\,e^u$ . The inverse map $x = \tfrac{u + v}{2}$ , $y = \tfrac{u - v}{2}$ has Jacobian $-\tfrac12$ , so $dx\,dy = \tfrac12\,du\,dv$ . The integrand is a product of a function of $u$ and a function of $v$ over a rectangle, so the integral splits into two single integrals,

\begin{aligned} &\iint_R (x - y)\,e^{x+y}\,dx\,dy \\[0.4em] &= \frac12 \int_0^2 e^u\,du \int_0^1 v\,dv \\[0.4em] &= \frac{e^2 - 1}{2} \cdot \frac12 = \frac{e^2 - 1}{4}. \end{aligned}

That is about $1.597$ .

A student integrates $x + 2y$ over the region $R$ given by $0 \le x + 2y \le 3$ and $0 \le x - y \le 1$ . With $u = x + 2y$ and $v = x - y$ , the student computes the forward Jacobian $\partial(u, v)/\partial(x, y) = -3$ , writes $dx\,dy = 3\,du\,dv$ and gets $\tfrac{27}{2}$ . Find the mistake and the correct value.

Show the answer

The factor must come from the inverse map, which expresses $x$ and $y$ through $u$ and $v$ . Solving the two equations gives $x = \tfrac{u + 2v}{3}$ and $y = \tfrac{u - v}{3}$ , so

\begin{aligned} \frac{\partial(x, y)}{\partial(u, v)} &= \tfrac13 \cdot \bigl(-\tfrac13\bigr) - \tfrac23 \cdot \tfrac13 \\[0.4em] &= -\frac13, \end{aligned}

the reciprocal of $-3$ , and $dx\,dy = \tfrac13\,du\,dv$ . The region becomes $0 \le u \le 3$ , $0 \le v \le 1$ , and the integrand is $u$ , so

\begin{aligned} &\iint_R (x + 2y)\,dx\,dy \\[0.4em] &= \frac13 \int_0^3 \int_0^1 u\,dv\,du = \frac32. \end{aligned}

The student’s factor $3$ is the area scale of the forward map, which sends $R$ , with area $1$ , onto a rectangle with area $3$ . Using it makes the answer $9$ times too large, since $\tfrac{27}{2} = 9 \cdot \tfrac32$ . The area gives a quick check. The corners of $R$ are $(0, 0)$ , $(1, 1)$ , $(\tfrac53, \tfrac23)$ and $(\tfrac23, -\tfrac13)$ , and its area is $1 = \tfrac13 \cdot 3$ .

The parallelogram $P$ has corners $(0, 0)$ , $(1, 2)$ , $(3, 3)$ and $(2, 1)$ . Build linear coordinates in which $P$ is the unit square, and evaluate the integral of $x$ over $P$ . What goes wrong without the absolute value?

Show the answer

The sides of $P$ at the origin are the vectors $(1, 2)$ and $(2, 1)$ , so every point of $P$ is $(x, y) = s\,(1, 2) + t\,(2, 1)$ with $0 \le s \le 1$ and $0 \le t \le 1$ , that is, $x = s + 2t$ and $y = 2s + t$ . Solving for the new coordinates gives $s = \tfrac{2y - x}{3}$ and $t = \tfrac{2x - y}{3}$ , so $P$ is the parallelogram region $0 \le 2y - x \le 3$ , $0 \le 2x - y \le 3$ . The map from $(s, t)$ to $(x, y)$ has Jacobian $1 \cdot 1 - 2 \cdot 2 = -3$ , so $dx\,dy = 3\,ds\,dt$ and

\begin{aligned} &\iint_P x\,dx\,dy \\[0.4em] &= 3 \int_0^1 \int_0^1 (s + 2t)\,ds\,dt \\[0.4em] &= 3\Bigl(\frac12 + 1\Bigr) = \frac92. \end{aligned}

As a check, $P$ has area $3$ and its center is $(\tfrac32, \tfrac32)$ , so the integral of $x$ is $3 \cdot \tfrac32 = \tfrac92$ . Without the absolute value the factor would be $-3$ and the result $-\tfrac92$ , a negative number for the integral of a function that is positive inside $P$ . The sign of $-3$ only says that the map from the square to $P$ reverses orientation.

Build coordinates from the boundaries

Key terms

Parallelogram region

Linear change of variables

Inverse map

Jacobian determinant

Area scale factor

Absolute value of the Jacobian

Check yourself

How this lesson connects

Builds on

Used later in