5.7 Change of Variables in Multiple Integrals

Big idea

Polar coordinates worked because a hard region became a rectangle and an extra factor of $r$ appeared to pay for the distortion. Spherical coordinates worked the same way with a factor of $\rho^2\sin\varphi$. Both factors came from a geometric argument specific to that one coordinate system. This section replaces those arguments with a single formula that produces the right factor for any substitution you invent.

The setup generalizes the substitution rule from one variable. There, a change $x = g(u)$ converted $dx$ into $g'(u)\, du$, and the derivative measured how much the substitution stretched lengths. In two variables a substitution takes a pair $(u,v)$ to a pair $(x,y)$, so the local stretching is not a single number but the action of a two-by-two matrix of partial derivatives. What a matrix does to areas is measured by its determinant, and that determinant is the Jacobian.

The local picture is worth holding on to. A small rectangle in the $uv$-plane with sides $\Delta u$ and $\Delta v$ maps to a small curved patch. Near a point the transformation is well approximated by its linear part, so that patch is nearly a parallelogram spanned by the two partial derivative vectors, and its area is the absolute value of the determinant times $\Delta u\, \Delta v$. Summing over cells gives the change of variables formula.

The practical use runs backwards from the formula. You rarely begin with a transformation; you begin with an awkward region or integrand and design one that fixes it. If the region is bounded by four curves on which some expression is constant, name those expressions $u$ and $v$ and the region becomes a rectangle. If the integrand is a function of one of them, the same naming simplifies it too.

Decoder

If a transformation is one-to-one and continuously differentiable and maps a region $S$ in the $uv$-plane onto a region $R$ in the $xy$-plane, then the integral over $R$ equals the integral over $S$ of the integrand rewritten in $u$ and $v$, multiplied by the absolute value of the Jacobian determinant.

The Jacobian is a local area-scaling factor: at each point it says by what factor the map inflates a tiny patch of area there. For a linear transformation that factor is constant and could be pulled out front. For everything else it varies from point to point and belongs inside the integral.

The absolute value is there because area has no sign and determinants do. A transformation that reflects the plane, or that lists the coordinates in the opposite rotational order, has a negative determinant, yet the patches it produces still have positive area. In one variable the same issue exists and is handled invisibly: a decreasing substitution flips the limits, and flipping the limits changes the sign. In two and three variables the limits are not ordered in that way, so the sign has to be taken explicitly.

The condition to check is one-to-one. Different points of $S$ must land on different points of $R$, or the region gets covered twice and the integral doubles. Failures on the boundary are tolerated, as in polar coordinates, where the edge $r = 0$ collapses to the origin and $\theta = 0$ and $\theta = 2\pi$ name the same ray. A curve contributes nothing to an area integral, so the formula survives.

Definitions and results

Transformation. A map $T(u,v) = \big( x(u,v),\, y(u,v) \big)$ sends a region $S$ in the $uv$-plane to its image $R = T(S)$. It is one-to-one on $S$ if no two interior points share an image.

The Jacobian in two variables.

$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[6pt] \dfrac{\partial y}{\partial u} & \dfrac{\partial y}{\partial v} \end{vmatrix} = \frac{\partial x}{\partial u}\frac{\partial y}{\partial v} - \frac{\partial x}{\partial v}\frac{\partial y}{\partial u}$$

The change of variables formula.

$$ \iint_R f(x,y)\, dA = \iint_S f\big( x(u,v), y(u,v) \big) \left| \frac{\partial(x,y)}{\partial(u,v)} \right| du\, dv $$

Reciprocal Jacobians. The Jacobian of the inverse transformation is the reciprocal of the Jacobian of the transformation. When $u$ and $v$ are given explicitly in terms of $x$ and $y$ but solving back is messy, compute $\frac{\partial(u,v)}{\partial(x,y)}$ and invert it.

Polar coordinates as a special case. With $x = r\cos\theta$ and $y = r\sin\theta$,

$$ \frac{\partial(x,y)}{\partial(r,\theta)} = \begin{vmatrix} \cos\theta & -r\sin\theta \\ \sin\theta & r\cos\theta \end{vmatrix} = r\cos^2\theta + r\sin^2\theta = r $$

so $dA = r\, dr\, d\theta$, the factor derived geometrically earlier.

The Jacobian in three variables. For $T(u,v,w) = (x,y,z)$ the Jacobian is the determinant of the three-by-three matrix of partial derivatives, and

$$ \iiint_R f\, dV = \iiint_S f \left| \frac{\partial(x,y,z)}{\partial(u,v,w)} \right| du\, dv\, dw $$

Spherical coordinates as a special case. Expanding the determinant for $x = \rho\sin\varphi\cos\theta$, $y = \rho\sin\varphi\sin\theta$, $z = \rho\cos\varphi$ along the bottom row leaves two brackets, $\rho^2\sin\varphi\cos\varphi$ and $\rho\sin^2\varphi$, and the total is

$$ \cos\varphi \cdot \rho^2 \sin\varphi\cos\varphi + \rho\sin\varphi \cdot \rho\sin^2\varphi = \rho^2\sin\varphi\big( \cos^2\varphi + \sin^2\varphi \big) = \rho^2\sin\varphi $$

which is nonnegative for $0 \le \varphi \le \pi$, so the absolute value can be dropped.

Designing a substitution. Read the boundary curves of $R$. If two expressions each take two constant values along them, set $u$ and $v$ equal to those expressions and $S$ becomes a rectangle.

Worked examples

Straightening a tilted square

Evaluate $\iint_R (x+y)\, dA$ where $R$ is bounded by the lines $x+y=1$, $x+y=3$, $x-y=0$ and $x-y=2$.

The boundaries name the substitution: let $u = x+y$ and $v = x-y$, so $S$ is the rectangle $1 \le u \le 3$, $0 \le v \le 2$. Solving gives $x = \tfrac{u+v}{2}$ and $y = \tfrac{u-v}{2}$, so

$$ \frac{\partial(x,y)}{\partial(u,v)} = \begin{vmatrix} 1/2 & 1/2 \\ 1/2 & -1/2 \end{vmatrix} = -\frac{1}{2}, \qquad \left| \frac{\partial(x,y)}{\partial(u,v)} \right| = \frac{1}{2} $$

The determinant came out negative and the absolute value is what enters the integral. The integrand is simply $u$:

$$ \int_0^2 \int_1^3 u \cdot \frac{1}{2}\, du\, dv = \frac{1}{2} \cdot 2 \cdot \left[ \frac{u^2}{2} \right]_1^3 = 4 $$

Check it. The rectangle $S$ has area $4$, so $R$ has area $4 \cdot \tfrac{1}{2} = 2$. The integrand $x+y$ averages $2$ over $R$, since $u$ runs uniformly from $1$ to $3$, and $2 \cdot 2 = 4$.

A triangle and an impossible integrand

Evaluate $\iint_R e^{(x-y)/(x+y)}\, dA$ where $R$ is the triangle with vertices $(0,0)$, $(1,0)$ and $(0,1)$.

In rectangular coordinates the inner integral is hopeless. Set $u = x-y$ and $v = x+y$, so the integrand is $e^{u/v}$ and $x = \tfrac{u+v}{2}$, $y = \tfrac{v-u}{2}$. The Jacobian is

$$ \begin{vmatrix} 1/2 & 1/2 \\ -1/2 & 1/2 \end{vmatrix} = \frac{1}{4} + \frac{1}{4} = \frac{1}{2} $$

Now the region: $x+y \le 1$ says $v \le 1$, $x \ge 0$ says $u \ge -v$, and $y \ge 0$ says $u \le v$. So $S$ is the triangle $0 \le v \le 1$, $-v \le u \le v$, with $v$ integrated last:

$$ \int_0^1 \int_{-v}^{v} e^{u/v}\, \frac{1}{2}\, du\, dv = \frac{1}{2}\int_0^1 v\Big[ e^{u/v} \Big]_{-v}^{v} dv = \frac{e - e^{-1}}{2}\int_0^1 v\, dv = \frac{e - e^{-1}}{4} $$

That is about $0.588$. The triangle has area $\tfrac{1}{2}$ and the exponent ranges from $-1$ to $1$, so the integrand lies between $e^{-1} \approx 0.37$ and $e \approx 2.72$. The implied average of about $1.18$ sits inside that range.

Stretching a disk into an ellipse

Evaluate $\iint_R \left( 1 - \dfrac{x^2}{9} - \dfrac{y^2}{4} \right) dA$ where $R$ is the region $\dfrac{x^2}{9} + \dfrac{y^2}{4} \le 1$.

Take $x = 3u$ and $y = 2v$. Then $S$ is the unit disk $u^2+v^2 \le 1$ and the Jacobian is the constant

$$ \begin{vmatrix} 3 & 0 \\ 0 & 2 \end{vmatrix} = 6 $$

The integrand becomes $1 - u^2 - v^2$, and polar coordinates finish the job:

$$ 6 \int_0^{2\pi} \int_0^1 \big( 1 - r^2 \big) r\, dr\, d\theta = 6 \cdot 2\pi \left( \frac{1}{2} - \frac{1}{4} \right) = 3\pi $$

Two checks. Integrand $1$ would give $6\pi$, the area of an ellipse with semiaxes $3$ and $2$, matching $\pi a b$. And $3\pi$ is a volume under a dome of height $1$ over a region of area $6\pi$, so it had to be smaller; exactly half is the paraboloid rule again.

Curved boundaries and an implicit Jacobian

Evaluate $\iint_R xy\, dA$ where $R$ is the first-quadrant region bounded by the hyperbolas $xy = 1$, $xy = 4$ and the lines $y = x$, $y = 4x$.

The boundaries suggest $u = xy$ and $v = y/x$, giving the rectangle $1 \le u \le 4$, $1 \le v \le 4$. Solving for $x$ and $y$ is possible but the reciprocal rule is faster:

$$ \frac{\partial(u,v)}{\partial(x,y)} = \begin{vmatrix} y & x \\ -y/x^2 & 1/x \end{vmatrix} = \frac{y}{x} + \frac{y}{x} = 2v $$

so the Jacobian we need is $\tfrac{1}{2v}$, positive on $S$. The integrand is $u$:

$$ \int_1^4 \int_1^4 u \cdot \frac{1}{2v}\, du\, dv = \left( \int_1^4 u\, du \right)\left( \int_1^4 \frac{dv}{2v} \right) = \frac{15}{2} \cdot \frac{\ln 4}{2} = \frac{15 \ln 2}{2} $$

About $5.20$. The same setup with integrand $1$ gives the area of $R$ as $3\ln 2 \approx 2.08$, so $xy$ averages about $2.5$ on $R$, between the boundary values $1$ and $4$ as it must be.

Practice

Look at the boundary curves first. Name the expressions that are constant along them, compute the Jacobian from whichever direction is easier, take its absolute value, and only then set up the integral over the new region.

Quiz

Five items on computing Jacobians, applying the change of variables formula in two and three variables, and choosing a substitution that simplifies a region.