Math Core

In single-variable calculus, substitution turns a hard integral into an easy one, and it comes with a factor: $dx = g'(u)\,du$ . Multiple integrals have substitutions too. Polar, cylindrical and spherical coordinates are special cases. The general tool lets you pick new variables that fit the region, and the correction factor is called the Jacobian.

Transformations

A transformation $T(u, v) = (x, y)$ is a rule that sends points of a $uv$ -plane to points of the $xy$ -plane:

x = g(u, v), \qquad y = h(u, v).

The idea is to find a region $S$ in the $uv$ -plane that $T$ maps onto the region $R$ you care about, choosing $T$ so that $S$ is simple, ideally a rectangle.

For example, polar coordinates are the transformation $x = r\cos\theta$ , $y = r\sin\theta$ . It maps the rectangle $0 \le r \le 1$ , $0 \le \theta \le 2\pi$ in the $r\theta$ -plane onto the unit disk.

How area stretches

A transformation stretches and distorts area. A tiny rectangle with sides $\Delta u$ and $\Delta v$ is carried to an approximate parallelogram with sides

\left\langle \frac{\partial x}{\partial u}, \frac{\partial y}{\partial u}\right\rangle\Delta u \quad\text{and}\quad \left\langle \frac{\partial x}{\partial v}, \frac{\partial y}{\partial v}\right\rangle\Delta v.

The area of a parallelogram spanned by two vectors is the absolute value of a $2 \times 2$ determinant. That determinant is the Jacobian.

Definition

Jacobian

The Jacobian of the transformation $x = g(u, v)$ , $y = h(u, v)$ is

\frac{\partial(x, y)}{\partial(u, v)} = \begin{vmatrix} \dfrac{\partial x}{\partial u} & \dfrac{\partial x}{\partial v} \\[2mm] \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}.

Near each point, $T$ multiplies areas by $\left|\dfrac{\partial(x, y)}{\partial(u, v)}\right|$ .

For polar coordinates, $\dfrac{\partial(x, y)}{\partial(r, \theta)} = \cos\theta \cdot r\cos\theta - (-r\sin\theta)\sin\theta = r$ . That is exactly the $r$ in $dA = r\,dr\,d\theta$ .

Change of variables in a double integral

If $T$ is a one-to-one transformation (except possibly on the boundary) with continuous partial derivatives that maps $S$ onto $R$ , then

\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.

Common mistake

Use the absolute value of the Jacobian. A Jacobian of $-7$ means areas are multiplied by $7$ (and orientation flips). Dropping the absolute value gives a negative area and the wrong sign.

Worked example: Computing a Jacobian

Find the Jacobian of $x = u^2 - v^2$ , $y = 2uv$ .

Solution. The partial derivatives are $x_u = 2u$ , $x_v = -2v$ , $y_u = 2v$ , $y_v = 2u$ :

\frac{\partial(x, y)}{\partial(u, v)} = (2u)(2u) - (-2v)(2v) = 4u^2 + 4v^2.

Choosing the substitution from the region

Often the region is bounded by lines or curves of the form $\text{(expression)} = \text{constant}$ . Let each expression be a new variable. Then the region becomes a rectangle in the new variables.

When you define $u$ and $v$ in terms of $x$ and $y$ , you can solve for $x$ and $y$ , or use the shortcut

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

The region R is the square bounded by x − y = 0, x − y = 2, x + y = 1 and x + y = 3.Open in grapher →

Worked example: A tilted square

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

Solution. The integrand and the boundaries both involve $x + y$ and $x - y$ , so let $u = x + y$ and $v = x - y$ . The region becomes the rectangle $1 \le u \le 3$ , $0 \le v \le 2$ .

Solving gives $x = \dfrac{u + v}{2}$ and $y = \dfrac{u - v}{2}$ , so

\frac{\partial(x, y)}{\partial(u, v)} = \frac{1}{2}\left(-\frac{1}{2}\right) - \frac{1}{2} \cdot \frac{1}{2} = -\frac{1}{2}, \qquad \left|\frac{\partial(x, y)}{\partial(u, v)}\right| = \frac{1}{2}.

Then

\iint_R (x + y)\,e^{x - y}\,dA = \int_0^2 \int_1^3 u\,e^v \cdot \frac{1}{2}\,du\,dv = \frac{1}{2} \cdot 4 \cdot \left(e^2 - 1\right) = 2\left(e^2 - 1\right).

Worked example: Integrating over an ellipse

Evaluate $\displaystyle\iint_R x^2\,dA$ , where $R$ is the region inside the ellipse $\dfrac{x^2}{9} + \dfrac{y^2}{4} = 1$ .

Solution. Stretch a circle into the ellipse with $x = 3u$ , $y = 2v$ . Then $\dfrac{x^2}{9} + \dfrac{y^2}{4} = u^2 + v^2$ , so the region becomes the unit disk $S$ . The Jacobian is $3 \cdot 2 - 0 \cdot 0 = 6$ :

\iint_R x^2\,dA = \iint_S 9u^2 \cdot 6\,du\,dv = 54\iint_S u^2\,du\,dv.

In polar coordinates on the unit disk, $\displaystyle\iint_S u^2 = \int_0^{2\pi} \int_0^1 r^2\cos^2\theta \cdot r\,dr\,d\theta = \pi \cdot \frac{1}{4} = \frac{\pi}{4}$ . So the answer is $54 \cdot \dfrac{\pi}{4} = \dfrac{27\pi}{2}$ .

The same substitution with $f = 1$ shows that the ellipse $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} \le 1$ has area $ab \cdot \pi = \pi ab$ .

Three variables

For $x = g(u, v, w)$ , $y = h(u, v, w)$ , $z = k(u, v, w)$ , the Jacobian is a $3 \times 3$ determinant of partial derivatives, and

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

For spherical coordinates this determinant works out to $\rho^2\sin\phi$ , which is where that volume element comes from. The stretch $x = au$ , $y = bv$ , $z = cw$ has Jacobian $abc$ and maps the unit ball onto the ellipsoid $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} + \dfrac{z^2}{c^2} \le 1$ , so the ellipsoid has volume $\dfrac{4}{3}\pi abc$ .

Tip

To check a Jacobian, test it on area. If $T$ maps the unit square to a parallelogram, the parallelogram's area should equal $|J|$ when $J$ is constant.

Practice

Practice 1

Find the Jacobian $\dfrac{\partial(x, y)}{\partial(u, v)}$ of the transformation $x = 2u + v$ , $y = u - 3v$ .