Change of Variables & Jacobians

1. Linear Transformations & the Determinant

A linear map $T(u,v) = (au + bv,\; cu + dv)$ sends the unit square to a parallelogram. The absolute value of the determinant $|\det A| = |ad - bc|$ tells you exactly how much area stretches (or shrinks).

Drag the sliders to reshape the matrix and watch the parallelogram change.

2. Nonlinear Stretching & the Jacobian

For a nonlinear map $(x,y) = T(u,v)$, the stretching varies from point to point. At each point, the best linear approximation is the Jacobian matrix:

Zoom into a tiny cell of the $(u,v)$-grid. Under the nonlinear map, it becomes a tiny parallelogram whose area ratio is $|\det J|$ evaluated at that point.

In this example, the map $(r, \theta) \to (r\cos\theta,\; r\sin\theta)$ sends the rectangular $(r,\theta)$-grid to the familiar polar grid of circles and rays. The Jacobian is $|\det J| = r$, so cells farther from the origin get stretched more.

3. Jacobian Heatmap

Visualize $\left|\frac{\partial(x,y)}{\partial(u,v)}\right|$ over the entire parameter domain. Bright regions stretch area; dark regions compress it.

4. Polar Coordinates: Where the $r$ Comes From

In polar coordinates, $x = r\cos\theta$, $y = r\sin\theta$. The Jacobian is:

So $dA = r\,dr\,d\theta$. At small $r$, wedge-shaped cells are tiny; at large $r$, they're big. The factor of $r$ accounts for this radial stretching.

5. The Change-of-Variables Formula

Putting it all together:

6. Worked Example - Linear Substitution with Jacobian

When the integrand has a pattern like $f(x-y,\, x+y)$, nature is telling us the substitution. The expressions $x - y$ and $x + y$ appear together so naturally that we should just name them.

The problem

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

Step 1: Choose the substitution

Let $u = x - y$ and $v = x + y$. Then $x = (u + v)/2$ and $y = (v - u)/2$. The integrand becomes simply $e^{u/v}$ - much cleaner.

Step 2: Compute the Jacobian

We need $\left|\dfrac{\partial(x,y)}{\partial(u,v)}\right|$:

So $dA = dx\,dy = \tfrac{1}{2}\,du\,dv$.

Step 3: Transform the region

Always transform the boundary, not just the integrand. Sketch both regions. Let's check each vertex of the $xy$-square under $(u,v) = (x-y,\; x+y)$:

• $(1,0) \mapsto (1, 1)$
• $(0,1) \mapsto (-1, 1)$
• $(-1,0) \mapsto (-1, -1)$
• $(0,-1) \mapsto (1, -1)$

The tilted square in $xy$ maps to an axis-aligned square $-1 \le u \le 1$, $-1 \le v \le 1$ in $uv$. That's the whole point of this substitution - it un-tilts the region.

Step 4: Evaluate

The inner integral (with respect to $u$, treating $v$ as constant):

$$\int_{-1}^{1} e^{u/v}\,du = v\,e^{u/v}\Big|_{u=-1}^{u=1} = v\!\left(e^{1/v} - e^{-1/v}\right)$$

So the full integral becomes:

Notice that $g(v) = v(e^{1/v} - e^{-1/v})$ is an even function of $v$ (check: replacing $v$ with $-v$ gives $-v(e^{-1/v} - e^{1/v}) = v(e^{1/v} - e^{-1/v})$). So the integral equals $\displaystyle\int_0^{1} v(e^{1/v} - e^{-1/v})\,dv$, which can be evaluated by the substitution $w = 1/v$. We won't pursue the final closed form here - the key lesson is how the Jacobian and region transformation work together.

7. Worked Example - Polar Coordinates on an Annulus

Polar coordinates really shine when the region is a polar rectangle. An annulus (ring), a wedge, or a sector - these regions have constant $r$-limits and constant $\theta$-limits, so there's no $\theta$-dependence in the $r$-bounds. The integral just factors.

The problem

Evaluate $\displaystyle\iint_R \frac{dA}{(x^2 + y^2)^{3/2}}$, where $R$ is the annulus $1 \le x^2 + y^2 \le 4$.

Step 1: Convert to polar

With $x = r\cos\theta$, $y = r\sin\theta$, the integrand is $(x^2 + y^2)^{-3/2} = (r^2)^{-3/2} = r^{-3}$. Including the Jacobian factor $r$ from $dA = r\,dr\,d\theta$, the integrand becomes:

Step 2: Identify the limits

The annulus $1 \le x^2 + y^2 \le 4$ is simply $1 \le r \le 2$ with a full revolution $0 \le \theta \le 2\pi$. Both limits are constant - no $\theta$-dependence in the $r$-bounds and no $r$-dependence in the $\theta$-bounds. The integral factors cleanly.

Step 3: Evaluate

Sanity check

The area of the annulus is $\pi(2^2 - 1^2) = 3\pi$. So the average value of the integrand over the region is $\pi / 3\pi = 1/3$. Meanwhile, the integrand $r^{-3}$ ranges from $1$ (at $r = 1$) down to $1/8$ (at $r = 2$). An average of $1/3$ sits between these extremes, weighted toward the inner boundary where the integrand is larger and the area elements are smaller. That checks out.

Common trap: When the region IS a polar rectangle (annulus, wedge, sector), the limits are constant - no $\theta$-dependence in the $r$-limits. If you find yourself writing $r$-limits that depend on $\theta$, double-check whether you've set up the region correctly.

8. Cylindrical Coordinates

We've been setting up triple integrals in Cartesian. But what if the region has a circular cross-section - a cylinder, a cone, a pipe? Wrestling with $x^2 + y^2 \le 4$ in Cartesian is painful. We'd have to write $-\sqrt{4 - x^2} \le y \le \sqrt{4 - x^2}$ and deal with square roots in every limit.

Polar coordinates turn circles into rectangles in 2D. Cylindrical coordinates do the same thing in 3D - just stack a $z$-axis on top of the polar plane:

The Jacobian for this transformation is exactly the polar Jacobian (the $z$-coordinate passes through unchanged):

Common trap: Don't forget the extra $r$ in $dV = r\,dr\,d\theta\,dz$. The Jacobian contributes this factor. Missing it is the #1 cylindrical-coordinate mistake.

Worked Example

Evaluate $\displaystyle\iiint_R (x^2 + y^2)\,z\;dV$ over the solid cylinder $x^2 + y^2 \le 4$, $0 \le z \le 3$.

Before we compute, let's predict the answer. The average value of $r^2 = x^2 + y^2$ over a disk of radius 2 is: the disk has area $4\pi$, and $\int_0^{2\pi}\!\int_0^2 r^2 \cdot r\,dr\,d\theta = 2\pi\cdot\frac{r^4}{4}\Big|_0^2 = 8\pi$, so the average $r^2 = 8\pi / 4\pi = 2$. The average $z$ over $[0,3]$ is $3/2$. The volume of the cylinder is $\pi(2)^2(3) = 12\pi$. So the integral should be roughly $2 \cdot \frac{3}{2} \cdot 12\pi = 36\pi$. Let's check...

In cylindrical coordinates, $x^2 + y^2 = r^2$, and the region becomes:

• $0 \le r \le 2$
• $0 \le \theta \le 2\pi$
• $0 \le z \le 3$

The integral factors cleanly:

Our prediction was exactly right. The factoring trick works whenever the integrand separates into $f(r)\cdot g(\theta) \cdot h(z)$ and the region is a product $[r_1, r_2] \times [\theta_1, \theta_2] \times [z_1, z_2]$.