Chapter 19

Change of Variables

19.1A Map Between Two Planes

We have spent some time understanding polar and spherical changes of variables, but these are merely the beginning of a rich collection of coordinate changes. There are elliptical coordinates, which are helpful when a problem involves ellipses, hyperbolic coordinates for problems involving hyperbolas, and many more.

Because coordinates help simplify a problem by making the bounds easier, many of you will see lots of different coordinate systems in your future, especially in physics and engineering. In this one semester course we will not have the time nor need to deep dive into many specific examples, but we will take a brief look at the general theory.

When we computed double integrals in polar coordinates, we used the change of variables

𝑥 = 𝑟 c o s 𝜃, 𝑦 = 𝑟 s i n 𝜃

to rewrite a region in the $𝑥 𝑦$ -plane in terms of new coordinates $𝑟$ and $𝜃$ . In many examples, this change of variables dramatically simplified both the region of integration and the integrand. The key idea is that if we understand the geometry of a region better in some new coordinate system, we can rewrite the integral accordingly, as long as we account for how area is distorted by the change of coordinates.

This motivates the more general question: given a change of variables

𝑥 = 𝑔 (𝑢, 𝑣), 𝑦 = ℎ (𝑢, 𝑣),

can we express a double integral over a region $𝑅$ in the $𝑥 𝑦$ -plane as an integral over a region $𝑆$ in the $𝑢 𝑣$ -plane? To answer this, we need to understand how small area elements transform under the map $(𝑢, 𝑣) \mapsto (𝑥, 𝑦)$ .

Definition 19.1 (A Change-of-Variables Map). A change of variables is a map

𝑇 : 𝑆 \subseteq ℝ 2 ⟶ ℝ 2, 𝑇 (𝑢, 𝑣) = (𝑔 (𝑢, 𝑣), ℎ (𝑢, 𝑣)) = (𝑥 (𝑢, 𝑣), 𝑦 (𝑢, 𝑣)) .

The map takes points and regions in the new $𝑢 𝑣$ -coordinates to points and regions in the old $𝑥 𝑦$ -coordinates.

A change of variables is therefore not just a symbolic substitution: it bends one coordinate grid into another. The figure begins with the familiar polar map, but its menu includes the other maps used later in the chapter.

Figure 19.1 A change of variables is a map, and what it does to a grid is bend it. Run the bend and watch straight lines curve, one family into circles and the other into rays — the coordinate curves from the last chapter, arriving by a different route. The thing to notice is not that the lines move but that the cells stop being equal: near the origin they are slivers, far out they are wide. Whatever we do with these coordinates will have to account for that.

19.1.1A Motivating Example

Suppose we are integrating over the region $𝑅$ in the first quadrant bounded by the lines $𝑦 = 𝑥$ , $𝑦 = 2 𝑥$ , and the hyperbolas $𝑥 𝑦 = 1$ , $𝑥 𝑦 = 2$ . This region is awkward to describe in Cartesian coordinates, but if we define new variables

𝑢 = 𝑥 𝑦, 𝑣 = 𝑦 𝑥,

then the boundaries become $𝑢 = 1$ , $𝑢 = 2$ , $𝑣 = 1$ , and $𝑣 = 2$ ---a rectangle in the $𝑢 𝑣$ -plane!

This illustrates the power of a good change of variables: a complicated region in $𝑥 𝑦$ becomes a simple rectangle in $𝑢 𝑣$ . We can't yet evaluate an integral in these coordinates, though, because we need to understand how area is affected by the change of variables.

19.2The Jacobian as an Area Scale

The two grids below sit side by side so we can follow the same small cell before and after the map. The image cells do not all have the same area, so each one must carry its own scale factor.

Figure 19.2 The two grids sit side by side, and the lit cell is the same cell in both. On the left every cell is the same rectangle

Δ 𝑢 Δ 𝑣

; on the right that one rectangle has become a sliver near the origin or a wide tile far out. Follow the count one cell at a time and compare the pair before reading the total: a sum over cells counts area only if each term is weighted by how much its own cell grew, and that weight is the determinant. The second number is what you get by forgetting it, and it is not the area of anything. This is why

Let $(𝑥, 𝑦) = (𝑔 (𝑢, 𝑣), ℎ (𝑢, 𝑣))$ be a smooth change of variables. To compute how area changes, consider a small rectangle in the $𝑢 𝑣$ -plane with corners at $(𝑢, 𝑣)$ , $(𝑢 + Δ 𝑢, 𝑣)$ , $(𝑢, 𝑣 + Δ 𝑣)$ , and $(𝑢 + Δ 𝑢, 𝑣 + Δ 𝑣)$ . The image of this rectangle under the transformation is approximately a parallelogram in the $𝑥 𝑦$ -plane spanned by the vectors

𝜕 (𝑥, 𝑦) 𝜕 𝑢 Δ 𝑢 a n d 𝜕 (𝑥, 𝑦) 𝜕 𝑣 Δ 𝑣 .

The figure isolates one cell and follows these two derivative vectors as they move through the grid.

Figure 19.3 And here is where that weight comes from. The counting is over, so the patch no longer steps from cell to cell — it glides, and the construction follows it smoothly. Two coordinate lines cross at the point, the map carries them to two coordinate curves crossing at its image, and the patch's two edges — a step

Δ 𝑢

along one line, a step

Δ 𝑣

along the other — become the tangent vectors

⃗ 𝑟 𝑢 Δ 𝑢

and

⃗ 𝑟 𝑣 Δ 𝑣

to those curves. The dashed parallelogram they span is drawn over the patch's true curved image, and they agree. Its area is a determinant, which is why one shows up in the integral at all; and because the tangent vectors turn and stretch continuously as the point moves, that determinant is a function of position, ready to be integrated.

The area of this parallelogram is given by the magnitude of the determinant

∣ 𝜕 (𝑥, 𝑦) 𝜕 (𝑢, 𝑣) ∣ = ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ 𝜕 𝑥 𝜕 𝑢 𝜕 𝑥 𝜕 𝑣 𝜕 𝑦 𝜕 𝑢 𝜕 𝑦 𝜕 𝑣 ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ ∣ .

This determinant is called the Jacobian of the transformation. It measures how a small area element $𝑑 𝑢 𝑑 𝑣$ is stretched or compressed when mapped to the $𝑥 𝑦$ -plane:

𝑑 𝐴 = | d e t 𝐷 𝑇 (𝑢, 𝑣) | 𝑑 𝑢 𝑑 𝑣 = ∣ 𝜕 (𝑥, 𝑦) 𝜕 (𝑢, 𝑣) ∣ 𝑑 𝑢 𝑑 𝑣 .

The absolute value is essential. A negative determinant means that the map reverses orientation, but geometric area is still positive.

Theorem 19.2 (Change of Variables in a Double Integral). Suppose $𝑇 : 𝑆 \to 𝑅$ is differentiable and one-to-one on the relevant interior, with $d e t 𝐷 𝑇 \neq 0$ there, and suppose $𝑓$ is continuous. Then

\iint 𝑅 𝑓 (𝑥, 𝑦) 𝑑 𝑥 𝑑 𝑦 = \iint 𝑆 𝑓 (𝑥 (𝑢, 𝑣), 𝑦 (𝑢, 𝑣)) ∣ 𝜕 (𝑥, 𝑦) 𝜕 (𝑢, 𝑣) ∣ 𝑑 𝑢 𝑑 𝑣 .

Notice the direction of the determinant: the formula requires the Jacobian of the map from the new variables to the old variables,

(𝑢, 𝑣) ⟼ (𝑥, 𝑦) .

If the new variables are initially given as functions of $𝑥, 𝑦$ , we must first solve for $𝑥$ and $𝑦$ in terms of $𝑢, 𝑣$ before forming this determinant.

19.2.1Recovering the Polar Area Element

For the polar map

𝑇 (𝑟, 𝜃) = (𝑟 c o s 𝜃, 𝑟 s i n 𝜃),

we have

𝐷 𝑇 (𝑟, 𝜃) = (c o s 𝜃 - 𝑟 s i n 𝜃 s i n 𝜃 𝑟 c o s 𝜃) .

Therefore

∣ 𝜕 (𝑥, 𝑦) 𝜕 (𝑟, 𝜃) ∣ = ∣ 𝑟 c o s 2 𝜃 + 𝑟 s i n 2 𝜃 ∣ = 𝑟,

because $𝑟 \geq 0$ . Thus the general theory recovers exactly the area element from the last chapter:

𝑑 𝐴 = 𝑟 𝑑 𝑟 𝑑 𝜃 .

19.2.2A Complete Change of Variables

Let us return to our earlier example, where

𝑢 = 𝑥 𝑦, 𝑣 = 𝑦 𝑥 .

To apply the change of variables, we solve for $𝑥$ and $𝑦$ in terms of $𝑢$ and $𝑣$ . Since $𝑦 = 𝑣 𝑥$ , we substitute into $𝑢 = 𝑥 𝑦$ to get

𝑢 = 𝑥 (𝑣 𝑥) = 𝑣 𝑥 2 ⟹ 𝑥 = \sqrt 𝑢 𝑣, 𝑦 = 𝑣 \sqrt 𝑢 𝑣 = \sqrt 𝑢 𝑣 .

We restrict to the first quadrant, so $𝑥, 𝑦 > 0$ and the positive square roots give a one-to-one map.

Now, to compute the transformed integral, we must also calculate the Jacobian determinant $∣ 𝜕 (𝑥, 𝑦) 𝜕 (𝑢, 𝑣) ∣$ . We differentiate:

𝜕 𝑥 𝜕 𝑢 = 1 2 \sqrt 𝑢 𝑣, 𝜕 𝑥 𝜕 𝑣 = - 1 2 𝑣 \sqrt 𝑢 𝑣,

𝜕 𝑦 𝜕 𝑢 = 12 \sqrt 𝑣 𝑢, 𝜕 𝑦 𝜕 𝑣 = 12 \sqrt 𝑢 𝑣 .

Now we compute the determinant:

∣ 𝜕 (𝑥, 𝑦) 𝜕 (𝑢, 𝑣) ∣ = ∣ ∣ ∣ ∣ ∣ 1 2 \sqrt 𝑢 𝑣 - 1 2 𝑣 \sqrt 𝑢 𝑣 12 \sqrt 𝑣 𝑢 12 \sqrt 𝑢 𝑣 ∣ ∣ ∣ ∣ ∣ = 1 4 𝑣 + 1 4 𝑣 = 1 2 𝑣 .

This tells us our new area element:

𝑑 𝐴 = ∣ 𝜕 (𝑥, 𝑦) 𝜕 (𝑢, 𝑣) ∣ 𝑑 𝑢 𝑑 𝑣 = 1 2 𝑣 𝑑 𝑢 𝑑 𝑣 .

We can assemble all of this into a full example.

Example 19.3 (A Change of Variables). Compute $\iint 𝑅 𝑥 2 𝑑 𝐴$ , where $𝑅$ is the first-quadrant region bounded by $𝑦 = 𝑥$ , $𝑦 = 2 𝑥$ , $𝑥 𝑦 = 1$ , and $𝑥 𝑦 = 2$ .

Setting $𝑢 = 𝑥 𝑦$ and $𝑣 = 𝑦 / 𝑥$ converts our bounds to the region

𝑆 = {(𝑢, 𝑣) ∣ 1 \leq 𝑢 \leq 2, 1 \leq 𝑣 \leq 2}

in the $𝑢 𝑣$ -plane.

Our new bounds are constants, but to convert the integrand from $𝑥, 𝑦$ to $𝑢, 𝑣$ , we need to solve for $𝑥, 𝑦$ in our coordinate change. This gives $𝑥 = \sqrt 𝑢 / 𝑣$ and $𝑦 = \sqrt 𝑢 𝑣$ . Thus, our integrand is $𝑥 2 = 𝑢 / 𝑣$ .

We can also use this to find the area element

𝑑 𝐴 = 1 2 𝑣 𝑑 𝑢 𝑑 𝑣 .

All together, then,

\iint 𝑅 𝑥 2 𝑑 𝐴 = \int 21 \int 21 𝑢 𝑣 1 2 𝑣 𝑑 𝑢 𝑑 𝑣 = 12 \int 21 \int 21 𝑢 𝑣 2 𝑑 𝑢 𝑑 𝑣 = 12 (\int 21 𝑢 𝑑 𝑢) (\int 21 1 𝑣 2 𝑑 𝑣) = 12 (32) (12) = 38 .

19.3Area Elements on a Surface

The same geometric idea extends beyond maps between two planes. Suppose a map

⃗ 𝑟 (𝑢, 𝑣) = ⟨ 𝑥 (𝑢, 𝑣), 𝑦 (𝑢, 𝑣), 𝑧 (𝑢, 𝑣) ⟩

sends a region in the $𝑢 𝑣$ -plane into three-dimensional space. A small parameter rectangle is again approximated by a parallelogram, now spanned by

⃗ 𝑟 𝑢 𝑑 𝑢 a n d ⃗ 𝑟 𝑣 𝑑 𝑣 .

There is no two-dimensional determinant for vectors in $ℝ 3$ , but the magnitude of their cross product is exactly the area of the parallelogram they span. Thus the corresponding area scale is

𝑑 𝑆 = ∥ ⃗ 𝑟 𝑢 \times ⃗ 𝑟 𝑣 ∥ 𝑑 𝑢 𝑑 𝑣 .

The first preview revisits the sphere and cylinder from the coordinate-systems chapter. Their familiar scale factors now come from one common construction.

Figure 19.4 Now send the rectangle into space. Nothing about the construction changes — the patch is still spanned by two tangent vectors and its area is still the area of the parallelogram they make — except that two vectors in space need a cross product rather than a determinant, and the cross product is that determinant one dimension up. Watch the patch travel from pole to pole and shrink away to nothing: that is the same

s i n 𝜙

the volume element carried. On the cylinder nothing changes at all, which is what a constant area element looks like.

For a graph $𝑧 = 𝑔 (𝑥, 𝑦)$ , use the parameterization

⃗ 𝑟 (𝑥, 𝑦) = ⟨ 𝑥, 𝑦, 𝑔 (𝑥, 𝑦) ⟩ .

Then

⃗ 𝑟 𝑥 = ⟨ 1, 0, 𝑔 𝑥 ⟩, ⃗ 𝑟 𝑦 = ⟨ 0, 1, 𝑔 𝑦 ⟩,

𝑑 𝑆 = ∥ ⃗ 𝑟 𝑥 \times ⃗ 𝑟 𝑦 ∥ 𝑑 𝑥 𝑑 𝑦 = \sqrt 1 + 𝑔 2 𝑥 + 𝑔 2 𝑦 𝑑 𝑥 𝑑 𝑦 .

Figure 19.5 The case the next chapter computes with, and the one where the scale factor answers an obvious question: how much surface is there above a given patch of ground? Flat ground gives exactly

1

— a patch of floor with a patch of surface directly over it. Every slope beyond that costs extra surface, and

\sqrt 1 + 𝑔 2 𝑥 + 𝑔 2 𝑦

is the price. The wash on the floor is that price at every point, so the bright regions are the steep ones.

This is only the geometric bridge. The next chapter develops parameterized surfaces carefully and uses this area element to integrate scalar fields over them.