5.1 Polar Coordinates and the Jacobian

A double integral is a refined sum over cells. In rectangular coordinates the cells are congruent little squares, each of area $Δ x Δ y$ , and nobody thinks twice. But regions with a center — disks, rings, anything round — beg to be chopped along rings and rays instead, and then something changes: the cells are not congruent. Every cell in one ring matches its neighbors, but the rings themselves widen as they move outward. The area element must record that widening, and the figure lets you measure it before any formula claims it.

A flat disk meshed by a coarse polar grid of 12 rings and 12 wedges, whose cells visibly widen toward the rim, beneath a dome-shaped paraboloid carrying the same coarse polar mesh. A hidden Cartesian wireframe of the same paraboloid, with equal square cells, can be toggled on for contrast. — Two coarse wireframe objects: a flat disk at $z equals 0$ meshed by a polar grid of rings and wedges (resolution 12, so the cells are countable), and the dome $z = 4 - r^{2}$ carrying the same coarse polar mesh. A third object — a Cartesian-mesh copy of the dome, with identical square cells — waits hidden in the object list for the contrast step.

Explore

Orbit until you look straight down the $z$ -axis, so the flat grid reads like a dartboard. Count the cells in the innermost ring and in the outermost. Same count — same size?
Each cell is approximately a rectangle, one side running radially and one side running along the arc. Moving from center to rim, which side length stays fixed and which one grows?
Write, before touching anything, a formula for the approximate area of a cell at radius $r$ with radial step $Δ r$ and angular step $Δ θ$ .
Now interrogate the mesh: retype the flat grid's resolution field from 12 to 24, then 64, then 128. At every fineness, cells within one ring stay congruent, and cell area keeps growing in proportion to $r$ — compare a cell near $r equals 1$ with one near $r equals 2$ at each level and watch the ratio stay locked near $2$ . The factor of $r$ is a property of polar coordinates, not of any particular mesh.
Set the resolution back to 12. Reveal the hidden Cartesian-mesh dome with its eye toggle (hide the two polar objects for a clean look) and orbit straight down: every Cartesian cell is an identical square, so $d A = d x d y$ needs no correction. Explain why summing $f \cdot Δ r Δ θ$ over the polar cells — with no $r$ — would overweight the cells near the center.
Restore the polar objects. The colored dome is the integrand $f = 4 - r^{2}$ riding the same grid: each cell contributes height times $r Δ r Δ θ$ of volume. Set up the volume both ways — the example below finishes the story.

One cell, measured exactly

The cell between radii $r$ and $r + Δ r$ , spanning an angle $Δ θ$ , is a slice of a ring. Its exact area is the difference of two circular sectors:

$\frac{1}{2} (r + Δ r)^{2} Δ θ - \frac{1}{2} r^{2} Δ θ = (r + \frac{Δ r}{2}) Δ r Δ θ,$

which is $r Δ r Δ θ$ up to a vanishing correction — arc length $r Δ θ$ times radial step $Δ r$ , exactly what steps 2 and 3 measured. Refining the mesh turns the sum $∑ f times r Δ r Δ θ$ into the integral $∬ f r d r d θ$ . The extra $r$ is the local area-scaling of the map $(r, θ) \mapsto (r \cos θ, r \sin θ)$ — its Jacobian, a name the next section grows into a general tool.

r

The Jacobian determinant of the polar map ✓ Computed · mojocas 0.1.0 ✓ Agrees with the text

The area-scaling factor of $(r, θ) \mapsto (r \cos θ, r \sin θ)$ , computed from the map itself rather than read off the ring-slice picture. It is the same $r$ the mesh above made you count, and the same $r$ that turns $d A$ into $r d r d θ$ . ( $t$ stands for $θ$ .)

Example 1 The volume under the dome, both ways

Find the volume of the solid below $z = 4 - x^{2} - y^{2}$ and above the disk $x squared plus y squared less than or equal to 4$ .

In rectangular coordinates the volume is

$V = \int_{- 2}^{2} \int_{- \sqrt{4 - x^{2}}}^{\sqrt{4 - x^{2}}} (4 - x^{2} - y^{2}) d y d x,$

with square roots in the limits and worse ahead — legal, but miserable. In polar coordinates the region is $0 less than or equal to r less than or equal to 2$ , $0 less than or equal to θ less than or equal to 2 π$ , the integrand is $4 minus r squared$ , and the area element supplies the $r$ you counted on the mesh:

$V = \int_{0}^{2 π} \int_{0}^{2} (4 - r^{2}) r d r d θ = \int_{0}^{2 π} {[2 r^{2} - \frac{r^{4}}{4}]}_{0}^{2} d θ = \int_{0}^{2 π} (8 - 4) d θ = 8 π .$

Note where the $r$ did its work: it turned $open parenthesis 4 minus r squared close parenthesis d r$ into $open parenthesis 4 r minus r cubed close parenthesis d r$ , an antiderivative you can finish in your head — and it weighted each ring by its true share of area, which is what makes the answer correct and not merely easy.

An original work of XYZ Homework, built around interactive XYZ 3D figures. Aligned to OpenStax Calculus Volume 3 (Strang & Herman), © OpenStax (Rice University), licensed CC BY-NC-SA 4.0; no OpenStax content is reproduced, and this work is not affiliated with or endorsed by OpenStax or Rice University. License: CC-BY-NC-SA-4.0.