Visual walkthrough — Double integrals in polar coordinates — Jacobian r

Read every step in order. Each step tells you WHAT we drew, WHY we drew it, and points you at the PICTURE that carries the idea.

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Step 1 — What is a point in polar language?

WHAT. Instead of "go right by $x$, up by $y$", we say "walk out a distance $r$ from the centre, in a direction that makes angle $\theta$ with the rightward axis."

WHY. We are about to build tiny tiles out of radii and angles, so we first need to know that two numbers — one length, one angle — pin down any point. That is the whole polar idea.

PICTURE. Below, the point $P$ sits at distance $r$ (the yellow ruler) along a ray tilted by angle $\theta$ (the red wedge) from the positive $x$-axis (white).

Why $\cos$ for $x$ and $\sin$ for $y$, and not the other way? Because $\cos \theta$ is defined as the adjacent-over-hypotenuse ratio — the horizontal fraction of the arrow — and $\sin \theta$ the vertical fraction. The horizontal shadow is horizontal, so it wears $\cos$.

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Step 2 — What does "hold $r$ fixed" and "hold $\theta$ fixed" each draw?

WHAT. We freeze one coordinate at a time and trace what the other one sweeps.

WHY. A Cartesian tile is bounded by the grid lines "$x=$ const" and "$y=$ const". To build the polar tile we need its two families of boundary curves. So we ask: what curve is "$r=$ const"? What curve is "$\theta =$ const"?

PICTURE. Fixing $r$ (yellow) traces a circle — every point at the same distance. Fixing $\theta$ (red) traces a ray — every point in the same direction.

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Step 3 — Build one tiny polar tile

WHAT. Pick a starting distance $r$ and starting angle $\theta$. Nudge the distance by a tiny amount $dr$ and the angle by a tiny amount $d\theta$. The four boundary curves — two circles, two rays — fence off one small tile.

WHY. A double integral is a sum of $f(\text{point})\times (\text{tile area})$ over millions of tiny tiles. To convert the integral to polar we must know the area of one polar tile in terms of $dr$ and $d\theta$. So we isolate exactly one.

PICTURE. The shaded tile is bounded by the inner circle (radius $r$), the outer circle (radius $r+dr$), and the two rays at angles $\theta$ and $\theta +d\theta$.

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Step 4 — Measure the two sides: the straight one and the curved one

WHAT. We measure the tile's two kinds of side.

• The radial side (pointing outward) is just the gap between the two circles: length $dr$.
• The arc side (running along a circle) is a slice of a circle's circumference.

WHY. Area of a nearly-rectangular tile is (one side) $\times$ (adjacent side). We have $dr$ already; we need the arc length. This is the one place a tool from outside enters — the arc-length rule — so we state it plainly.

PICTURE. The straight side $dr$ is yellow; the curved bottom side, of length $rd\theta$, is blue. Read the labels right on the tile.

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Step 5 — Multiply the sides to get the tile's area

WHAT. For a tile so small that the two arcs are almost the same length, it is essentially a rectangle with sides $dr$ and $rd\theta$. Its area is their product.

WHY. This is the answer we came for: the area element $dA$ in polar coordinates.

$\underset{\text{tile area}}{\underbrace{dA}}\approx \underset{\text{straight side, length}}{\underbrace{(dr)}}\times \underset{\text{curved side, arc}}{\underbrace{(rd\theta )}}=rdrd\theta .$

Term by term:

• $dA$ — the small piece of area we are integrating over.
• $dr$ — the radial thickness of the tile.
• $rd\theta$ — the arc width of the tile; the $r$ is the arc-length magnifier from Step 4.
• The product $rdrd\theta$ — thickness times width = area.

PICTURE. The tile is flattened into an honest rectangle so you can see the multiplication $\text{area}=(\text{height }dr)\times (\text{width }rd\theta )$.

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Step 6 — But the two arcs are NOT equal. Does the difference matter?

WHAT. We steel-man our own laziness. The outer arc is $(r+dr)d\theta$, longer than the inner arc $rd\theta$. The true tile is a thin curved trapezoid, not a rectangle. Let's compute its exact area and see how much we cheated.

WHY. If the "leftover" were the same size as $rdrd\theta$, our formula would be wrong. We must prove the leftover is negligibly smaller.

The exact tile is the difference of two circular sectors:

$$d{A}_{\text{exact}}=\underset{\text{big sector}}{\underbrace{\frac{1}{2}(r+dr{)}^{2}d\theta }}-\underset{\text{small sector}}{\underbrace{\frac{1}{2}{r}^{2}d\theta }}=\frac{1}{2}d\theta [(r+dr{)}^2-r^2].$$

Expand $(r+dr{)}^2-r^2=2rdr+(dr{)}^2$:

$$d{A}_{\text{exact}}=\underset{\text{the term we kept}}{\underbrace{rdrd\theta }}+\underset{\text{the leftover}}{\underbrace{\frac{1}{2}(dr{)}^{2}d\theta }}.$$

WHY the leftover dies. The kept term has one factor of $dr$; the leftover has two factors $(dr{)}^2$. As tiles shrink, $dr\to 0$, and $(dr{)}^2$ shrinks far faster than $dr$ — like an ant next to an elephant. In the limiting sum it contributes nothing.

PICTURE. The kept rectangle (green) versus the tiny leftover sliver (red) hugging the outer corner. The red piece is visibly a "second-order crumb."

Recall Where each symbol in the exact formula came from

$\frac{1}{2}{R}^{2}d\theta$ is the area of a pie sector of radius $R$ and angle $d\theta$ ::: because the sector is the fraction $\frac{d\theta }{2\pi }$ of the full disk $\pi R^2$, giving $\frac{d\theta }{2\pi }\cdot \pi R^2=\frac{1}{2}{R}^{2}d\theta$. The term $rdrd\theta$ ::: the first-order piece we keep — our area element. The term $\frac{1}{2}(dr{)}^{2}d\theta$ ::: the second-order crumb we discard as $dr\to 0$.

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Step 7 — The algebra check: the Jacobian determinant says the same thing

WHAT. The picture gave $r$. A machine — the Jacobian Determinant — must agree, or one of us is wrong. We compute it.

WHY. The general Change of Variables Theorem says that when you rewrite $(x,y)$ in new variables $(r,\theta )$, the area stretches by $\left|\det J\right|$, where $J$ collects how $x$ and $y$ each respond to a nudge in $r$ and in $\theta$. It is the algebraic twin of "measure the sides and multiply."

$$J=\left|\begin{array}{cc}\frac{\partial x}{\partial r}& \frac{\partial x}{\partial \theta }\\ \frac{\partial y}{\partial r}& \frac{\partial y}{\partial \theta }\end{array}\right|=\left|\begin{array}{cc}\cos \theta & -r\sin \theta \\ \sin \theta & r\cos \theta \end{array}\right|.$$

Reading each entry:

• $\partial x/\partial r=\cos \theta$ — nudge the distance, the $x$-shadow moves by $\cos \theta$.
• $\partial x/\partial \theta =-r\sin \theta$ — nudge the angle, the $x$-shadow swings back by $r\sin \theta$.
• $\partial y/\partial r=\sin \theta$ — nudge the distance, the $y$-shadow moves by $\sin \theta$.
• $\partial y/\partial \theta =r\cos \theta$ — nudge the angle, the $y$-shadow rises by $r\cos \theta$.

Now the determinant (cross-multiply, subtract):

$$\det J=\cos \theta (r\cos \theta )-(-r\sin \theta )(\sin \theta )=r{\cos }^2\theta +r{\sin }^2\theta =r\underset{=1}{\underbrace{({\cos }^2\theta +{\sin }^2\theta )}}=r.$$

Since $r\ge 0$, $\left|\det J\right|=r$. Same $r$. Two independent roads, one answer.

PICTURE. The two column-vectors of $J$ drawn as the actual edges of the tile: the "$\partial /\partial r$" arrow of length $1$ and the "$\partial /\partial \theta$" arrow of length $r$. The area of the parallelogram they span is the determinant — and it equals $r$.

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Step 8 — Edge and degenerate cases (do not skip)

WHAT. We check the formula where things get weird: at the origin, and for a full-circle sweep.

WHY. A rule you cannot survive at its boundaries is a trap. Cover every case.

Case A — at the origin, $r=0$. Here $dA=0\cdot drd\theta =0$. The tile has zero area. Does that make sense? Yes: at the very centre, all the rays meet at a point, so the "arc width" $rd\theta =0$. The tile is a degenerate sliver of no area. The $r$ correctly kills it.

Case B — full angular sweep, $\theta :0\to 2\pi$. Integrating the angle over a full turn must reproduce familiar circle facts. Ring of radii $r$ to $r+dr$: its area should be circumference $\times$ thickness $=2\pi rdr$. Check with our formula:

$${\int }_0^{2\pi }rdrd\theta =rdr{\int }_0^{2\pi }d\theta =2\pi rdr.✓$$

Case C — the whole disk of radius $a$. Summing all rings from $0$ to $a$:

$${\int }_0^{2\pi }{\int }_0^{a}rdrd\theta =2\pi {\left[\frac{r^2}{2}\right]}_0^a=2\pi \cdot \frac{a^2}{2}=\pi a^2.✓$$

The area of a disk falls out — the ultimate sanity check.

PICTURE. Three panels: (A) the shrinking tile vanishing at the centre; (B) one thin ring of area $2\pi rdr$; (C) the rings nested to fill the disk. Watch the tiles grow from centre to rim.

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The one-picture summary

Everything above, compressed: a fan of polar tiles from centre to rim. Same $dr$, same $d\theta$ everywhere — but the tiles visibly fatten outward, their arc width $rd\theta$ growing in lockstep with distance $r$. The label on each tile reads its own area, $rdrd\theta$.

Recall Feynman: the whole walkthrough in plain words

We wanted the size of a tiny tile in a plane described by "distance out" ($r$) and "angle around" ($\theta$). We froze one small tile: it's boxed by two circles and two rays. Its up-down side is just how much the distance changed, $dr$. Its along-the-circle side is an arc, and arcs on a circle are radius times angle, so it's $rd\theta$ — and that's the trick, the side grows because bigger circles have longer arcs for the same angle. Multiply the two sides: area $=dr\times rd\theta =rdrd\theta$. We worried the tile is really a curved trapezoid, but the "extra bit" involves $(dr{)}^2$, which is a crumb that vanishes when tiles get tiny. To be sure, we ran the algebra machine (the Jacobian determinant) and it spat out the same $r$. At the very centre the tiles shrink to nothing ($r=0$ kills them), a full turn rebuilds a ring of area $2\pi rdr$, and stacking rings rebuilds the disk's $\pi a^2$. The single word to remember: arc grows with radius, and that growth is the $r$.

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Flashcards

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Connections

• Double integrals in polar coordinates — Jacobian r — the parent; this page is its picture-book proof.
• Arc Length and Radian Measure — supplies the crucial $rd\theta$ arc.
• Jacobian Determinant — the algebraic twin that also gives $r$.
• Change of Variables Theorem — the general law our tile obeys.
• Area of Regions Bounded by Polar Curves — the $rdrd\theta$ tile in action.