Worked examples — Double integrals in polar coordinates — Jacobian r

This page is the drill floor for the parent topic. The parent proved why $dA=rdrd\theta$. Here we make sure you can survive every kind of problem an exam or the real world can throw — full-circle sweeps, angle-dependent boundaries, integrands that vanish, integrals that run to infinity, and one word problem.

Before any symbol appears, recall the only three facts you need:

Recall The three facts every example below uses

• $x=r\cos \theta$ and $y=r\sin \theta$ (a point's Cartesian coordinates from its distance $r$ and angle $\theta$).
• $x^2+y^2=r^2$ (Pythagoras on the little triangle the point makes with the axes).
• $dA=rdrd\theta$ (the tiny curved box has area = straight side $dr$ × arc side $rd\theta$).

Here $==r==$ means "how far the point is from the origin" and $==\theta ==$ means "the angle swept from the positive $x$-axis, counter-clockwise". Both are pictured in the first figure.

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The scenario matrix

Every polar double-integral problem lives in one of these cells. The examples below are numbered to cover each cell at least once.

Cell	What makes it distinct	Example
A. Full sweep, constant radius	$\theta :0\to 2\pi$, $r:0\to a$ (a disk)	Ex 1
B. Annulus (hole in the middle)	$r$ starts above $0$	Ex 2
C. Single quadrant / sector	$\theta$ in a limited range, sign of $\cos \theta ,\sin \theta$ matters	Ex 3
D. Angle-dependent $r$-limit	boundary like $r=2\cos \theta$	Ex 4
E. Degenerate / zero integrand	integrand vanishes at origin or by symmetry	Ex 5
F. Infinite / limiting region	$r\to \infty$, improper integral	Ex 6
G. Real-world word problem	mass, average, physical units	Ex 7
H. Exam twist (region "between two curves")	outer minus inner in $r$	Ex 8

We march A → H.

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Ex 1 — Cell A: the disk (the sanity anchor)

Forecast: you already know a disk's area is $\pi r^2$. With radius $3$ that should be $9\pi \approx 28.27$. Watch the machinery reproduce it.

1. Set up the region. $r$ runs from the centre out to the rim: $r:0\to 3$. The angle sweeps the whole circle: $\theta :0\to 2\pi$. Why this step? "Full sweep, constant radius" is exactly Cell A — both limits are constants.
2. Write the integral of $1$ with the Jacobian. $\text{Area}={\int }_0^{2\pi }{\int }_0^{3}1\cdot rdrd\theta .$ Why the $r$? Because $dA=rdrd\theta$. Drop it and you'd be adding up lengths, not areas.
3. Inner integral. ${\int }_0^{3}rdr={\left[\frac{r^2}{2}\right]}_0^3=\frac{9}{2}.$ Why inner first? The limits are constant, so we can do the $r$-integral independently.
4. Outer integral. ${\int }_0^{2\pi }\frac{9}{2}d\theta =\frac{9}{2}\cdot 2\pi =9\pi .$

Verify: $9\pi \approx 28.27$, matching $\pi \cdot 3^2$. Units: $rdr$ is (length)(length) = area — correct.

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Ex 2 — Cell B: the annulus (a hole in the middle)

Forecast: it's a big disk (radius 5) minus a small disk (radius 2): $\pi (25-4)=21\pi \approx 65.97$.

1. Region. $r:2\to 5$ (note: starts at $2$, not $0$ — that's the hole), $\theta :0\to 2\pi$. Why this step? Cell B differs from A only in the lower $r$-limit; the origin is excluded.
2. Integral. ${\int }_0^{2\pi }{\int }_2^{5}rdrd\theta .$
3. Inner. ${\int }_2^{5}rdr={\left[\frac{r^2}{2}\right]}_2^5=\frac{25-4}{2}=\frac{21}{2}.$ Why this step? Subtracting the two $r^2/2$ values is the "big disk minus small disk" idea, done automatically by the limits.
4. Outer. ${\int }_0^{2\pi }\frac{21}{2}d\theta =21\pi .$

Verify: $21\pi \approx 65.97=\pi (5^2-2^2)$. ✓ (See Area of Regions Bounded by Polar Curves for the general subtraction idea.)

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Ex 3 — Cell C: a single quadrant, signs matter

Forecast: $x$ is positive everywhere in the first quadrant, so the answer must be positive. Rough size: average $x$ near $1$, area $=\frac{1}{4}\pi (2^2)=\pi$, so guess something around $2$–$3$.

1. Convert the integrand. $x=r\cos \theta$. Why this step? Mistake 2 from the parent — you must convert $x$, not leave it as-is.
2. Assemble with the Jacobian. $\iint xdA={\int }_0^{\pi /2}{\int }_0^2(r\cos \theta )rdrd\theta ={\int }_0^{\pi /2}{\int }_0^2{r}^2\cos \theta drd\theta .$ Why $r^2$? One $r$ from $x=r\cos \theta$, one $r$ from the Jacobian.
3. Separate. The integrand factors into an $r$-part and a $\theta$-part: $\left({\int }_0^2{r}^{2}dr\right)\left({\int }_0^{\pi /2}\cos \theta d\theta \right).$ Why allowed? Because the limits are constant and the integrand is a product of a function of $r$ times a function of $\theta$.
4. Evaluate each. ${\int }_0^2{r}^{2}dr=\frac{8}{3}$, and ${\int }_0^{\pi /2}\cos \theta d\theta =[\sin \theta {]}_0^{\pi /2}=1$. Why $\cos \theta >0$ here? In the first quadrant $\theta \in [0,\pi /2]$, so $\cos \theta \ge 0$ — the sign of $x$ never flips.
5. Multiply. $\frac{8}{3}\cdot 1=\frac{8}{3}$.

Verify: $\frac{8}{3}\approx 2.67>0$, positive as forecast. If we had integrated over the full disk instead, symmetry ($x$ positive on the right, negative on the left) would cancel to $0$ — see Ex 5.

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Ex 4 — Cell D: angle-dependent radius $r=2\cos \theta$

The boundary $r=2\cos \theta$ is a full circle of radius $1$ centred at $(1,0)$ — it passes through the origin. Here the inner $r$-limit is a function of $\theta$, the trickiest exam case.

Forecast: it's a circle of radius $1$, so its area must be $\pi (1{)}^2=\pi \approx 3.14$.

1. Find the $\theta$-range. For $r=2\cos \theta \ge 0$ we need $\cos \theta \ge 0$, i.e. $\theta :-\frac{\pi }{2}\to \frac{\pi }{2}$. Why this step? $r$ is a distance and cannot be negative; when $\cos \theta <0$ the formula would demand a negative $r$, so those angles are excluded (they'd re-trace the same circle).
2. Set $r$-limits. For each fixed $\theta$, a ray from the origin enters at $r=0$ and leaves the circle at $r=2\cos \theta$. So $r:0\to 2\cos \theta$. Why this step? Cell D: the outer boundary bends with $\theta$, so its value is $2\cos \theta$, not a constant. Look at the red radius in the figure lengthening then shrinking as $\theta$ sweeps.
3. Integral. ${\int }_{-\pi /2}^{\pi /2}{\int }_0^{2\cos \theta }rdrd\theta .$
4. Inner. ${\int }_0^{2\cos \theta }rdr={\left[\frac{r^2}{2}\right]}_0^{2\cos \theta }=\frac{(2\cos \theta {)}^2}{2}=2{\cos }^2\theta .$ Why does the upper limit appear squared? Because $\int rdr=r^2/2$ and we plug the $\theta$-dependent top in.
5. Outer. Use ${\cos }^2\theta =\frac{1+\cos 2\theta }{2}$: ${\int }_{-\pi /2}^{\pi /2}2{\cos }^2\theta d\theta ={\int }_{-\pi /2}^{\pi /2}(1+\cos 2\theta )d\theta =[\theta +\frac{\sin 2\theta }{2}{]}_{-\pi /2}^{\pi /2}=\pi .$ Why the identity? ${\cos }^2$ has no elementary antiderivative by inspection; the double-angle identity turns it into something you can integrate.

Verify: $\pi \approx 3.14=\pi (1{)}^2$, the area of a radius-$1$ circle. ✓

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Ex 5 — Cell E: degenerate / vanishing integrand (symmetry gives $0$)

Forecast: the disk is symmetric about the $y$-axis. For every point with $x>0$ there's a mirror point with $x<0$. They cancel. Predict exactly $0$ — no computation needed.

1. Convert & assemble. ${\int }_0^{2\pi }{\int }_0^4(r\cos \theta )rdrd\theta =\left({\int }_0^4{r}^{2}dr\right)\left({\int }_0^{2\pi }\cos \theta d\theta \right).$ Why factor out? Same product structure as Ex 3.
2. The $\theta$ integral kills it. ${\int }_0^{2\pi }\cos \theta d\theta =[\sin \theta {]}_0^{2\pi }=\sin 2\pi -\sin 0=0.$ Why zero? Over a full turn $\cos \theta$ spends equal time positive and negative — the signed area under one full cosine cycle is zero. This is the analytic face of the mirror-symmetry.
3. Whole product = 0, regardless of the $r$-integral's value ($\frac{64}{3}$).

Verify: anything $\times 0=0$. Matches the symmetry forecast. This is the degenerate case: the Jacobian $r$ is present and correct, but the integrand's sign structure forces the answer to vanish.

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Ex 6 — Cell F: infinite region (the Gaussian, generalised)

Forecast: the parent showed ${\left({\int }_{-\infty }^{\infty }{e}^{-x^2}dx\right)}^2=\pi$, and this double integral is that square. So predict $\pi \approx 3.14$.

1. Polarise. $x^2+y^2=r^2$, $dA=rdrd\theta$, region $r:0\to \infty$, $\theta :0\to 2\pi$: ${\int }_0^{2\pi }{\int }_0^{\infty }{e}^{-r^2}rdrd\theta .$ Why polar? The integrand depends only on distance from origin — perfect circular symmetry (Cell F meets Cell A but with an infinite radius).
2. Substitute $u=r^2$, $du=2rdr$. ${\int }_0^{\infty }{e}^{-r^2}rdr=\frac{1}{2}{\int }_0^{\infty }{e}^{-u}du=\frac{1}{2}[-e^{-u}{]}_0^{\infty }=\frac{1}{2}(0-(-1))=\frac{1}{2}.$ Why does the improper integral converge? As $r\to \infty$, $e^{-r^2}\to 0$ faster than $r$ grows, so the tail contributes nothing — the "infinite" region has finite integral. The Jacobian $r$ is exactly the piece the substitution needs.
3. Angle. ${\int }_0^{2\pi }\frac{1}{2}d\theta =\pi .$

Verify: $\pi \approx 3.14$, matching the forecast and Gaussian Integral. ✓

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Ex 7 — Cell G: real-world word problem (mass of a disk)

Forecast: density grows outward, so most mass sits near the rim. If density were the constant average, mass would be $\bar{\sigma }\cdot \pi a^2$; expect a value a few times $k{a}^3$. Guess "tens of kg".

1. Mass = integral of density over area. $M={\iint }_R\sigma dA={\int }_0^{2\pi }{\int }_0^2(kr)rdrd\theta .$ Why this step? Mass is density times area, summed over the plate. Density $\sigma =kr$ already lives in polar form.
2. Two $r$'s again. $\sigma =kr$ contributes one $r$, the Jacobian another, giving $k{r}^2$: $M=k{\int }_0^{2\pi }{\int }_0^2{r}^{2}drd\theta .$
3. Inner. ${\int }_0^2{r}^{2}dr=\frac{8}{3}.$
4. Outer & constant. $M=k\cdot \frac{8}{3}\cdot 2\pi =\frac{16\pi k}{3}=\frac{16\pi \cdot 3}{3}=16\pi \approx 50.27\text{kg}.$

Verify — units: $[k]={\text{kg/m}}^3$, $\int r^{2}dr$ has units m³, times dimensionless $\int d\theta$ ⇒ kg. ✓ Numeric: $16\pi \approx 50.27\text{kg}$, "tens of kg" as forecast.

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Ex 8 — Cell H: exam twist, region between two curves

Forecast: $y=r\sin \theta \ge 0$ throughout the first quadrant, so the answer is positive. The $r$-integral runs $1\to 3$ (Cell B/H "outer minus inner"), the $\theta$-integral picks up $\sin \theta$.

1. Convert integrand & assemble. $y=r\sin \theta$, Jacobian $r$: ${\int }_0^{\pi /2}{\int }_1^3(r\sin \theta )rdrd\theta =\left({\int }_1^3{r}^{2}dr\right)\left({\int }_0^{\pi /2}\sin \theta d\theta \right).$ Why $r^2$? One $r$ from $y$, one from the Jacobian.
2. Radial part (outer minus inner). ${\int }_1^3{r}^{2}dr={\left[\frac{r^3}{3}\right]}_1^3=\frac{27-1}{3}=\frac{26}{3}.$ Why this step? The subtraction is the "between two curves" mechanic — big radius cubed minus small radius cubed.
3. Angular part. ${\int }_0^{\pi /2}\sin \theta d\theta =[-\cos \theta {]}_0^{\pi /2}=-\cos \frac{\pi }{2}+\cos 0=0+1=1.$ Why positive? $\sin \theta \ge 0$ across the first quadrant, so $y\ge 0$ everywhere — no cancellation, unlike Ex 5.
4. Multiply. $\frac{26}{3}\cdot 1=\frac{26}{3}.$

Verify: $\frac{26}{3}\approx 8.67>0$, positive as forecast. ✓

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Case-coverage self-check

Recall Did we hit every cell?

A (disk) ::: Ex 1 B (annulus) ::: Ex 2 C (sector, signs) ::: Ex 3 D (angle-dependent $r$-limit) ::: Ex 4 E (degenerate / zero by symmetry) ::: Ex 5 F (infinite region) ::: Ex 6 G (word problem, units) ::: Ex 7 H (between two curves, exam twist) ::: Ex 8

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Connections

• Double Integrals in Polar Coordinates — Jacobian r — the parent this drill supports.
• Change of Variables Theorem — why converting everything (Ex 3, 5, 8) is mandatory.
• Jacobian Determinant — the source of the $r$ every example carries.
• Arc Length and Radian Measure — why the arc side is $rd\theta$.
• Gaussian Integral — Ex 6 in full.
• Area of Regions Bounded by Polar Curves — Ex 2, 4, 8 are area-style setups.
• Triple Integrals in Cylindrical and Spherical Coordinates — the next step up in dimensions.