Round problems want round coordinates. A radar dish sweeps out a circle, a sprinkler waters
a disc of lawn, a pizza is cut into wedges, an archery target is a stack of rings — none of
these shapes cares about north-south, east-west axes. Yet if you try to compute something
over a disc or a ring using plain x, y — say, the total rainfall
collected by a circular pond — you're stuck with square-root limits like
y = \pm\sqrt{a^2 - x^2} before you've even looked at the
integrand.
A
double integral
over a disk or a wedge is a nightmare in x, y. Circular regions
beg for polar coordinates, where a disk is just a rectangle:
x = r\cos\theta, \qquad y = r\sin\theta, \qquad 0 \le r \le a, \quad 0 \le \theta \le 2\pi.
But you cannot simply swap dA = dx\, dy for
dr\, d\theta. There is an extra factor of
r — and seeing exactly where it comes from is the whole point.
The area element: where the extra r comes from
In Cartesian coordinates a tiny tile is a rectangle of sides
dx and dy, area
dA = dx\, dy. The polar grid is different: it is made of circular
arcs and radial spokes, so a tiny "polar rectangle" is a curved patch. Let us measure it.
Step 1 — name the two sides of the polar patch. Hold a corner at radius
r and angle \theta, then nudge each
coordinate. Increasing r by dr moves
you outward along a spoke — a straight radial side of length
\text{radial side} = dr.
Step 2 — measure the arc side. Increasing
\theta by d\theta sweeps along a
circular arc of radius r. Arc length is radius times angle:
\text{arc side} = r\, d\theta.
Step 3 — multiply the two sides. The patch is, to first order, a tiny
rectangle with perpendicular sides dr and
r\, d\theta, so its area is their product:
dA = (dr)\,(r\, d\theta) = r\, dr\, d\theta.
That stray r is the Jacobian of the polar map —
it accounts for the fact that patches far from the origin (large
r) are fatter than patches near it, because the arc side
r\, d\theta grows with r. The
change-of-variables theorem
will derive this same r from a determinant.
The polar conversion in full
Replace x, y by r\cos\theta, r\sin\theta
in the integrand, and replace dA by
r\, dr\, d\theta:
\iint_R f(x, y)\, dA = \iint_S f(r\cos\theta,\, r\sin\theta)\; r\, dr\, d\theta.
Worked example — \displaystyle \iint_D (x^2+y^2)\, dA over the disc of radius a
In Cartesian coordinates this integral is already unpleasant before you've integrated
anything: the limits alone are -a \le x \le a,
-\sqrt{a^2-x^2} \le y \le \sqrt{a^2-x^2}. In polar it takes three
lines.
Step 1 — convert. Since x^2+y^2 = r^2, the
integrand r^2 picks up one more factor of
r from the area element dA = r\, dr\, d\theta:
\iint_D (x^2+y^2)\, dA = \int_0^{2\pi} \!\int_0^a r^2 \cdot r\, dr\, d\theta = \int_0^{2\pi}\!\int_0^a r^3\, dr\, d\theta.
Step 2 — the inner integral is a plain power rule:
\int_0^a r^3\, dr = \left[\frac{r^4}{4}\right]_0^a = \frac{a^4}{4}.
Step 3 — the outer integral just multiplies by the full turn:
\int_0^{2\pi} \frac{a^4}{4}\, d\theta = \frac{\pi a^4}{2}.
Three short lines, no square roots anywhere — compare that with the Cartesian version,
which needs a trigonometric substitution just to get off the ground. The integral is not
idle bookkeeping, either: for a flat spinning plate of unit density,
\iint_D (x^2+y^2)\, dA is exactly its
polar moment of inertia — the number that says how hard the plate is to
spin up or slow down.
Worked example — volume under a paraboloid bowl
A satellite dish, a parabolic mirror and a rain-catching bowl are all slices of the surface
z = h - \dfrac{h}{a^2}(x^2+y^2) — a paraboloid that peaks at
height h in the centre and tapers to z = 0
at the rim x^2+y^2 = a^2. How much rainwater does the bowl hold?
Step 1 — set up in polar. The base is the disc r \le a,
and the height function becomes a simple function of r alone:
V = \iint_D \left(h - \frac{h}{a^2} r^2\right) dA = \int_0^{2\pi}\!\int_0^a \left(hr - \frac{h}{a^2} r^3\right) dr\, d\theta.
Step 2 — the inner integral, term by term with the power rule:
\int_0^a \left(hr - \frac{h}{a^2} r^3\right) dr = \left[\frac{hr^2}{2} - \frac{hr^4}{4a^2}\right]_0^a = \frac{ha^2}{2} - \frac{ha^2}{4} = \frac{ha^2}{4}.
Step 3 — the outer integral multiplies by 2\pi:
V = \int_0^{2\pi} \frac{ha^2}{4}\, d\theta = \frac{\pi a^2 h}{2}.
That is exactly half the volume of the cylinder that just encloses the
bowl (\pi a^2 h) — a paraboloid bowl always holds half of
whatever can it is inscribed in, however wide or tall you make it.
Worked example — \displaystyle \iint_R e^{-(x^2+y^2)}\, dA over the disk of radius a
Step 1 — convert the integrand. Since
x^2 + y^2 = r^2, the awkward exponential becomes a function of
r alone:
e^{-(x^2+y^2)} = e^{-r^2}.
Step 2 — set up the polar integral. The disk is the polar rectangle
0 \le r \le a, 0 \le \theta \le 2\pi,
and we must include the r\, dr\, d\theta:
\iint_R e^{-(x^2+y^2)}\, dA = \int_0^{2\pi} \!\int_0^a e^{-r^2}\; r\, dr\, d\theta.
Step 3 — the inner integral, where the extra r earns its
keep. Substitute u = r^2, so
du = 2r\, dr — the leftover r is
exactly what the substitution needs:
\int_0^a e^{-r^2}\, r\, dr = \frac{1}{2}\int_0^{a^2} e^{-u}\, du = \frac{1}{2}\left(1 - e^{-a^2}\right).
Step 4 — the outer integral. The inner result is constant in
\theta, so the \theta integral just
multiplies by 2\pi:
\int_0^{2\pi} \frac{1}{2}\left(1 - e^{-a^2}\right) d\theta = \pi\left(1 - e^{-a^2}\right).
Step 5 — let the disk fill the plane. Send
a \to \infty; the term e^{-a^2} \to 0:
\iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\, dA = \pi.
A clean \pi from an integral that has no elementary
antiderivative in Cartesian form. This single fact unlocks the famous Gaussian integral.
Let R be a region described in polar coordinates by
\alpha \le \theta \le \beta and
r_1(\theta) \le r \le r_2(\theta), and let
f be continuous on R. Then:
-
The substitution is
x = r\cos\theta, y = r\sin\theta,
with x^2 + y^2 = r^2.
-
The area element carries an extra
r: dA = r\, dr\, d\theta
(the Jacobian of the polar map).
-
The integral becomes
\iint_R f\, dA = \int_\alpha^\beta \!\int_{r_1(\theta)}^{r_2(\theta)} f(r\cos\theta, r\sin\theta)\; r\, dr\, d\theta.
-
A full disk of radius a is the polar
rectangle 0 \le r \le a,
0 \le \theta \le 2\pi.
The bell-curve integral \int_{-\infty}^{\infty} e^{-x^2}\, dx
resists every one-variable trick — e^{-x^2} has no elementary
antiderivative. The escape, due to Gauss's
trick, is to go up a dimension. Call the unknown value I and
square it, writing the two copies in different dummy variables:
I^2 = \left(\int_{-\infty}^{\infty} e^{-x^2} dx\right)\!\left(\int_{-\infty}^{\infty} e^{-y^2} dy\right) = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\, dA.
That right-hand side is exactly the polar integral we just evaluated to
\pi. Hence
I^2 = \pi \quad\Longrightarrow\quad I = \int_{-\infty}^{\infty} e^{-x^2}\, dx = \sqrt{\pi}.
A one-dimensional integral that defeats single-variable calculus falls instantly once you
let it breathe in two dimensions and switch to polar. The whole of statistics —
normalising the normal distribution — rests on this \sqrt{\pi}.
Three slips account for nearly every mistake with polar double integrals:
-
Forgetting the extra r. The single most
common error: writing dA = dr\, d\theta and dropping the
Jacobian factor entirely. Skip it, and every answer is wrong by a factor that gets
worse the further the region sits from the origin.
-
x^2+y^2 = r^2, not r.
It's easy to remember that \sqrt{x^2+y^2} becomes
r, and then absent-mindedly do the same thing to
x^2+y^2 itself, which is really r^2.
Check which one you actually have before you substitute.
-
The \theta limits describe a wedge, not a
distance. 0 \le \theta \le \pi/2 is a quarter-turn
slice of the plane — it says nothing about how far out the region reaches. That job
belongs entirely to the r limits.
Air-traffic radar, sonar and old rotating-beam lighthouses all sweep out circles, and
every one of them is drawn on a display ruled with concentric rings and radial spokes —
a polar grid, never a Cartesian one. That isn't decoration: the sweeping beam naturally
reports "something at distance r, bearing
\theta", which is exactly the language polar coordinates
speak. A pizza cut into equal wedges is the same idea in the kitchen — each slice is a
tiny polar rectangle, wider at the crust (large r) than at the
tip (small r). That extra width at the crust is the factor
r in dA = r\, dr\, d\theta, made
edible.