Double Integrals in Polar Coordinates

Convert a double integral over a disk or annulus to polar coordinates and evaluate it.

The core idea. Some regions are miserable in (x, y) and easy in polar: disks, rings, wedges, anything built from circles.

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The conversion rests on three substitutions and one factor everyone forgets:

x = r\cos\theta, \qquad y = r\sin\theta, \qquad x^2 + y^2 = r^2

Polar substitutions

$r$ distance from the origin; the radial coordinate.
$\theta$ angle from the positive x-axis; the angular coordinate.
$x = r\cos\theta,\; y = r\sin\theta$ Cartesian-to-polar conversion.
$x^2 + y^2 = r^2$ the identity that collapses circular integrands.

And the area element that goes with it:

dA = r\, dr\, d\theta

Area element in polar coordinates

$dA = r\,dr\,d\theta$ the area element; the r is the stretch factor, not optional.

That r is not optional. dA is not dr dθ — it's r dr dθ. Forgetting it is the single most common error in this section. Geometrically, a polar "rectangle" is a curved patch that's wider the farther out you go, and the r accounts for that stretch.

A polar region is described by an angle sweep and a radial reach:

D = \{(r,\theta) \;:\; \alpha \le \theta \le \beta,\; h_1(\theta) \le r \le h_2(\theta)\}

Polar region

$[\alpha,\beta]$ the angular sweep of the region.
$[h_1(\theta), h_2(\theta)]$ inner and outer radial bounds.

The integral over that region, converted to polar:

\iint_D f(x,y)\, dA = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,\, r\sin\theta)\; r\, dr\, d\theta

Double integral in polar coordinates

The integrand gets converted too — every x and y becomes its polar form — and the r rides along at the end.

Area in polar:

A(D) = \iint_D 1 \, dA = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} r\, dr\, d\theta

Area in polar coordinates

The tell — when to reach for polar: the region is a disk, ring, or wedge, or the integrand contains x² + y². That combination collapses to r² and usually makes an ugly integral clean. The classic is e^(x²+y²), impossible in Cartesian, trivial as e^(r²).

Worked example

\iint_D (x^2 + y^2)\, dA, \qquad D: x^2 + y^2 \le 4

$x^2 + y^2 \le 4 \;\Longrightarrow\; 0 \le \theta \le 2\pi,\; 0 \le r \le 2$
Convert the region. x² + y² ≤ 4 is a disk of radius 2 centered at the origin. So θ sweeps 0 to 2π, and r runs 0 to 2.
$x^2 + y^2 = r^2$
Convert the integrand.
$\int_0^{2\pi} \int_0^2 r^2 \cdot r\, dr\, d\theta = \int_0^{2\pi} \int_0^2 r^3\, dr\, d\theta$
Set up — and attach the r. That r² · r = r³ is the whole point — the forgotten r turns r² into r³.
$\int_0^2 r^3\, dr = \frac{r^4}{4}\bigg|_0^2 = \frac{16}{4} = 4$
Inner integral.
$\int_0^{2\pi} 4\, d\theta = 4\theta\big|_0^{2\pi} = 8\pi$
Outer integral. Try that in Cartesian and you'd be fighting square-root bounds over a circular region — polar makes it three clean lines.

Answer:

8\pi

How to solve one of these

1.
Recognize it's a polar problem. Circular region, or $x^2+y^2$ in the integrand. If neither, polar probably won't help.
2.
Convert the region to $r$ and $\theta$. Find the angle sweep $[\alpha,\beta]$ and the radial bounds. A full disk is $\theta \in [0, 2\pi]$ ; a wedge is narrower.
3.
Convert the integrand. Replace every $x$ with $r\cos\theta$ , every $y$ with $r\sin\theta$ , and $x^2+y^2$ with $r^2$ .
4.
Attach the $r$. Write $r\,dr\,d\theta$ , never $dr\,d\theta$ . Multiply it into the integrand before integrating.
5.
Evaluate inner, then outer, same as always. Substitute the bounds.
6.
Sanity-check the geometry. If the answer is an area, does it match $\pi r^2$ roughly? A wildly off number usually means a missing $r$ or a bad bound.

\iint_D f(x,y)\, dA = \int_\alpha^\beta \int_{h_1(\theta)}^{h_2(\theta)} f(r\cos\theta,\, r\sin\theta)\; r\, dr\, d\theta

Convert the region to r and θ bounds, replace every x and y with its polar form, and multiply by r dr dθ — the r rides along at the end and is never optional.

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