# Double Integrals: Iterated Integrals, Regions and Polar Form

> Evaluate double integrals step by step: iterated integrals, Fubini's theorem, curved regions, switching the order and polar coordinates.

Source: https://practicecalculus.com/blog/double-integrals
Updated: 2026-10-20

## Key takeaways

- A double integral $\iint_R f(x, y)\,dA$ adds up $f$ over a region, and for $f \ge 0$ it is the volume under the surface $z = f(x, y)$.
- You evaluate it as an iterated integral: integrate the inner variable while holding the outer one constant, then integrate the result.
- For a non-rectangular region, the inner limits are functions of the outer variable, and the outer limits are always constants.
- In polar coordinates $dA = r\,dr\,d\theta$. Never forget the extra factor of $r$.

A single integral adds up a function along an interval. A double integral adds one up over a whole region in the plane, which lets you find volumes, masses and averages over 2D shapes. The surprise is that you never need a new technique to evaluate one: you just do two ordinary integrals in a row. The hard part is getting the limits right.

This guide walks through iterated integrals and Fubini's theorem, integrals over curved regions, switching the order of integration, and polar coordinates, with every step shown.

## What is a double integral?

Chop a region $R$ in the $xy$-plane into tiny rectangles of area $\Delta A$. Over each one, build a thin column of height $f(x, y)$. The **double integral** is the limit of the total volume of those columns as the rectangles shrink:

$$\iint_R f(x, y)\,dA = \lim_{\Delta A \to 0} \sum f(x_i, y_i)\,\Delta A$$

It is the 2D version of a Riemann sum. When $f \ge 0$ it is the **volume** under $z = f(x, y)$ and above $R$. When $f = 1$ it is simply the **area** of $R$.

## Iterated integrals and Fubini's theorem

You almost never use the limit definition. Instead you evaluate an **iterated integral**, one variable at a time. **Fubini's theorem** says that for a continuous $f$ on the rectangle $R = [a, b] \times [c, d]$,

$$\iint_R f(x, y)\,dA = \int_a^b \int_c^d f(x, y)\,dy\,dx = \int_c^d \int_a^b f(x, y)\,dx\,dy$$

Read the iterated integral from the inside out. In $\int_a^b \int_c^d f\,dy\,dx$, integrate with respect to $y$ first, treating $x$ as a constant (just like a [partial derivative](/blog/partial-derivatives) in reverse). The result is a function of $x$ alone. Then integrate that from $a$ to $b$.

### Worked example 1: a rectangle, both orders

Evaluate $\iint_R (x + 2y)\,dA$ where $R = [0, 2] \times [0, 3]$, meaning $0 \le x \le 2$ and $0 \le y \le 3$.

**Step 1. Inner integral in $y$, holding $x$ constant.**

$$\int_0^3 (x + 2y)\,dy = \Big[xy + y^2\Big]_{y=0}^{y=3} = 3x + 9$$

**Step 2. Outer integral in $x$.**

$$\int_0^2 (3x + 9)\,dx = \Big[\tfrac{3}{2}x^2 + 9x\Big]_0^2 = 6 + 18 = 24$$

**Step 3. Check with the other order.** Inner in $x$: $\int_0^2 (x + 2y)\,dx = 2 + 4y$. Outer in $y$: $\int_0^3 (2 + 4y)\,dy = 6 + 18 = 24$. Same answer, as Fubini promises.

> **Tip:** if the integrand factors as $g(x)\,h(y)$ and the region is a rectangle, the double integral splits into a product: $\left(\int_a^b g(x)\,dx\right)\left(\int_c^d h(y)\,dy\right)$.

## Double integrals over non-rectangular regions

Most regions are bounded by curves. There are two standard ways to describe them.

**Type I (vertical strips):** $a \le x \le b$ and $g_1(x) \le y \le g_2(x)$. Integrate $y$ first:

$$\iint_R f\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y)\,dy\,dx$$

**Type II (horizontal strips):** $c \le y \le d$ and $h_1(y) \le x \le h_2(y)$. Integrate $x$ first:

$$\iint_R f\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y)\,dx\,dy$$

The golden rule: **the outer limits are constants, and the inner limits may depend only on the outer variable.**

How to set up the limits:

1. Sketch the region and label every boundary curve.
2. Choose the order. For $dy\,dx$, draw a vertical line through the region: it enters at the bottom curve $g_1(x)$ and leaves at the top curve $g_2(x)$.
3. The outer limits are the smallest and largest $x$ in the whole region.

### Worked example 2: a region under a curve

Evaluate $\iint_R xy\,dA$, where $R$ is bounded by $y = \sqrt{x}$, $y = 0$ and $x = 4$.

```graph
title: 'The region bounded by y = √x, the x-axis and the line x = 4'
caption: 'As a Type I region: $0 \le x \le 4$ and $0 \le y \le \sqrt{x}$. As a Type II region: $0 \le y \le 2$ and $y^2 \le x \le 4$.'
x: [-0.5, 5]
y: [-0.5, 2.6]
functions:
    - expr: sqrt(x)
      label: y = √x
shade: {expr: sqrt(x), from: 0, to: 4}
vertical:
    - {x: 4, label: 'x = 4'}
points:
    - [4, 2, '(4, 2)']
```

**Step 1. Describe the region.** A vertical line at $x$ enters at $y = 0$ and leaves at $y = \sqrt{x}$, for $x$ from $0$ to $4$.

**Step 2. Inner integral in $y$.**

$$\int_0^{\sqrt{x}} xy\,dy = x\left[\frac{y^2}{2}\right]_0^{\sqrt{x}} = x \cdot \frac{x}{2} = \frac{x^2}{2}$$

**Step 3. Outer integral in $x$.**

$$\int_0^4 \frac{x^2}{2}\,dx = \left[\frac{x^3}{6}\right]_0^4 = \frac{64}{6} = \frac{32}{3}$$

## How to switch the order of integration

Switching the order means describing the **same region** with horizontal strips instead of vertical ones (or vice versa). You cannot just swap the limits. You have to redraw.

**Step 1. Recover the region** from the current limits and sketch it.

**Step 2. Find the constant range of the new outer variable.**

**Step 3. For a typical value of it, find where the new inner variable enters and leaves.** Solve each boundary curve for the inner variable.

Redo example 2 in the order $dx\,dy$. The region runs from $y = 0$ to $y = 2$ (since $\sqrt{4} = 2$). A horizontal line at height $y$ enters at the curve, where $x = y^2$, and leaves at $x = 4$:

$$\int_0^2 \int_{y^2}^{4} xy\,dx\,dy = \int_0^2 y\cdot\frac{16 - y^4}{2}\,dy = \int_0^2 \left(8y - \frac{y^5}{2}\right)dy = 16 - \frac{64}{12} = \frac{32}{3}$$

Same answer, which is a great way to check your limits.

### Worked example 3: switching to make it possible

Evaluate $\displaystyle\int_0^1 \int_y^1 e^{x^2}\,dx\,dy$.

The inner integral $\int e^{x^2}\,dx$ has no elementary antiderivative, so in this order you are stuck. Switch.

**Step 1. Read off the region.** $0 \le y \le 1$ and $y \le x \le 1$. That is the triangle below the line $y = x$, above $y = 0$, left of $x = 1$.

```graph
title: 'The triangular region under y = x between x = 0 and x = 1'
caption: 'The same triangle described two ways: $0 \le y \le 1,\ y \le x \le 1$ or $0 \le x \le 1,\ 0 \le y \le x$.'
x: [-0.3, 1.5]
y: [-0.3, 1.3]
functions:
    - expr: x
      label: y = x
shade: {expr: x, from: 0, to: 1}
vertical:
    - {x: 1, label: 'x = 1'}
```

**Step 2. Describe it with vertical strips.** $x$ runs from $0$ to $1$, and for each $x$, $y$ runs from $0$ up to the line $y = x$.

**Step 3. Integrate in the new order.**

$$\int_0^1 \int_0^x e^{x^2}\,dy\,dx = \int_0^1 x e^{x^2}\,dx$$

**Step 4. Finish with [u-substitution](/blog/u-substitution).** With $u = x^2$, $du = 2x\,dx$:

$$\int_0^1 x e^{x^2}\,dx = \left[\tfrac{1}{2}e^{x^2}\right]_0^1 = \frac{e - 1}{2}$$

The inner $y$ integral produced the factor $x$ that made the substitution work. That is the classic reason to switch.

## Double integrals in polar coordinates

When the region is a disk, a ring or a wedge, rectangular limits get messy. Switch to polar coordinates:

$$x = r\cos\theta, \quad y = r\sin\theta, \quad x^2 + y^2 = r^2, \quad dA = r\,dr\,d\theta$$

$$\iint_R f(x, y)\,dA = \int_\alpha^\beta \int_{r_1}^{r_2} f(r\cos\theta, r\sin\theta)\,r\,dr\,d\theta$$

The extra $r$ is there because a small polar patch has width $dr$ and arc length $r\,d\theta$.

### Worked example 4: volume under a paraboloid

Find the volume under $z = 4 - x^2 - y^2$ and above the $xy$-plane.

**Step 1. Find the region.** The surface meets $z = 0$ where $x^2 + y^2 = 4$, a disk of radius 2. In polar: $0 \le r \le 2$, $0 \le \theta \le 2\pi$.

**Step 2. Convert the integrand.** $4 - x^2 - y^2 = 4 - r^2$.

**Step 3. Set up with $dA = r\,dr\,d\theta$.**

$$V = \int_0^{2\pi} \int_0^2 (4 - r^2)\,r\,dr\,d\theta$$

**Step 4. Inner integral.**

$$\int_0^2 (4r - r^3)\,dr = \left[2r^2 - \frac{r^4}{4}\right]_0^2 = 8 - 4 = 4$$

**Step 5. Outer integral.** $\int_0^{2\pi} 4\,d\theta = 8\pi$.

The volume is $8\pi$. In rectangular coordinates this would need $\sqrt{4 - x^2}$ limits and a much harder integral.

## Common mistakes with double integrals

1. **Forgetting the $r$ in polar.** Without it, example 4 gives $\int_0^{2\pi}\int_0^2 (4 - r^2)\,dr\,d\theta = \frac{32\pi}{3}$, which is wrong.
2. **Variables in the outer limits.** If an outer limit contains $x$ or $y$, the setup is broken. Outer limits are always numbers.
3. **Swapping limits without redrawing.** Changing $\int_0^1\int_y^1 \,dx\,dy$ to $\int_0^1\int_x^1 \,dy\,dx$ describes a different triangle.
4. **Mismatched $dx\,dy$ order.** The inner differential must match the inner limits.
5. **Upper and lower curves reversed.** If $g_1(x)$ is above $g_2(x)$, you get a negative area. Check with a sketch.

## Quick reference: setting up double integrals

| Region | Setup |
| --- | --- |
| Rectangle $[a, b] \times [c, d]$ | $\int_a^b \int_c^d f\,dy\,dx$ (either order) |
| Type I: $g_1(x) \le y \le g_2(x)$ | $\int_a^b \int_{g_1(x)}^{g_2(x)} f\,dy\,dx$ |
| Type II: $h_1(y) \le x \le h_2(y)$ | $\int_c^d \int_{h_1(y)}^{h_2(y)} f\,dx\,dy$ |
| Disk of radius $a$ | $\int_0^{2\pi} \int_0^a f\,r\,dr\,d\theta$ |
| Area of $R$ | $\iint_R 1\,dA$ |
| Average value of $f$ on $R$ | $\frac{1}{\text{Area}(R)}\iint_R f\,dA$ |

## Practice the setup, not just the integration

The integration in a double integral is usually Calculus 1 and 2 material. What takes practice is reading a region and writing the right limits in the right order. The interactive lesson below gives you regions and integrals to set up, checks your limits before you integrate, and shows exactly which boundary went wrong.

## Frequently asked questions

### What does a double integral represent?

A double integral $\iint_R f(x, y)\,dA$ adds up the values of $f$ over a two-dimensional region $R$. When $f \ge 0$ it gives the volume of the solid under the surface $z = f(x, y)$ and above $R$. With $f = 1$ it gives the area of $R$.

### Does the order of integration matter in a double integral?

For a continuous function, the value does not depend on the order (that is Fubini's theorem), but the limits do. Over a rectangle you can swap $dx\,dy$ and $dy\,dx$ freely. Over a general region you must redraw the region and rewrite the limits when you swap.

### How do you switch the order of integration?

Sketch the region from the given limits, then describe the same region the other way: find the constant range of the new outer variable, and for each value find where the new inner variable enters and leaves the region. Those become the new limits.

### When should you use polar coordinates for a double integral?

Use polar coordinates when the region is a disk, an annulus or a sector, or when the integrand contains $x^2 + y^2$. Substitute $x = r\cos\theta$, $y = r\sin\theta$ and $dA = r\,dr\,d\theta$.

### Why is there an extra r in polar double integrals?

A small polar patch is not a square. It is a piece of a ring with radial width $dr$ and arc length $r\,d\theta$, so its area is $r\,dr\,d\theta$. Patches farther from the origin are bigger, and the factor $r$ accounts for that.