Do I integrate dx first or dy first in a double integral?

Either order works for rectangular regions, but for non-rectangular regions, one order often has simpler bounds. Sketch your region, identify which bounds should be constants, and make the inner variable depend on the outer one if needed. The choice is yours—just pick the one that is easier to compute.

What does dA mean in a double integral?

The notation dA represents an infinitesimal area element. When you write \iint_R f(x,y) \, dA, you are summing up the product of the function value and the tiny area piece at each location, which gives you the total accumulation. In Cartesian coordinates, dA = dx \, dy.

Is the double integral of a constant just the constant times the area of R?

Yes! If f(x,y) = c (a constant), then \iint_R c \, dA = c \cdot \text{Area}(R). This is why \iint_R dA alone gives you the area of the region without integrating any function over it.

Can a double integral be negative?

Yes, if the function f(x,y) takes negative values over part or all of the region R. The double integral sums positive contributions where f > 0 and negative contributions where f < 0, so the final result can be negative overall.

Double Integral

Compute the total accumulation of a quantity over a 2D region—like volume under a surface or mass—by integrating a function of two variables over an area.

\iint_R f(x,y)\,dA

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What each symbol means

What Double Integral takes

Symbol	Meaning
$f$	The function $f(x,y)$ represents the quantity at each point (height, density, temperature); treating it as single-variable loses information about how the quantity changes in both directions.
$x$	One independent variable representing a coordinate in the plane; swapping it with $y$ changes the order of integration and your result.
$y$	The other independent variable representing a coordinate in the plane; omitting it reduces the problem from 2D to 1D.
$R$	The region over which you integrate, defined by bounds on $x$ and $y$ ; forgetting to specify it leaves your integral ambiguous.

When to use it

When you need to find a total amount (volume, mass, probability) distributed across a 2D region.

Level

Usually taught in: Calculus III

Worked examples

1. Volume under a surface over a rectangle

Problem

Find the volume under

f(x,y) = 3x + 2y

over the rectangular region

0 \le x \le 2

0 \le y \le 1

$\iint_R (3x + 2y) \, dA = \int_0^2 \int_0^1 (3x + 2y) \, dy \, dx$
Set up the double integral with constant bounds for both variables.
$\int_0^1 (3x + 2y) \, dy = \left[3xy + y^2\right]_0^1 = 3x + 1$
Integrate with respect to $y$ first, treating $x$ as a constant.
$\int_0^2 (3x + 1) \, dx = \left[\frac{3x^2}{2} + x\right]_0^2 = 6 + 2 = 8$
Now integrate with respect to $x$ to get the total volume.

Answer: $V = 8 \text{ cubic units}$

We integrated first with respect to $y$ (treating $x$ as constant), then with respect to $x$ . This rectangular region allowed us to use simple constant bounds for both integrals.

2. Double integral with negative bounds and negative result

Problem

Evaluate

\iint_R (2x - y) \, dA

where

R

is the rectangle

-1 \le x \le 1

0 \le y \le 2

$\iint_R (2x - y) \, dA = \int_{-1}^1 \int_0^2 (2x - y) \, dy \, dx$
Set up the double integral with the given bounds.
$\int_0^2 (2x - y) \, dy = \left[2xy - \frac{y^2}{2}\right]_0^2 = 4x - 2$
Integrate with respect to $y$ , treating $x$ as a constant.
$\int_{-1}^1 (4x - 2) \, dx = \left[2x^2 - 2x\right]_{-1}^1 = (2 - 2) - (2 + 2) = -4$
Integrate with respect to $x$ ; the negative bounds and subtraction in the function produce a negative result.

Answer: $\iint_R (2x - y) \, dA = -4$

The negative bounds on $x$ and the negative result show that double integrals can handle regions with negative coordinates and produce negative outputs. The subtraction in $2x - y$ means parts of the region contribute negatively to the total.

3. Volume of paint on a construction site

Problem

A construction company is painting a rectangular site measuring 10 meters by 8 meters. The paint thickness (in millimeters) at position

(x, y)

is given by

h(x,y) = 0.5x + 0.3y

. Find the total volume of paint (in cubic meters) needed to cover the entire site.

$\iint_R (0.0005x + 0.0003y) \, dA \text{ where } 0 \le x \le 10, 0 \le y \le 8$
The thickness is $0.5x + 0.3y$ mm, which converts to $0.0005x + 0.0003y$ meters. Set up the integral with these bounds.
$\int_0^8 (0.0005x + 0.0003y) \, dy = \left[0.0005xy + 0.00015y^2\right]_0^8 = 0.004x + 0.0096$
Integrate with respect to $y$ first.
$\int_0^{10} (0.004x + 0.0096) \, dx = \left[0.002x^2 + 0.0096x\right]_0^{10} = 0.2 + 0.096 = 0.296$
Integrate with respect to $x$ to find the total volume in cubic meters.

Answer: $V = 0.296 \text{ m}^3$

Double integrals appear naturally in applied problems where a quantity varies across a 2D region. Here, integrating the varying thickness over the rectangular site gives the total paint volume needed.

Common mistakes

Where Double Integral usually goes wrong

The mistake	Why it is wrong	The fix
Writing $\int_a^b \int_c^d f(x,y) \, dx \, dy$ without specifying the region $R$ .	The bounds and order of integration depend on the shape and geometry of $R$ ; what works for a rectangle differs from a triangle or circle.	Always specify the region $R$ explicitly by describing its bounds in terms of $x$ and $y$ , or by sketching it first to determine the correct limits of integration.
Treating $\iint_R f(x,y) \, dA$ as $\left(\int_R f(x) \, dx\right) \cdot \left(\int_R f(y) \, dy\right)$ (multiplying two separate single integrals).	The double integral is not a product; it is a unified sum of the function values over the 2D region.	Compute the double integral by iterating two single integrals: integrate with respect to one variable first, then with respect to the other, keeping the bounds correct for each step.
Writing $\iint_R f(x,y)$ without the differential $dA$ (or $dx \, dy$).	The differential $dA$ indicates that you are accumulating over infinitesimal area elements; without it, the notation is incomplete and does not define a valid integral.	Always write $\iint_R f(x,y) \, dA$ or $\int \int_R f(x,y) \, dx \, dy$ to clearly show the infinitesimal area element.

Tips and when to use something else

Set up the region $R$ first by sketching or describing its bounds; this determines the limits for both integrals.
Choose the order of integration (dx dy or dy dx) based on which one keeps your bounds simpler; sometimes one order requires piecewise bounds while the other uses constants.
For a non-rectangular region (like a circle or triangle), express the bounds on the inner variable as functions of the outer variable.
To find the area of a region $R$ without integrating a function, use $\iint_R dA$ ; do not confuse this with Area Between Curves, which finds area between two 1D graphs instead of over a 2D region.

Related formulas

Definite Integral PropertiesTwo key properties of definite integrals: reversing the bounds negates the integral, and integrating over a single point gives zero.Riemann SumRiemann Sum approximates a definite integral by dividing the area under a curve into rectangles; it connects intuition to formal integration.Area Between CurvesFind the area of a region between two curves by computing the definite integral of their difference over a specified interval.Average Value of a FunctionFind the mean height of a function over an interval with the average value formula; use it for real-world average rates of change over time.

Frequently asked questions

Do I integrate $dx$ first or $dy$ first in a double integral?: Either order works for rectangular regions, but for non-rectangular regions, one order often has simpler bounds. Sketch your region, identify which bounds should be constants, and make the inner variable depend on the outer one if needed. The choice is yours—just pick the one that is easier to compute.
What does $dA$ mean in a double integral?: The notation $dA$ represents an infinitesimal area element. When you write $\iint_R f(x,y) \, dA$ , you are summing up the product of the function value and the tiny area piece at each location, which gives you the total accumulation. In Cartesian coordinates, $dA = dx \, dy$ .
Is the double integral of a constant just the constant times the area of $R$ ?: Yes! If $f(x,y) = c$ (a constant), then $\iint_R c \, dA = c \cdot \text{Area}(R)$ . This is why $\iint_R dA$ alone gives you the area of the region without integrating any function over it.
Can a double integral be negative?: Yes, if the function $f(x,y)$ takes negative values over part or all of the region $R$ . The double integral sums positive contributions where $f > 0$ and negative contributions where $f < 0$ , so the final result can be negative overall.

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Reviewed 2026-09-18