Double Integrals

A double integral adds up the values of f(x, y) over a region R in the plane. Over a rectangle R = [a, b] × [c, d], you evaluate double integrals as iterated integrals: integrate with respect to one variable while holding the other fixed, then integrate that result. When f ≥ 0, the double integral is the volume under the surface.

Before this lesson: Functions of Several Variables, Partial Derivatives and Fundamental Theorem of Calculus, Part 2

What a double integral measures

In Calculus 1, a definite integral adds up thin rectangles under a curve. A double integral does the same thing one dimension up. Cut the rectangle R into small pieces of area ΔA, build a thin box on each piece with height f(x, y), and add the box volumes. As the pieces shrink, these Riemann sums approach the double integral.

You almost never use the sums to compute. Instead, slice the solid. Hold x fixed and integrate f(x, y) in y: that gives the area of one slice. Then integrate those slice areas in x to stack the slices into the whole volume.

Double integral formula over a rectangle

\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

This is Fubini’s theorem: if f is continuous on R = [a, b] × [c, d], both iterated integrals equal the double integral. The notation dA stands for a small piece of area.

1. Read the integral from the inside out. The inner differential (dy or dx) goes with the inner limits.
2. Do the inner integral. Treat the other variable as a constant, the way you do for partial derivatives.
3. Substitute the inner limits. The result should contain only the outer variable.
4. Do the outer integral with the outer limits.

Double integrals examples

Example 1: Integrate y first

Evaluate the iterated integral \int_0^2 \int_1^3 (x + 2y)\,dy\,dx.

\int_1^3 (x + 2y)\,dy = \Big[xy + y^2\Big]_{y=1}^{y=3}

= (3x + 9) - (x + 1)

= 2x + 8

\int_0^2 (2x + 8)\,dx = \Big[x^2 + 8x\Big]_0^2

= (4 + 16) - 0

= 20

Answer: 20. If x + 2y is a height in meters over a region measured in meters, the solid holds 20 m³.

Example 2: Both orders give the same answer

Evaluate \iint_R x y^2\,dA over R = [0, 3] × [1, 2], first with x on the inside and then with y on the inside.

\int_0^3 x y^2\,dx = y^2 \Big[\dfrac{x^2}{2}\Big]_0^3

= \dfrac{9y^2}{2}

\int_1^2 \dfrac{9y^2}{2}\,dy = \dfrac{9}{2}\Big[\dfrac{y^3}{3}\Big]_1^2

= \dfrac{9}{2} \cdot \dfrac{7}{3}

= \dfrac{21}{2}

\int_1^2 x y^2\,dy = x \Big[\dfrac{y^3}{3}\Big]_1^2

= \dfrac{7x}{3}

\int_0^3 \dfrac{7x}{3}\,dx = \dfrac{7}{3}\Big[\dfrac{x^2}{2}\Big]_0^3

= \dfrac{7}{3} \cdot \dfrac{9}{2}

= \dfrac{21}{2}

Answer: 21/2 both ways, as Fubini’s theorem promises.

Example 3: Choose the easier order

Evaluate \iint_R y e^{xy}\,dA over R = [0, 1] × [0, 2].

\iint_R y e^{xy}\,dA = \int_0^2 \int_0^1 y e^{xy}\,dx\,dy

\int_0^1 y e^{xy}\,dx = \Big[e^{xy}\Big]_{x=0}^{x=1}

= e^{y} - 1

\int_0^2 (e^y - 1)\,dy = \Big[e^y - y\Big]_0^2

= (e^2 - 2) - (1 - 0)

= e^2 - 3 \approx 4.389

Answer: e² − 3 ≈ 4.389. Both orders are correct, but one is far shorter. Look for the order that makes the inner integral easy.

Example 4: Volume under a surface

Find the volume of the solid under z = 16 − x² − 2y² and above R = [0, 2] × [0, 2].

V = \int_0^2 \int_0^2 (16 - x^2 - 2y^2)\,dy\,dx

\int_0^2 (16 - x^2 - 2y^2)\,dy = \Big[16y - x^2 y - \dfrac{2y^3}{3}\Big]_0^2

= 32 - 2x^2 - \dfrac{16}{3}

= \dfrac{80}{3} - 2x^2

\int_0^2 \Big(\dfrac{80}{3} - 2x^2\Big)dx = \Big[\dfrac{80x}{3} - \dfrac{2x^3}{3}\Big]_0^2

= \dfrac{160}{3} - \dfrac{16}{3}

= 48

Answer: the volume is 48 cubic units.

Example 5: Water in a sloped pool

A pool is 20 m long and 10 m wide. Its depth is d(x, y) = 1 + 0.1x meters, where x is the distance from the shallow end. How much water does it hold when full, and what is its average depth?

V = \int_0^{20} \int_0^{10} (1 + 0.1x)\,dy\,dx

\int_0^{10} (1 + 0.1x)\,dy = (1 + 0.1x)\Big[y\Big]_0^{10}

= 10 + x

\int_0^{20} (10 + x)\,dx = \Big[10x + \dfrac{x^2}{2}\Big]_0^{20}

= 200 + 200

= 400

\text{Area}(R) = 20 \cdot 10 = 200

d_{\text{avg}} = \dfrac{400}{200} = 2

Answer: the pool holds 400 m³ of water, and its average depth is 2 m.

Common mistakes

Related methods

Product shortcut

When f(x, y) = g(x)h(y) and R is a rectangle, the double integral splits into two single integrals: \iint_R g(x)h(y)\,dA = \Big(\int_a^b g(x)\,dx\Big)\Big(\int_c^d h(y)\,dy\Big). For example, over [0, π] × [0, π/2], the integral of sin x cos y is (2)(1) = 2.

Average value of a function of two variables

Like the average value of a function in Calculus 1, divide by the size of the region: f_{\text{avg}} = \dfrac{1}{\text{Area}(R)} \iint_R f(x, y)\,dA. Example 5 used this to find the average depth.

Area as a double integral

With f(x, y) = 1, the double integral gives the area of R. This matters more in the next lesson, on double integrals over general regions, where regions have curved edges.

Practice problems

1. Evaluate \int_0^1 \int_0^2 (x + y)\,dy\,dx.
2. Evaluate \int_0^2 \int_0^1 6x^2 y\,dy\,dx.
3. Evaluate \iint_R (4 - x - y)\,dA over R = [0, 2] × [0, 1].
4. Evaluate \int_0^{\pi} \int_0^{\pi/2} \sin(x + y)\,dy\,dx.
5. Evaluate \iint_R x^2 y\,dA over R = [0, 3] × [1, 2] with x on the inside.
6. Evaluate \iint_R x e^{xy}\,dA over R = [0, 1] × [0, 1]. Choose the easier order.
7. Evaluate \iint_R e^{x + 2y}\,dA over R = [0, 1] × [0, 1] with the product shortcut.
8. Find the volume under the plane z = 6 − x − 2y and above R = [0, 2] × [0, 1].
9. Find the average value of f(x, y) = x² + y² over R = [0, 2] × [0, 2].
10. A forest is 4 km long and 3 km wide. Its tree density is ρ(x, y) = 50 + 10x + 20y trees per km². How many trees does it have?

\int_0^2 (x + y)\,dy = \Big[xy + \dfrac{y^2}{2}\Big]_0^2

= 2x + 2

\int_0^1 (2x + 2)\,dx = \Big[x^2 + 2x\Big]_0^1

= 1 + 2

= 3

\int_0^1 6x^2 y\,dy = 6x^2 \Big[\dfrac{y^2}{2}\Big]_0^1

= 3x^2

\int_0^2 3x^2\,dx = \Big[x^3\Big]_0^2

= 8

\int_0^1 (4 - x - y)\,dy = \Big[4y - xy - \dfrac{y^2}{2}\Big]_0^1

= 4 - x - \dfrac{1}{2}

= \dfrac{7}{2} - x

\int_0^2 \Big(\dfrac{7}{2} - x\Big)dx = \Big[\dfrac{7x}{2} - \dfrac{x^2}{2}\Big]_0^2

= 7 - 2

= 5

\int_0^{\pi/2} \sin(x + y)\,dy = \Big[-\cos(x + y)\Big]_{y=0}^{y=\pi/2}

= -\cos\Big(x + \dfrac{\pi}{2}\Big) + \cos x

= \sin x + \cos x

\int_0^{\pi} (\sin x + \cos x)\,dx = \Big[-\cos x + \sin x\Big]_0^{\pi}

= (1 + 0) - (-1 + 0)

= 2

\int_0^3 x^2 y\,dx = y \Big[\dfrac{x^3}{3}\Big]_0^3

= 9y

\int_1^2 9y\,dy = \Big[\dfrac{9y^2}{2}\Big]_1^2

= 18 - \dfrac{9}{2}

= \dfrac{27}{2}

\int_0^1 \int_0^1 x e^{xy}\,dy\,dx

\int_0^1 x e^{xy}\,dy = \Big[e^{xy}\Big]_{y=0}^{y=1}

= e^x - 1

\int_0^1 (e^x - 1)\,dx = \Big[e^x - x\Big]_0^1

= (e - 1) - (1 - 0)

= e - 2 \approx 0.718

\iint_R e^{x}e^{2y}\,dA = \Big(\int_0^1 e^x\,dx\Big)\Big(\int_0^1 e^{2y}\,dy\Big)

\int_0^1 e^x\,dx = e - 1

\int_0^1 e^{2y}\,dy = \dfrac{e^2 - 1}{2}

(e - 1) \cdot \dfrac{e^2 - 1}{2} \approx 5.489

\int_0^1 (6 - x - 2y)\,dy = \Big[6y - xy - y^2\Big]_0^1

= 5 - x

\int_0^2 (5 - x)\,dx = \Big[5x - \dfrac{x^2}{2}\Big]_0^2

= 10 - 2

= 8

\int_0^2 (x^2 + y^2)\,dy = \Big[x^2 y + \dfrac{y^3}{3}\Big]_0^2

= 2x^2 + \dfrac{8}{3}

\int_0^2 \Big(2x^2 + \dfrac{8}{3}\Big)dx = \Big[\dfrac{2x^3}{3} + \dfrac{8x}{3}\Big]_0^2

= \dfrac{16}{3} + \dfrac{16}{3}

= \dfrac{32}{3}

f_{\text{avg}} = \dfrac{32/3}{4} = \dfrac{8}{3}

N = \int_0^4 \int_0^3 (50 + 10x + 20y)\,dy\,dx

\int_0^3 (50 + 10x + 20y)\,dy = \Big[50y + 10xy + 10y^2\Big]_0^3

= 150 + 30x + 90

= 240 + 30x

\int_0^4 (240 + 30x)\,dx = \Big[240x + 15x^2\Big]_0^4

= 960 + 240

= 1200

Double integrals FAQ

What is the difference between a double integral and an iterated integral?

A double integral is defined as a limit of Riemann sums over a region. An iterated integral is how you compute it, one variable at a time. Fubini’s theorem says they are equal for continuous functions on a rectangle.

Does the order of integration matter in a double integral?

Over a rectangle with a continuous function, both orders give the same answer. The order can still change how much work it takes, as Example 3 shows.

What does a double integral represent?

If f ≥ 0, it is the volume under the surface z = f(x, y) and above R. If f is 1, it is the area of R. If f is a density, it is a total amount, such as mass or population.

How do you find the average value of a function of two variables?

Integrate f over R and divide by the area of R. For a pool, the volume divided by the floor area is the average depth.

Summary

• A double integral adds f(x, y) dA over a region; when f ≥ 0, it is a volume.
• Over a rectangle, compute it as an iterated integral, from the inside out.
• In the inner integral, treat the other variable as a constant.
• Fubini’s theorem: both orders give the same value, so pick the easier one.
• If f(x, y) = g(x)h(y), split it into a product of two single integrals.
• Average value = double integral ÷ area of R.

Previous lesson: Lagrange Multipliers

Next lesson: Double Integrals over General Regions

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