Double Integrals over General Regions

Double integrals over general regions handle regions with curved or slanted edges. For a region where y runs from a bottom curve g₁(x) to a top curve g₂(x) as x goes from a to b, integrate y first between those curves, then x from a to b. The inner limits can be functions; the outer limits must be numbers.

Before this lesson: Double Integrals and Area Between Curves

From rectangles to curved regions

Over a rectangle, every vertical strip runs between the same two numbers. Over a curved region D, a vertical strip at x starts on a bottom curve and stops on a top curve. So the inner limits become functions of x.

The outer integral then sweeps the strip across the region. Its limits are the smallest and largest x in D, so they are plain numbers. The final answer is a number with no variables left.

Type I and Type II regions

A Type I region lies between two curves of x. A Type II region lies between two curves of y.

\text{Type I: } \iint_D f\,dA = \int_a^b \int_{g_1(x)}^{g_2(x)} f(x, y)\,dy\,dx

\text{Type II: } \iint_D f\,dA = \int_c^d \int_{h_1(y)}^{h_2(y)} f(x, y)\,dx\,dy

1. Sketch the region and find where the boundary curves meet.
2. Choose an order. For dy dx, draw a vertical strip: it enters D on the bottom curve and leaves on the top curve. For dx dy, draw a horizontal strip from the left curve to the right curve.
3. Write the inner limits as those curves. Write the outer limits as the numbers where the strips start and stop.
4. Integrate from the inside out, as in double integrals over rectangles.

Double integrals over general regions examples

Example 1: A Type I region

Evaluate \iint_D x\,dA, where D is the region between y = x² and y = 2x.

x^2 = 2x

x(x - 2) = 0

x = 0 \text{ or } x = 2

\iint_D x\,dA = \int_0^2 \int_{x^2}^{2x} x\,dy\,dx

\int_{x^2}^{2x} x\,dy = x\Big[y\Big]_{x^2}^{2x}

= x(2x - x^2)

= 2x^2 - x^3

\int_0^2 (2x^2 - x^3)\,dx = \Big[\dfrac{2x^3}{3} - \dfrac{x^4}{4}\Big]_0^2

= \dfrac{16}{3} - 4

= \dfrac{4}{3}

Answer: 4/3.

Example 2: A Type II region for area

Find the area of the region bounded by the parabola x = y² and the line x = y + 2.

y^2 = y + 2

y^2 - y - 2 = 0

(y - 2)(y + 1) = 0

y = 2 \text{ or } y = -1

A = \int_{-1}^{2} \int_{y^2}^{y + 2} 1\,dx\,dy

\int_{y^2}^{y + 2} 1\,dx = y + 2 - y^2

\int_{-1}^{2} (y + 2 - y^2)\,dy = \Big[\dfrac{y^2}{2} + 2y - \dfrac{y^3}{3}\Big]_{-1}^{2}

= \Big(2 + 4 - \dfrac{8}{3}\Big) - \Big(\dfrac{1}{2} - 2 + \dfrac{1}{3}\Big)

= \dfrac{10}{3} - \Big(-\dfrac{7}{6}\Big)

= \dfrac{9}{2}

Answer: the area is 9/2. In dy dx order, the bottom edge changes from the parabola to the line at x = 1, so you would need two integrals. Type II needs only one.

Example 3: Reverse the order of integration

Evaluate \int_0^2 \int_{y/2}^{1} e^{x^2}\,dx\,dy.

\int e^{x^2}\,dx

0 \le y \le 2, \quad \dfrac{y}{2} \le x \le 1

0 \le x \le 1, \quad 0 \le y \le 2x

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

\int_0^{2x} e^{x^2}\,dy = 2x e^{x^2}

\int_0^1 2x e^{x^2}\,dx = \Big[e^{x^2}\Big]_0^1

= e - 1 \approx 1.718

Answer: e − 1 ≈ 1.718. Reversing the order turned an impossible inner integral into an easy one.

Example 4: A triangle from its corners

Evaluate \iint_D y\,dA, where D is the triangle with corners (0, 0), (2, 0) and (0, 4).

y = 4 - 2x

\iint_D y\,dA = \int_0^2 \int_0^{4 - 2x} y\,dy\,dx

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

= \dfrac{(4 - 2x)^2}{2}

= \dfrac{16 - 16x + 4x^2}{2}

= 8 - 8x + 2x^2

\int_0^2 (8 - 8x + 2x^2)\,dx = \Big[8x - 4x^2 + \dfrac{2x^3}{3}\Big]_0^2

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

= \dfrac{16}{3}

Answer: 16/3.

Example 5: Volume of a concrete wedge

A concrete block fills the space under the plane z = 6 − 2x − y and above the floor in the first octant, in feet. How much concrete is in the block?

2x + y = 6

0 \le x \le 3, \quad 0 \le y \le 6 - 2x

V = \int_0^3 \int_0^{6 - 2x} (6 - 2x - y)\,dy\,dx

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

= (6 - 2x)^2 - \dfrac{(6 - 2x)^2}{2}

= \dfrac{(6 - 2x)^2}{2}

= \dfrac{36 - 24x + 4x^2}{2}

= 18 - 12x + 2x^2

\int_0^3 (18 - 12x + 2x^2)\,dx = \Big[18x - 6x^2 + \dfrac{2x^3}{3}\Big]_0^3

= 54 - 54 + 18

= 18

Answer: 18 ft³ of concrete. The block is a pyramid with a triangular base of area 9 ft² and height 6 ft, and the volume of a pyramid formula gives (1/3)(9)(6) = 18 ft³ too.

Common mistakes

Related methods

Area between curves as a double integral

With f = 1, the inner integral of a Type I region is just top − bottom. That is the area between curves formula from Calculus 1. Example 2 is the “right minus left” version.

Splitting a region

If neither order gives one integral, cut D into pieces that are each Type I or Type II, and add the integrals.

Round regions

For disks and other round regions, the limits are much simpler in polar coordinates. That is the next lesson, double integrals in polar coordinates.

Practice problems

1. Evaluate \int_0^1 \int_0^{x} (x + y)\,dy\,dx.
2. Evaluate \int_1^2 \int_0^{y} xy\,dx\,dy.
3. Evaluate \iint_D y\,dA, where D is the region between y = x² and y = x.
4. Use a double integral to find the area between y = 4 − x² and the x-axis.
5. Evaluate \iint_D x\,dA, where D is the region with 0 ≤ y ≤ 2 and 0 ≤ x ≤ y².
6. Reverse the order and evaluate \int_0^1 \int_{x}^{1} \sin(y^2)\,dy\,dx.
7. Reverse the order and evaluate \int_0^4 \int_{\sqrt{x}}^{2} e^{y^3}\,dy\,dx.
8. Evaluate \iint_D xy\,dA, where D is the triangle with corners (0, 0), (1, 0) and (1, 3).
9. Find the volume under z = x + y over the triangle with corners (0, 0), (2, 0) and (2, 2).
10. A garden bed covers the region between y = x² and y = 4, in meters. The soil depth is 0.2 + 0.1y meters. How much soil is in the bed?

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

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

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

\int_0^1 \dfrac{3x^2}{2}\,dx = \Big[\dfrac{x^3}{2}\Big]_0^1

= \dfrac{1}{2}

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

= \dfrac{y^3}{2}

\int_1^2 \dfrac{y^3}{2}\,dy = \Big[\dfrac{y^4}{8}\Big]_1^2

= 2 - \dfrac{1}{8}

= \dfrac{15}{8}

x^2 = x \Rightarrow x = 0 \text{ or } x = 1

\int_{x^2}^{x} y\,dy = \Big[\dfrac{y^2}{2}\Big]_{x^2}^{x}

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

\int_0^1 \dfrac{x^2 - x^4}{2}\,dx = \dfrac{1}{2}\Big[\dfrac{x^3}{3} - \dfrac{x^5}{5}\Big]_0^1

= \dfrac{1}{2}\Big(\dfrac{1}{3} - \dfrac{1}{5}\Big)

= \dfrac{1}{2} \cdot \dfrac{2}{15}

= \dfrac{1}{15}

4 - x^2 = 0 \Rightarrow x = \pm 2

A = \int_{-2}^{2} \int_0^{4 - x^2} 1\,dy\,dx

\int_0^{4 - x^2} 1\,dy = 4 - x^2

\int_{-2}^{2} (4 - x^2)\,dx = \Big[4x - \dfrac{x^3}{3}\Big]_{-2}^{2}

= \Big(8 - \dfrac{8}{3}\Big) - \Big(-8 + \dfrac{8}{3}\Big)

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

= \dfrac{32}{3}

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

= \dfrac{y^4}{2}

\int_0^2 \dfrac{y^4}{2}\,dy = \Big[\dfrac{y^5}{10}\Big]_0^2

= \dfrac{32}{10}

= \dfrac{16}{5}

0 \le x \le 1, \quad x \le y \le 1

0 \le y \le 1, \quad 0 \le x \le y

\int_0^1 \int_0^{y} \sin(y^2)\,dx\,dy

\int_0^{y} \sin(y^2)\,dx = y\sin(y^2)

\int_0^1 y\sin(y^2)\,dy = \Big[-\dfrac{\cos(y^2)}{2}\Big]_0^1

= -\dfrac{\cos 1}{2} + \dfrac{1}{2}

= \dfrac{1 - \cos 1}{2} \approx 0.230

0 \le x \le 4, \quad \sqrt{x} \le y \le 2

0 \le y \le 2, \quad 0 \le x \le y^2

\int_0^2 \int_0^{y^2} e^{y^3}\,dx\,dy

\int_0^{y^2} e^{y^3}\,dx = y^2 e^{y^3}

\int_0^2 y^2 e^{y^3}\,dy = \Big[\dfrac{e^{y^3}}{3}\Big]_0^2

= \dfrac{e^8 - 1}{3} \approx 993.3

0 \le x \le 1, \quad 0 \le y \le 3x

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

= \dfrac{9x^3}{2}

\int_0^1 \dfrac{9x^3}{2}\,dx = \Big[\dfrac{9x^4}{8}\Big]_0^1

= \dfrac{9}{8}

0 \le x \le 2, \quad 0 \le y \le x

\int_0^{x} (x + y)\,dy = x^2 + \dfrac{x^2}{2}

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

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

= 4

-2 \le x \le 2, \quad x^2 \le y \le 4

\int_{x^2}^{4} (0.2 + 0.1y)\,dy = \Big[0.2y + 0.05y^2\Big]_{x^2}^{4}

= (0.8 + 0.8) - (0.2x^2 + 0.05x^4)

= 1.6 - 0.2x^2 - 0.05x^4

\int_{-2}^{2} (\ldots)\,dx = 2\int_0^2 (1.6 - 0.2x^2 - 0.05x^4)\,dx

= 2\Big[1.6x - \dfrac{0.2x^3}{3} - 0.01x^5\Big]_0^2

= 2\Big(3.2 - \dfrac{1.6}{3} - 0.32\Big)

= \dfrac{352}{75} \approx 4.69

Double integrals over general regions FAQ

What is the difference between Type I and Type II regions?

A Type I region lies between two curves y = g₁(x) and y = g₂(x), so you integrate y first. A Type II region lies between x = h₁(y) and x = h₂(y), so you integrate x first.

How do you reverse the order of integration?

Read the region from the old limits and sketch it. Then describe the same region with strips in the other direction, and write new limits from the sketch. Never just swap the limits.

How do you find the limits of integration for a double integral?

Draw a strip across the region in the direction of the inner variable. Where it enters and leaves gives the inner limits. Where the strips start and stop gives the outer limits.

Can a double integral find the area of a region?

Yes. Integrate f = 1 over the region: the result is its area. This matches the area between curves from Calculus 1.

Summary

• Over a general region, the inner limits are curves and the outer limits are numbers.
• Type I: vertical strips, dy dx. Type II: horizontal strips, dx dy.
• Find intersection points and test a point to see which curve is on top or on the right.
• To reverse the order, sketch the region and describe it again; never just swap the limits.
• Choose the order that gives one integral and an easy inner integral.

Previous lesson: Double Integrals

Next lesson: Double Integrals in Polar Coordinates

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