Applications of Double Integrals

Applications of double integrals add up a quantity spread over a flat region. With a density ρ(x, y), the double integral of ρ gives the mass of a thin plate. Dividing its moments by the mass gives the center of mass. The double integral of √(1 + f_x² + f_y²) gives the area of a surface z = f(x, y).

Before this lesson: Double Integrals in Polar Coordinates, Center of Mass and Partial Derivatives

Adding up mass, balance and area

Think of a thin metal plate shaped like a region D. A small piece of area ΔA with density ρ(x, y) has mass about ρ ΔA. Adding all the pieces gives the total mass, a double integral.

To find the balance point, weight each piece’s position by its mass. These sums are the moments. Divide each moment by the total mass to get the average position, which is the center of mass. If the density is constant, this point is the centroid.

Mass, center of mass and surface area formulas

m = \iint_D \rho(x, y)\,dA

M_y = \iint_D x\,\rho(x, y)\,dA, \qquad M_x = \iint_D y\,\rho(x, y)\,dA

\bar{x} = \dfrac{M_y}{m}, \qquad \bar{y} = \dfrac{M_x}{m}

S = \iint_D \sqrt{1 + f_x^2 + f_y^2}\,dA

M_y is the moment about the y-axis, so it uses x, the distance from that axis. M_x uses y. In the surface area formula, f_x and f_y are the partial derivatives of the surface, and D is the shadow of the surface in the xy-plane.

Applications of double integrals examples

Example 1: Mass of a plate

A square plate covers [0, 2] × [0, 2], in meters. Its density is ρ(x, y) = 1 + x + y kilograms per square meter. Find its mass.

m = \int_0^2 \int_0^2 (1 + x + y)\,dy\,dx

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

= 2 + 2x + 2

= 4 + 2x

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

= 8 + 4

= 12

Answer: the plate has a mass of 12 kg.

Example 2: Center of mass of the same plate

Find the center of mass of the plate in Example 1.

M_y = \int_0^2 \int_0^2 x(1 + x + y)\,dy\,dx

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

= 2x + 2x^2 + 2x

= 4x + 2x^2

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

= 8 + \dfrac{16}{3}

= \dfrac{40}{3}

\bar{x} = \dfrac{40/3}{12}

= \dfrac{10}{9}

\bar{y} = \dfrac{10}{9}

Answer: the center of mass is (10/9, 10/9) ≈ (1.11, 1.11). It sits a little above and right of the center (1, 1), toward the heavier corner.

Example 3: A triangle with changing density

A thin plate is the triangle with corners (0, 0), (1, 0) and (0, 1). Its density is ρ = x. Find its center of mass.

m = \int_0^1 \int_0^{1 - x} x\,dy\,dx

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

\int_0^1 (x - x^2)\,dx = \dfrac{1}{2} - \dfrac{1}{3}

= \dfrac{1}{6}

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

\int_0^{1 - x} x^2\,dy = x^2 - x^3

\int_0^1 (x^2 - x^3)\,dx = \dfrac{1}{3} - \dfrac{1}{4}

= \dfrac{1}{12}

\bar{x} = \dfrac{1/12}{1/6} = \dfrac{1}{2}

M_x = \int_0^1 \int_0^{1 - x} xy\,dy\,dx

\int_0^{1 - x} xy\,dy = \dfrac{x(1 - x)^2}{2}

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

\int_0^1 \dfrac{x - 2x^2 + x^3}{2}\,dx = \dfrac{1}{2}\Big(\dfrac{1}{2} - \dfrac{2}{3} + \dfrac{1}{4}\Big)

= \dfrac{1}{24}

\bar{y} = \dfrac{1/24}{1/6} = \dfrac{1}{4}

Answer: the center of mass is (1/2, 1/4). The uniform triangle balances at (1/3, 1/3), but this plate is heavier on the right, so the balance point moves right.

Example 4: Centroid of a half disk

Find the centroid of the upper half of the disk x² + y² ≤ 9, with constant density ρ = 1.

m = \dfrac{1}{2}\pi(3)^2 = \dfrac{9\pi}{2}

\bar{x} = 0

M_x = \int_0^{\pi} \int_0^3 (r\sin\theta)\,r\,dr\,d\theta

\int_0^3 r^2\,dr = 9

\int_0^{\pi} 9\sin\theta\,d\theta = 9 \cdot 2

= 18

\bar{y} = \dfrac{18}{9\pi/2}

= \dfrac{4}{\pi} \approx 1.273

Answer: the centroid is (0, 4/π) ≈ (0, 1.273). In general, a half disk of radius a balances at height 4a/(3π).

Example 5: Glass for a bowl-shaped skylight

A skylight has the shape of the surface z = x² + y² from z = 0 up to z = 4, in meters. How much glass does it need?

f_x = 2x, \quad f_y = 2y

\sqrt{1 + 4x^2 + 4y^2} = \sqrt{1 + 4r^2}

x^2 + y^2 \le 4 \Rightarrow 0 \le r \le 2

S = \int_0^{2\pi} \int_0^2 \sqrt{1 + 4r^2}\,r\,dr\,d\theta

\int_0^2 \sqrt{1 + 4r^2}\,r\,dr = \Big[\dfrac{1}{12}(1 + 4r^2)^{3/2}\Big]_0^2

= \dfrac{17^{3/2} - 1}{12}

\int_0^{2\pi} \dfrac{17^{3/2} - 1}{12}\,d\theta = \dfrac{\pi}{6}\Big(17\sqrt{17} - 1\Big)

\dfrac{\pi}{6}\Big(17\sqrt{17} - 1\Big) \approx 36.18

Answer: about 36.18 m² of glass. The flat disk underneath has area 4π ≈ 12.57 m², so the curved surface needs almost three times as much.

Common mistakes

Related methods

Center of mass with single integrals

For a plate with constant density, the center of mass lesson in Calculus 2 uses single integrals of strips. Double integrals give the same answers and also handle density that changes with both x and y.

Average value and totals

The same idea gives totals of anything with a density: people per square kilometer, rainfall per square meter or charge per square centimeter. Dividing a total by the area gives the average value, as in double integrals.

Moments of inertia

In physics, the moments of inertia I_x = ∬ y² ρ dA and I_y = ∬ x² ρ dA measure how hard a plate is to spin about each axis.

Practice problems

1. Find the mass of the rectangle [0, 3] × [0, 2] with density ρ = xy.
2. Find the mass of the disk x² + y² ≤ 4 with density ρ = √(x² + y²).
3. Find the center of mass of the square [0, 1] × [0, 1] with density ρ = x.
4. Find the centroid of the region under y = x² for 0 ≤ x ≤ 1.
5. Find the centroid of the quarter disk x² + y² ≤ 4 in the first quadrant.
6. Find the area of the part of the plane z = 2x + 2y + 1 above [0, 1] × [0, 2].
7. Find the area of the part of the plane z = 6 − 2x − 3y in the first octant.
8. Find the area of the part of z = x² + y² below z = 1.
9. A town is a disk of radius 5 km. Its population density is 3000 − 400r people per km² at distance r from the center. Estimate its population.
10. Find the mass of the triangle with corners (0, 0), (2, 0) and (0, 2) and density ρ = 1 + x.

m = \Big(\int_0^3 x\,dx\Big)\Big(\int_0^2 y\,dy\Big)

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

= 9

m = \int_0^{2\pi} \int_0^2 r \cdot r\,dr\,d\theta

\int_0^2 r^2\,dr = \dfrac{8}{3}

\int_0^{2\pi} \dfrac{8}{3}\,d\theta = \dfrac{16\pi}{3}

m = \iint x\,dA = \dfrac{1}{2}

M_y = \iint x^2\,dA = \dfrac{1}{3}

\bar{x} = \dfrac{1/3}{1/2} = \dfrac{2}{3}

M_x = \iint xy\,dA = \dfrac{1}{2} \cdot \dfrac{1}{2}

= \dfrac{1}{4}

\bar{y} = \dfrac{1/4}{1/2} = \dfrac{1}{2}

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

= \dfrac{1}{3}

M_y = \int_0^1 x \cdot x^2\,dx = \dfrac{1}{4}

\bar{x} = \dfrac{1/4}{1/3} = \dfrac{3}{4}

M_x = \int_0^1 \dfrac{x^4}{2}\,dx = \dfrac{1}{10}

\bar{y} = \dfrac{1/10}{1/3} = \dfrac{3}{10}

m = \dfrac{1}{4}\pi(2)^2 = \pi

M_y = \int_0^{\pi/2} \int_0^2 r\cos\theta \cdot r\,dr\,d\theta = 1 \cdot \dfrac{8}{3}

\bar{x} = \dfrac{8/3}{\pi} = \dfrac{8}{3\pi} \approx 0.849

\bar{y} = \dfrac{8}{3\pi}

f_x = 2, \quad f_y = 2

\sqrt{1 + 4 + 4} = 3

S = 3 \cdot \text{Area}(D) = 3 \cdot 2 = 6

f_x = -2, \quad f_y = -3

\sqrt{1 + 4 + 9} = \sqrt{14}

\text{Area}(D) = \dfrac{1}{2}(3)(2) = 3

S = 3\sqrt{14} \approx 11.22

\sqrt{1 + 4r^2}

\int_0^1 \sqrt{1 + 4r^2}\,r\,dr = \dfrac{5^{3/2} - 1}{12}

S = 2\pi \cdot \dfrac{5\sqrt{5} - 1}{12} = \dfrac{\pi}{6}\Big(5\sqrt{5} - 1\Big)

\dfrac{\pi}{6}\Big(5\sqrt{5} - 1\Big) \approx 5.330

P = \int_0^{2\pi} \int_0^5 (3000 - 400r)\,r\,dr\,d\theta

\int_0^5 (3000r - 400r^2)\,dr = \Big[1500r^2 - \dfrac{400r^3}{3}\Big]_0^5

= 37{,}500 - 16{,}666.7

= 20{,}833.3

2\pi \cdot 20{,}833.3 \approx 130{,}900

m = \int_0^2 \int_0^{2 - x} (1 + x)\,dy\,dx

\int_0^{2 - x} (1 + x)\,dy = (1 + x)(2 - x)

= 2 + x - x^2

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

= 4 + 2 - \dfrac{8}{3}

= \dfrac{10}{3}

Applications of double integrals FAQ

How do you find mass with a double integral?

Integrate the density over the region: m = ∬ ρ(x, y) dA. If the density is constant, the mass is the density times the area.

How do you find the center of mass of a lamina?

A lamina is a thin flat plate. Find its mass m and its moments M_y = ∬ xρ dA and M_x = ∬ yρ dA. Then x̄ = M_y/m and ȳ = M_x/m.

What is the difference between the centroid and the center of mass?

The centroid is the center of mass when the density is constant, so it depends only on the shape. With changing density, the center of mass moves toward the heavier part.

What is the surface area formula with a double integral?

For z = f(x, y) over a region D, S = ∬ √(1 + f_x² + f_y²) dA. The square root measures how much the surface tilts compared with the flat region below it.

Summary

• Mass: m = ∬ ρ dA.
• Moments: M_y = ∬ xρ dA and M_x = ∬ yρ dA.
• Center of mass: (x̄, ȳ) = (M_y/m, M_x/m). Use symmetry when you can.
• With constant density, the center of mass is the centroid.
• Surface area: S = ∬ √(1 + f_x² + f_y²) dA over the shadow D.

Previous lesson: Double Integrals in Polar Coordinates

Next lesson: Triple Integrals

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