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Applications of Integration

7.6 Moments, Centers of Mass, and Centroids

Focus on moments of point masses, centroid of a lamina, moments about different axes.

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Ideas in this lesson

01Moments of point massesWorked steps · five questions02Centroid of a laminaWorked steps · five questions03Moments about different axesWorked steps · five questions
Idea 1 of 3

Moments of point masses

A moment combines mass with signed distance from an axis. Divide the total moment by total mass for the center of mass.

Worked example 1

Masses 11 and 22 sit at x=0x = 0 and x=3x = 3. Find the center of mass.
  1. m=1+2=3m = 1 + 2 = 3
  2. M=1⋅0+2(2+1)=6M = 1\cdot 0 + 2(2 + 1) = 6
  3. xx̄ =63=2= \dfrac{6}{3} = 2

Answer: 22

Worked example 2

Masses 11 and 33 sit at x=0x = 0 and x=4x = 4. Find the moment about 00.
  1. m=1+3=4m = 1 + 3 = 4
  2. M=1⋅0+3(3+1)=12M = 1\cdot 0 + 3(3 + 1) = 12
  3. M=12M = 12

Answer: 1212

Idea 2 of 3

Centroid of a lamina

For uniform density, use area moments divided by area. Density cancels in the centroid calculation.

Worked example 1

The uniform region below y=xy = x on [0,2][0, 2] is a triangle. Find its centroid xx-coordinate.
  1. A=∫02x dx=42A = \int_{0}^{2} x\,dx = \dfrac{4}{2}
  2. My=∫02x2 dx=83M_y = \int_{0}^{2} x^{2}\,dx = \dfrac{8}{3}; Mx=12∫02x2 dx=86M_x = \dfrac{1}{2}\int_{0}^{2} x^{2}\,dx = \dfrac{8}{6}
  3. xx̄ =MyA=43= \dfrac{M_y}{A} = \dfrac{4}{3}; ȳ =MxA=23= \dfrac{M_x}{A} = \dfrac{2}{3}; requested value =43≈1.33= \dfrac{4}{3} \approx 1.33

Answer: 43≈1.33\dfrac{4}{3} \approx 1.33

Worked example 2

The uniform region below y=xy = x on [0,3][0, 3] is a triangle. Find its centroid yy-coordinate.
  1. A=∫03x dx=92A = \int_{0}^{3} x\,dx = \dfrac{9}{2}
  2. My=∫03x2 dx=273M_y = \int_{0}^{3} x^{2}\,dx = \dfrac{27}{3}; Mx=12∫03x2 dx=276M_x = \dfrac{1}{2}\int_{0}^{3} x^{2}\,dx = \dfrac{27}{6}
  3. xx̄ =MyA=63= \dfrac{M_y}{A} = \dfrac{6}{3}; ȳ =MxA=33= \dfrac{M_x}{A} = \dfrac{3}{3}; requested value =1= 1

Answer: 11

Idea 3 of 3

Moments about different axes

MyM_y uses horizontal distance xx; MxM_x uses vertical distance. Match each moment to the correct coordinate.

Worked example 1

The uniform region below y=xy = x on [0,2][0, 2] is a triangle. Find its centroid xx-coordinate.
  1. A=∫02x dx=42A = \int_{0}^{2} x\,dx = \dfrac{4}{2}
  2. My=∫02x2 dx=83M_y = \int_{0}^{2} x^{2}\,dx = \dfrac{8}{3}; Mx=12∫02x2 dx=86M_x = \dfrac{1}{2}\int_{0}^{2} x^{2}\,dx = \dfrac{8}{6}
  3. xx̄ =MyA=43= \dfrac{M_y}{A} = \dfrac{4}{3}; ȳ =MxA=23= \dfrac{M_x}{A} = \dfrac{2}{3}; requested value =43≈1.33= \dfrac{4}{3} \approx 1.33

Answer: 43≈1.33\dfrac{4}{3} \approx 1.33

Worked example 2

The uniform region below y=xy = x on [0,3][0, 3] is a triangle. Find its centroid yy-coordinate.
  1. A=∫03x dx=92A = \int_{0}^{3} x\,dx = \dfrac{9}{2}
  2. My=∫03x2 dx=273M_y = \int_{0}^{3} x^{2}\,dx = \dfrac{27}{3}; Mx=12∫03x2 dx=276M_x = \dfrac{1}{2}\int_{0}^{3} x^{2}\,dx = \dfrac{27}{6}
  3. xx̄ =MyA=63= \dfrac{M_y}{A} = \dfrac{6}{3}; ȳ =MxA=33= \dfrac{M_x}{A} = \dfrac{3}{3}; requested value =1= 1

Answer: 11

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §7.6,p7.6, p. 486486. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.