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Applications of Integration

7.4 Arc Length and Surfaces of Revolution

Focus on arc length, surface of revolution, straight-line verification.

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Ideas in this lesson

01Arc lengthWorked steps · five questions02Surface of revolutionWorked steps · five questions03Straight-line verificationWorked steps · five questions
Idea 1 of 3

Arc length

Use ∫ 1+(y′)2dx\int\,\sqrt{1 + (y')^{2}} dx for a smooth graph y=f(x)y = f(x). The derivative is squared before adding 11.

Worked example 1

Find the arc length of y=x+1y = x + 1 from x=0x = 0 to x=2x = 2.
  1. y′=1y' = 1; 1+(y′)2=2\sqrt{1 + (y')^{2}} = \sqrt{2}
  2. L=∫02 2 dx=2 2=2 2≈2.83L = \int_{0}^{2}\,\sqrt{2}\,dx = 2\,\sqrt{2} = 2\,\sqrt{2} \approx 2.83

Answer: 2 2≈2.832\,\sqrt{2} \approx 2.83

Worked example 2

Find the arc length of y=2x+1y = 2x + 1 from x=0x = 0 to x=3x = 3.
  1. y′=2y' = 2; 1+(y′)2=5\sqrt{1 + (y')^{2}} = \sqrt{5}
  2. L=∫03 5 dx=3 5=3 5≈6.71L = \int_{0}^{3}\,\sqrt{5}\,dx = 3\,\sqrt{5} = 3\,\sqrt{5} \approx 6.71

Answer: 3 5≈6.713\,\sqrt{5} \approx 6.71

Idea 2 of 3

Surface of revolution

Multiply the arc-length element by the circumference 2πr2\pi r measured from the axis of rotation.

Worked example 1

Rotate y=xy = x on [0,2][0, 2] about the xx-axis. Find the curved surface area divided by π\pi .
  1. y=xy = x; y′=1y' = 1
  2. Sπ=∫022(1)x 2 dx\dfrac{S}{\pi } = \int_{0}^{2} 2(1)x\,\sqrt{2}\,dx
  3. Sπ=1(2)22=4 2≈5.66\dfrac{S}{\pi } = 1(2)\sqrt[2]{2} = 4\,\sqrt{2} \approx 5.66

Answer: 4 2≈5.664\,\sqrt{2} \approx 5.66

Worked example 2

Rotate y=2xy = 2x on [0,3][0, 3] about the xx-axis. Find the curved surface area divided by π\pi .
  1. y=2xy = 2x; y′=2y' = 2
  2. Sπ=∫032(2)x 5 dx\dfrac{S}{\pi } = \int_{0}^{3} 2(2)x\,\sqrt{5}\,dx
  3. Sπ=2(3)52=18 5≈40.25\dfrac{S}{\pi } = 2(3)\sqrt[2]{5} = 18\,\sqrt{5} \approx 40.25

Answer: 18 5≈40.2518\,\sqrt{5} \approx 40.25

Idea 3 of 3

Straight-line verification

For a line segment, the arc-length integral equals distance between its endpoints. Use this as a geometry check.

Worked example 1

Find the arc length of y=x+1y = x + 1 from x=0x = 0 to x=2x = 2.
  1. y′=1y' = 1; 1+(y′)2=2\sqrt{1 + (y')^{2}} = \sqrt{2}
  2. L=∫02 2 dx=2 2=2 2≈2.83L = \int_{0}^{2}\,\sqrt{2}\,dx = 2\,\sqrt{2} = 2\,\sqrt{2} \approx 2.83

Answer: 2 2≈2.832\,\sqrt{2} \approx 2.83

Worked example 2

Find the arc length of y=2x+1y = 2x + 1 from x=0x = 0 to x=3x = 3.
  1. y′=2y' = 2; 1+(y′)2=5\sqrt{1 + (y')^{2}} = \sqrt{5}
  2. L=∫03 5 dx=3 5=3 5≈6.71L = \int_{0}^{3}\,\sqrt{5}\,dx = 3\,\sqrt{5} = 3\,\sqrt{5} \approx 6.71

Answer: 3 5≈6.713\,\sqrt{5} \approx 6.71

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §7.4,p7.4, p. 466466. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.