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Applications of Integration

7.2 Volume: The Disk Method

Focus on disk volumes, washer volumes, surface area is not volume.

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Ideas in this lesson

01Disk volumesWorked steps · five questions02Washer volumesWorked steps · five questions03Surface area is not volumeWorked steps · five questions
Idea 1 of 3

Disk volumes

A disk’ss area is πR2\pi R^{2}. Integrate cross-sectional area with respect to the coordinate perpendicular to the slices.

Worked example 1

Rotate the region below y=xy = x on [0,2][0, 2] about the xx-axis. Find Vπ\dfrac{V}{\pi }.
  1. R=xR = x; cross-section =πx2= \pi x^{2}
  2. Vπ=∫02x2 dx\dfrac{V}{\pi } = \int_{0}^{2} x^{2}\,dx
  3. Vπ=233=83≈2.67\dfrac{V}{\pi } = \dfrac{2^{3}}{3} = \dfrac{8}{3} \approx 2.67

Answer: 83≈2.67\dfrac{8}{3} \approx 2.67

Worked example 2

Rotate the region below y=x2y = x^{2} on [0,3][0, 3] about the xx-axis. Find Vπ\dfrac{V}{\pi }.
  1. R=x2R = x^{2}; cross-section =πx4= \pi x^{4}
  2. Vπ=∫03x4 dx\dfrac{V}{\pi } = \int_{0}^{3} x^{4}\,dx
  3. Vπ=355=2435≈48.6\dfrac{V}{\pi } = \dfrac{3^{5}}{5} = \dfrac{243}{5} \approx 48.6

Answer: 2435≈48.6\dfrac{243}{5} \approx 48.6

Idea 2 of 3

Washer volumes

Subtract the inner disk: π(R2−r2)\pi (R^{2} - r^{2}). Squaring a difference of radii is incorrect.

Worked example 1

Rotate the region between y=5y = 5 and y=1y = 1 on [0,2][0, 2] about the xx-axis. Find Vπ\dfrac{V}{\pi }.
  1. R=5R = 5; r=1r = 1
  2. Aπ=R2−r2=24\dfrac{A}{\pi } = R^{2} - r^{2} = 24
  3. Vπ=2[52−12]=48\dfrac{V}{\pi } = 2[5^{2} - 1^{2}] = 48

Answer: 4848

Worked example 2

Rotate the region between y=6y = 6 and y=2y = 2 on [0,3][0, 3] about the xx-axis. Find Vπ\dfrac{V}{\pi }.
  1. R=6R = 6; r=2r = 2
  2. Aπ=R2−r2=32\dfrac{A}{\pi } = R^{2} - r^{2} = 32
  3. Vπ=3[62−22]=96\dfrac{V}{\pi } = 3[6^{2} - 2^{2}] = 96

Answer: 9696

Idea 3 of 3

Surface area is not volume

Rotation can also create a curved surface. Surface area uses circumference times arc-length element, rather than disk area.

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

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §7.2,p7.2, p. 446446. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.