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Applications of Integration

7.3 Volume: The Shell Method

Focus on cylindrical shells, shifted rotation axes, choose shells or disks.

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Ideas in this lesson

01Cylindrical shellsWorked steps · five questions02Shifted rotation axesWorked steps · five questions03Choose shells or disksWorked steps · five questions
Idea 1 of 3

Cylindrical shells

A thin shell contributes 2π(2\pi ( radius )()( height )()( thickness )). Identify all three before integrating.

Worked example 1

Rotate the region below y=xy = x on [0,2][0, 2] about the yy-axis. Find Vπ\dfrac{V}{\pi }.
  1. Radius =x= x; height =x= x
  2. Vπ=2∫02x2 dx\dfrac{V}{\pi } = 2\int_{0}^{2} x^{2}\,dx
  3. Vπ=2⋅233=163≈5.33\dfrac{V}{\pi } = \dfrac{2 \cdot 2^{3}}{3} = \dfrac{16}{3} \approx 5.33

Answer: 163≈5.33\dfrac{16}{3} \approx 5.33

Worked example 2

Rotate the region below y=x2y = x^{2} on [0,3][0, 3] about the yy-axis. Find Vπ\dfrac{V}{\pi }.
  1. Radius =x= x; height =x2= x^{2}
  2. Vπ=2∫03x3 dx\dfrac{V}{\pi } = 2\int_{0}^{3} x^{3}\,dx
  3. Vπ=2⋅344=812≈40.5\dfrac{V}{\pi } = \dfrac{2 \cdot 3^{4}}{4} = \dfrac{81}{2} \approx 40.5

Answer: 812≈40.5\dfrac{81}{2} \approx 40.5

Idea 2 of 3

Shifted rotation axes

Measure radius from the actual axis of rotation. A shifted axis changes the radius even if the region stays fixed.

Worked example 1

Rotate the region under y=xy = x on [0,2][0, 2] about x=−1x = -1. Find Vπ\dfrac{V}{\pi }.
  1. Radius =x−(−1)=x+1= x - (-1) = x + 1; height =x= x
  2. Vπ=2∫02x2+x dx\dfrac{V}{\pi } = 2\int_{0}^{2} x^{2} + x\,dx
  3. Vπ=2[233+222]=283≈9.33\dfrac{V}{\pi } = 2[\dfrac{2^{3}}{3} + \dfrac{2^{2}}{2}] = \dfrac{28}{3} \approx 9.33

Answer: 283≈9.33\dfrac{28}{3} \approx 9.33

Worked example 2

Rotate the region under y=xy = x on [0,3][0, 3] about x=−1x = -1. Find Vπ\dfrac{V}{\pi }.
  1. Radius =x−(−1)=x+1= x - (-1) = x + 1; height =x= x
  2. Vπ=2∫03x2+x dx\dfrac{V}{\pi } = 2\int_{0}^{3} x^{2} + x\,dx
  3. Vπ=2[333+322]=27\dfrac{V}{\pi } = 2[\dfrac{3^{3}}{3} + \dfrac{3^{2}}{2}] = 27

Answer: 2727

Idea 3 of 3

Choose shells or disks

Choose a slicing direction that describes the region simply. Both valid methods must give the same volume.

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

Resource alignment: Larson & Edwards, Calculus of a Single Variable, Tenth Edition, §7.3,p7.3, p. 457457. Explanations, diagrams and numerical practice variants are original MathLab content aligned to this section.