Pappus, Composite and Advanced Solids — Worked Examples

These examples emphasize method selection: direct formula, centroid-based revolution, geometric decomposition, exact prismatoidal evaluation, or numerical integration from measured sections.

Example 1: Volume of an ellipsoid

An ellipsoidal tank has semi-axes a=4.00 ma=4.00\,\text{m}, b=3.00 mb=3.00\,\text{m}, and c=2.00 mc=2.00\,\text{m}. Determine its geometric volume.

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Example 2: Ring torus volume and surface area

A ring torus has major radius R=5.00 mR=5.00\,\text{m} and tube radius r=1.20 mr=1.20\,\text{m}. Determine its volume and surface area.

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Example 3: Derive torus volume using Pappus

A circle of radius 1.20 m1.20\,\text{m} is revolved about an external coplanar axis 5.00 m5.00\,\text{m} from the circle center. Use Pappus-Guldinus to determine the generated volume.

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Example 4: Derive torus surface area using Pappus

Use Pappus-Guldinus to determine the surface area generated when a circle of radius 1.20 m1.20\,\text{m} is revolved about the same external axis 5.00 m5.00\,\text{m} from its center.

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Example 5: Generate a sphere from a semicircular area using Pappus

A semicircular area of radius 3.00 m3.00\,\text{m} is revolved through 360∘360^\circ about its bounding diameter. Use Pappus-Guldinus to determine the generated volume.

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Example 6: Pappus surface theorem with a semicircular arc

A semicircular wire arc of radius r=2.00 mr=2.00\,\text{m} is revolved about a coplanar line parallel to its diameter and located 5.00 m5.00\,\text{m} from the diameter on the side opposite the arc. Determine the generated surface area.

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Example 7: Exact prismatoidal volume of a square frustum

A square frustum has end areas A1=100 m2A_1=100\,\text{m}^2 and A2=36.0 m2A_2=36.0\,\text{m}^2. The corresponding side lengths are therefore 10.0 m10.0\,\text{m} and 6.00 m6.00\,\text{m}. Its height is 9.00 m9.00\,\text{m}. Determine the volume using the prismatoidal formula.

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Example 8: Composite trapezoidal volume from cross-sections

Five equally spaced cross-sections are 10.0 m10.0\,\text{m} apart and have areas 20.020.0, 28.028.0, 35.035.0, 41.041.0, and 44.0 m244.0\,\text{m}^2. Estimate the enclosed volume using the composite trapezoidal rule.

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Example 9: Simpson's one-third estimate from the same sections

Use the five equally spaced areas from Example 8 to estimate volume with Simpson's 1/31/3 rule.

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Example 10: Reject Simpson's one-third rule for unequal spacing

Four measured cross-sections occur at stations 00, 88, 1818, and 30 m30\,\text{m}. Their areas are known, but the spacing is not uniform. Determine whether the equal-spacing composite Simpson's 1/31/3 formula from the lesson may be applied directly.

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Example 11: Composite storage tank capacity

A tank consists of a vertical cylinder of radius 2.00 m2.00\,\text{m} and height 5.00 m5.00\,\text{m} topped by a hemisphere of the same radius. Determine its total internal geometric capacity.

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Example 12: Subtractive composite solid

A rectangular concrete block measures 4.00 m×3.00 m×2.00 m4.00\,\text{m}\times3.00\,\text{m}\times2.00\,\text{m}. A cylindrical opening of radius 0.500 m0.500\,\text{m} is bored completely through the 2.00 m2.00\,\text{m} dimension. Determine the remaining concrete volume.

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Advanced-method check

Pappus requires the correct curve or area centroid and a valid axis of revolution. Simpson's 1/31/3 rule requires equal spacing and an even number of intervals. Composite surface area requires a separate exposed-surface inventory; it cannot be obtained by blindly adding component total areas.