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Volume by revolving ellipse about x=2a

Short answer

Using washers perpendicular to the axis x=2a with outer radius 2a+x₁ and inner radius 2a-x₁ gives V = π∫(R²-r²)dy = 8πa²∫√(1-y²/b²)dy = 4π²a²b cubic units.

Show that the volume of the solid formed by revolving the ellipse x²/a² + y²/b² = 1 about the line x = 2a is 4π²a²b cubic units.

Ellipse: x²/a² + y²/b² = 1. The line x = 2a is parallel to the y-axis and does not meet the ellipse (the ellipse reaches only x = a), so the solid is a ring (torus-like) shape. Use the washer method, slicing perpendicular to the axis of rotation (horizontal slices at height y).

At height y, the horizontal strip runs from x = −x₁ to x = +x₁ where x₁ = a√(1 − y²/b²). Distances from the line x = 2a:

  • outer radius R = 2a + x₁
  • inner radius r = 2a − x₁
V = π ∫_(−b)^b (R² − r²) dy
  = π ∫_(−b)^b 8a x₁ dy
  = 8πa² ∫_(−b)^b √(1 − y²/b²) dy

Since ∫(−b)^b √(b² − y²) dy = πb²/2 (the area of a semicircle of radius b), ∫(−b)^b √(1 − y²/b²) dy = (1/b)(πb²/2) = πb/2.

V = 8πa² · πb/2

V = 4π²a²b cubic units. Shown.

(Check by Pappus's theorem: area of ellipse πab times the path of its centre 2π(2a) gives 4π²a²b.)

✓ Answer checked by Claude Opus 5.5 (high effort)

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