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AP Calculus BC · Unit 8: Applications of Integration

8.10 Volume with Disc Method: Revolving Around Other Axes

Disc Method for Horizontal Axes Other Than the x-axis

When a region is revolved around a horizontal line y = k (not the x-axis), the disc's radius becomes the vertical distance between the curve and the line: r = f(x) − k. The volume is V = π∫[a,b] (f(x) − k)² dx. Always subtract the axis value before squaring — order matters for setting up correctly, though squaring removes sign issues.

Disc Method for Vertical Axes Other Than the y-axis

When revolving around a vertical line x = h, express the curve as x = g(y) and use radius r = g(y) − h. The volume becomes V = π∫[c,d] (g(y) − h)² dx, integrating with respect to y. This requires solving for x in terms of y first.

Setting Up the Radius Correctly

The radius is always the distance from the axis of revolution to the curve, so identify which is farther out. Sketch the region and axis to avoid sign errors. Since the radius is squared, a negative sign inside doesn't change the volume, but the setup must still reflect the correct geometric distance.

Key terms

Radius when revolving about y = k
r = f(x) − k, the vertical distance from the curve to the line y = k.
Radius when revolving about x = h
r = g(y) − h, the horizontal distance from the curve x = g(y) to the line x = h.
Volume formula, revolving about y = k
V = π∫[a,b] (f(x) − k)² dx
Volume formula, revolving about x = h
V = π∫[c,d] (g(y) − h)² dy
Why squaring the radius avoids sign issues
Because (f(x) − k)² = (k − f(x))², the order of subtraction doesn't affect the computed volume.

Try an AP-style question

The region bounded by y = x², y = 1, and x ≥ 0 is revolved about y = 1. What is the volume?

  1. π/5
  2. 8π/15
  3. 4π/15
  4. 2π/3
Show the answer

B. Radius 1 − x²: π∫₀¹ (1 − x²)² dx = π(1 − 2/3 + 1/5) = 8π/15.

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