The region bounded by y = x² and the x-axis from x = 0 to x = 2 is revolved about the x-axis. Which integral correctly gives the volume?
Aπ ∫₀² x² dx
Bπ ∫₀² x⁴ dx
C∫₀² x² dx
Dπ ∫₀² 2x dx
The disk method gives V = π ∫ₐᵇ [f(x)]² dx. Here f(x) = x², so [f(x)]² = (x²)² = x⁴. The most common error is forgetting to square the radius function — using x² instead of x⁴. The radius of each disk is f(x) = x², and volume requires πr², so squaring is essential. Forgetting π is another frequent mistake, but option C also drops the squaring.
Question 2 Multiple Choice
A region is bounded by y = 3 and the x-axis from x = 0 to x = 5, and revolved about the line y = 3 (not the x-axis). What is the radius of each disk cross-section?
A3, because the function value is 3
B0, because the curve lies on the axis of revolution
CThe radius varies with x — it equals f(x) − 3
D5, because the region extends to x = 5
When the curve y = 3 is revolved about the axis y = 3, the curve lies exactly on the axis of revolution. Every point on the curve has zero distance from the axis, so each disk has radius 0 — the solid is a degenerate flat disk with zero volume. The key principle: the radius is always the perpendicular distance from the axis of revolution to the curve, which here is |f(x) − 3| = |3 − 3| = 0. This is the situation where a washer method with inner radius > 0 would be needed for a different curve configuration.
Question 3 True / False
If the region between y = f(x) and the x-axis has a gap — meaning the curve does not touch the x-axis — the disk method still correctly gives the volume of the solid formed by revolving this region about the x-axis.
TTrue
FFalse
Answer: False
When there is a gap between the curve and the axis of revolution, each cross-section is not a solid disk but an annulus (washer) — a disk with a hole through the center. The washer method is required: V = π ∫ₐᵇ ([outer radius]² − [inner radius]²) dx. Applying the disk method in this situation would incorrectly include volume in the hole that doesn't exist in the actual solid.
Question 4 True / False
The volume formula for the disk method, V = π ∫ₐᵇ [f(x)]² dx, requires squaring f(x) because we are computing areas of circular cross-sections, not lengths.
TTrue
FFalse
Answer: True
Each thin cross-section perpendicular to the axis is a circle with radius f(x). The area of that circle is πr² = π[f(x)]². Multiplying by the infinitesimal thickness dx and integrating accumulates these circular areas into volume. The squaring is exactly the same πr² formula for circular area, applied infinitesimally. This is why forgetting the square is the most conceptually damaging error — it would be computing something proportional to arc length, not area.
Question 5 Short Answer
Explain in terms of circular cross-sections why the disk method formula is V = π ∫ₐᵇ [f(x)]² dx and not V = ∫ₐᵇ f(x) dx.
Think about your answer, then reveal below.
Model answer: When a region is revolved about the x-axis, each thin vertical strip of width dx sweeps out a disk. The disk's radius is f(x) — the height of the strip — and its thickness is dx. The volume of that thin disk is π·(radius)²·(thickness) = π[f(x)]²dx. The formula ∫f(x)dx would compute area of the original 2D region, not volume. The squaring comes from the cross-sectional area being πr², and the π is part of the circle area formula — both are essential.
The intuition is that area accumulates lengths (∫f(x)dx), while volume accumulates areas (∫π[f(x)]²dx). The transition from 2D to 3D adds one power of the radius — it's the difference between a line segment and the circle it sweeps when rotated.