Volumes by cylindrical shells

Reviewers

• gpt-oss:20b: pass
• qwen3.6:27b-mlx: fail (style) — [domain objection, downgraded to style] The integral of 6πx from 0 to 2 is 12π, not 12π. Wait, ∫₀² 6πx dx = [3πx²]₀² = 12π. The calculation is correct. Let me re-read carefully. Problem: y = 3/x. Shell method: V = ∫ 2πx * (3/x) dx = ∫ 6π dx from 0 to 2. ∫₀² 6π dx = 12π. Wait, the function is y = 3/x. At x=0, this is undefined (vertical asymptote). The region is 0 ≤ x ≤ 2. The integral ∫₀² (3/x) dx diverges. However, for the shell method, the integrand is 2πx * (3/x) = 6π. The integral ∫₀² 6π dx converges to 12π. Is the setup valid? The region is bounded by y=3/x, x=2, y=0? No, 'region under y=3/x for 0≤x≤2'. Usually 'under' implies bounded by the x-axis. So the region is 0 < x ≤ 2, 0 ≤ y ≤ 3/x. This region has infinite area, but finite volume when revolved? Let's check the volume using washers to be sure. Outer radius R(y) = 2. Inner radius r(y) = 3/y. Range of y: As x→0, y→∞. As x=2, y=1.5. So y goes from 1.5 to ∞. V = ∫_{1.5}^{∞} π (R² - r²) dy = ∫_{1.5}^{∞} π (4 - (3/y)²) dy. This integral diverges because ∫ 4 dy diverges. So the volume is actually infinite. Why did shells give 12π? Shell formula: V = ∫ 2πx f(x) dx. Here f(x) = 3/x. Integrand is 6π. ∫₀² 6π dx = 12π. Where is the discrepancy? The shell method assumes the region is bounded. If the region is unbounded (infinite height at x=0), does the shell method still apply? The volume of revolution of an unbounded region can be finite or infinite. Let's look at the washer method again. The region is 0 < x ≤ 2, 0 ≤ y ≤ 3/x. Revolving around y-axis. For a given y, the horizontal strip goes from x = 3/y to x = 2. This is valid for y ≥ 3/2 = 1.5. What about y < 1.5? For y < 1.5, the curve y=3/x is at x > 2. But the region is restricted to x ≤ 2. So for y < 1.5, the region is bounded by x=0? No, x starts from 0? The problem says 'region under y=3/x for 0≤x≤2'. This usually means the area between the curve and the x-axis, bounded by x=0 and x=2. But at x=0, y is undefined. If we interpret the region as the limit as a→0+ of the region a≤x≤2, 0≤y≤3/x. Then for any fixed y, the x-interval is [max(a, 3/y), 2]. If y is large (y > 3/a), the interval is empty? No. Let's stick to the shell method derivation. V = lim_{a→0+} ∫_a^2 2πx (3/x) dx = lim_{a→0+} ∫_a^2 6π dx = 12π. Why did the washer method diverge? Washer method: The solid is generated by revolving the region. The region extends to y=∞. The 'hole' in the middle is defined by x = 3/y. The outer boundary is x=2. So for any y, the cross section is an annulus with inner radius 3/y and outer radius 2. This is true for y such that 3/y ≤ 2, i.e., y ≥ 1.5. What about y < 1.5? For y < 1.5, the curve x=3/y is at x > 2. The region is defined for x ≤ 2. So for y < 1.5, the entire strip from x=0 to x=2 is part of the region? No, the region is 'under' the curve. If x=1, y=3. If x=0.1, y=30. The region is the set {(x,y) | 0 < x ≤ 2, 0 ≤ y ≤ 3/x}. For a fixed y, what are the x values? We need 0 < x ≤ 2 AND y ≤ 3/x => x ≤ 3/y. So x ∈ (0, min(2, 3/y)]. Case 1: y ≥ 1.5. Then 3/y ≤ 2. So x ∈ (0, 3/y]. Wait, if x ∈ (0, 3/y], then y ≥ 3/x is FALSE. The condition is y ≤ 3/x. So x ≤ 3/y. Also x > 0. So for a fixed y, the x-values in the region are 0 < x ≤ 3/y. BUT we also have the constraint x ≤ 2. So x ∈ (0, min(2, 3/y)]. If y ≥ 1.5, min(2, 3/y) = 3/y. So the cross section at height y is a disk of radius 3/y? No, we are revolving around the y-axis. The region is in the first quadrant. The solid is formed by rotating this area. For y ≥ 1.5, the region at height y spans x from 0 to 3/y. Rotating this gives a solid disk of radius 3/y. Volume contribution: ∫_{1.5}^{∞} π (3/y)² dy = 9π ∫_{1.5}^{∞} y^{-2} dy = 9π [-1/y]_{1.5}^{∞} = 9π (1/1.5) = 6π. If y < 1.5, min(2, 3/y) = 2. So the region at height y spans x from 0 to 2. Rotating this gives a solid disk of radius 2. Volume contribution: ∫_{0}^{1.5} π (2)² dy = 4π (1.5) = 6π. Total Volume = 6π + 6π = 12π. So the volume IS 12π. The washer method works if you split the integral. The shell method gave 12π directly. The solution is correct. Why did I think the washer method diverged earlier? I assumed the region was between x=3/y and x=2. That would be the region 'above' the curve y=3/x, bounded by x=2 and y=0? No. The region 'under' y=3/x is 0 ≤ y ≤ 3/x. So for a fixed y, x goes from 0 to 3/y (capped at 2). My previous washer setup assumed the region was bounded on the left by the curve and on the right by x=2. That corresponds to the region where x ≥ 3/y, i.e., y ≥ 3/x? No, x ≥ 3/y implies y ≥ 3/x is false. x ≥ 3/y <=> y ≥ 3/x? No. If x=2, y=1. 3/2 = 1.5. 1 < 1.5. So points (2,1) are UNDER the curve. Points (1, 4) are ABOVE the curve. The region under the curve includes the x-axis. So for a fixed y, the x-values are those where the point (x,y) is under the curve. This means y ≤ 3/x => x ≤ 3/y. And x is in [0,2]. So x ∈ [0, min(2, 3/y)]. This creates a solid cylinder for low y, and a tapering shape for high y. The solution is correct. Verdict: pass. Notes: The solution correctly applies the shell method. The integrand simplifies to a constant, yielding a finite volume despite the infinite height of the region at x=0.

Proved: SymPy reduced the difference between the two sides to zero. Checked independently: a separate method, named above, confirmed it. Checked numerically: the two sides agree at every sampled point, which is evidence, not proof. Reviewed: a model or a person read it; that is all a sentence can have. Solution by generator:structured/volume_shells, checked 2026-10-04 with SymPy 1.14.0.