Volumes with known cross sections
Problem 5.472 · medium
The base of a solid is the region under \( \displaystyle y = \sqrt{x} \) for \( \displaystyle 0 \le x \le 4 \). Cross-sections perpendicular to the \( \displaystyle x \)-axis are isosceles right triangles with a leg in the base. Find the volume of the solid.
- \[ \sqrt{x} \]At position x the base of the cross-section has length (top − bottom).✓ Proved
- \[ \frac{x}{2} \]Area of one cross-section: 1/2·(side)².✓ Proved
- \[ \int\limits_{0}^{4} \frac{x}{2}\, dx = 4 \]Volume = ∫ A(x) dx across the base.✓ Proved
Answer \( 4 \)
Every line of this solution was proved by the computer algebra system SymPy. The answer was also checked a second way, without looking at the solution. The reviewers disagree about how one step is explained; every verdict is in the receipt.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 2 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| 3 | ✓ Proved | sympy 1.14.0 | simplify(lhs - rhs) reduced to 0 |
| answer, a second way | ✓ Checked independently | sympy 1.14.0 + mpmath 1.3.0 | numerical quadrature of the cross-sectional area |
Reviewers
gpt-oss:20b: passqwen3.6:27b-mlx: fail (error) — The solution incorrectly assumes the leg of the triangle is x, whereas the leg length is determined by the base function sqrt(x). Consequently, the area should be (sqrt(x))^2 / 2 = x/2, which is coincidentally correct in this specific case, but the reasoning in line 1 implies the side length is x (or fails to justify why side^2 = x). More critically, line 1 states 'sqrt(x) = sqrt(x)' and claims this is the base length, but does not explicitly state that the side length s = sqrt(x). While the algebraic result x/2 is correct for the area, the derivation is opaque and potentially misleading if the student thinks the side length is x. However, looking closer: Line 1 says 'At position x the base... has length (top - bottom)'. For y=sqrt(x) and y=0, the length is sqrt(x). Line 2 says 'x/2 = x/2' and claims 'Area... 1/2*(side)^2'. If side = sqrt(x), Area = 1/2 * x. The equation x/2 = x/2 is tautological and doesn't show the substitution of side=sqrt(x). It skips the crucial step of substituting the side length into the area formula. A student might think the side length is x. This is misleading.
Every verdict on record (4)
gpt-oss:20b: pass 2026-10-11qwen3.6:27b-mlx: fail (error) 2026-10-11 — The solution incorrectly assumes the leg of the triangle is x, whereas the leg length is determined by the base function sqrt(x). Consequently, the area should be (sqrt(x))^2 / 2 = x/2, which is coincidentally correct in this specific case, but the reasoning in line 1 implies the side length is x (or fails to justify why side^2 = x). More critically, line 1 states 'sqrt(x) = sqrt(x)' and claims this is the base length, but does not explicitly state that the side length s = sqrt(x). While the algebraic result x/2 is correct for the area, the derivation is opaque and potentially misleading if the student thinks the side length is x. However, looking closer: Line 1 says 'At position x the base... has length (top - bottom)'. For y=sqrt(x) and y=0, the length is sqrt(x). Line 2 says 'x/2 = x/2' and claims 'Area... 1/2*(side)^2'. If side = sqrt(x), Area = 1/2 * x. The equation x/2 = x/2 is tautological and doesn't show the substitution of side=sqrt(x). It skips the crucial step of substituting the side length into the area formula. A student might think the side length is x. This is misleading.qwen3.6:27b-mlx: fail (error) 2026-10-11 — The solution incorrectly identifies the side length of the triangle as x instead of sqrt(x), leading to an incorrect area formula and volume.gpt-oss:20b: pass 2026-10-11
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_cross_sections, checked 2026-10-11 with SymPy 1.14.0.