Volumes by disks and washers
Problem 5.76 · easy
The region under \( \displaystyle y = x \) from \( \displaystyle x = 0 \) to \( \displaystyle x = 1 \) is revolved about the x-axis. Find the volume.
- Cross sections perpendicular to the x-axis are disks of radius f(x), with area πf(x)².
- \[ \int\limits_{0}^{1} \pi x^{2}\, dx = \frac{\pi}{3} \]V = ∫ A(x) dx.✓ Proved
Answer \( \frac{\pi}{3} \)
Lines: 1 proved, 1 not checked. The answer was also checked a second way, without looking at the solution. The explanation has not been reviewed yet.
The full receipt
| Line | Status | Checked by | Detail |
|---|---|---|---|
| 1 | Not checked | — | a sentence; read, not computed |
| 2 | ✓ 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 agrees |
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_disk_washer, checked 2026-09-26 with SymPy 1.14.0.