Volumes by cylindrical shells
Problem 5.35 · easy
The region under \( \displaystyle y = \sqrt{x} \) for \( \displaystyle 0 \le x \le 1 \) is revolved about the y-axis. Use shells to find the volume.
- A shell at x has radius x, height f(x) and thickness dx, so its volume is 2πx f(x) dx.
- \[ \int\limits_{0}^{1} 2 \pi x^{\frac{3}{2}}\, dx = \frac{4 \pi}{5} \]V = ∫ 2πx f(x) dx.✓ Proved
Answer \( \frac{4 \pi}{5} \)
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 | 20,000 thin shells (annular slabs) summed numerically give the same volume |
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 2026-09-26 with SymPy 1.14.0.