Using the washer method, find the volume (in ${\text{m}}^3$) of the solid obtained by rotating the region bounded by $y=\sqrt{x}$, the $x$‑axis and the vertical lines $x=0$ and $x=9$ about the $x$‑axis.

Step-by-step solution

1. Identify the outer radius $R(x)$ as the distance from the $x$‑axis to the curve $y=\sqrt{x}$, so $R(x)=\sqrt{x}$; there is no inner radius because the region touches the axis.
2. Write the washer volume element $dV=\pi R(x{)}^{2}dx=\pi (\sqrt{x}{)}^{2}dx=\pi xdx$.
3. Integrate $dV$ from $x=0$ to $x=9$: $V={\int }_0^9\pi xdx$.
4. Factor out $\pi$: $V=\pi {\int }_0^{9}xdx=\pi {\left[\frac{x^2}{2}\right]}_0^9$.
5. Evaluate the antiderivative: $\frac{9^2}{2}=\frac{81}{2}=40.5$.
6. Multiply by $\pi$: $V=40.5\pi =108\pi /(\text{since }40.5=81/2=108/2)$, thus $V=108\pi$ cubic metres.
7. State the final answer.

Concept tested: Volume by washer method

Published 10 September 2026 · How this solution was prepared and checked