A region is bounded by $y=\sqrt{x}$, the $x$‑axis and the vertical line $x=1$. Find the volume of the solid generated when this region is revolved about the line $x=3$ using the method of cylindrical shells. Express your answer in terms of $\pi$.

Step-by-step solution

1. For a vertical slice at position $x$, the height of the shell is $h=\sqrt{x}$ (the $y$‑value).
2. The radius of the shell is the horizontal distance from the slice to the axis $x=3$: $r=3-x$.
3. The elemental volume by the shell method is $\text{d}V=2\pi rhdx=2\pi (3-x)\sqrt{x}dx$.
4. Integrate $x$ from $0$ to $1$: $V=2\pi {\int }_0^1(3-x)\sqrt{x}dx$.
5. Write the integrand as $3x^{1/2}-x^{3/2}$ and integrate termwise: ${\int }_0^{1}3x^{1/2}dx=3\cdot \frac{2}{3}{x}^{3/2}{|}_0^1=2$, and ${\int }_0^1{x}^{3/2}dx=\frac{2}{5}{x}^{5/2}{|}_0^1=\frac{2}{5}$.
6. Thus ${\int }_0^1(3-x)\sqrt{x}dx=2-\frac{2}{5}=\frac{8}{5}$.
7. Multiply by $2\pi$: $V=2\pi \cdot \frac{8}{5}=\frac{16\pi }{5}.$

Concept tested: Volume by shell method

Published 11 September 2026 · How this solution was prepared and checked