Find the volume (in cubic metres) of the solid obtained by rotating the region bounded by $y=\sqrt{x}$, the $x$‑axis, and the vertical lines $x=0$ and $x=4$ about the horizontal line $y=5$ using the washer method.

Step-by-step solution

1. Step 1: For a slice at position $x$, the region extends from $y=0$ to $y=\sqrt{x}$. Rotating about $y=5$ gives a washer with outer radius $R=5-0=5$ and inner radius $r=5-\sqrt{x}$.
2. Step 2: Washer area $A(x)=\pi (R^2-r^2)=\pi \left[25-(5-\sqrt{x}{)}^2\right]=\pi \left[25-\left(25-10\sqrt{x}+x\right)\right]=\pi (10\sqrt{x}-x)$.
3. Step 3: Integrate volume $V={\int }_0^{4}A(x)dx=\pi {\int }_0^4(10\sqrt{x}-x)dx$.
4. Step 4: Compute integrals: ${\int }_0^4\sqrt{x}dx=\frac{2}{3}{x}^{3/2}{|}_0^4=\frac{16}{3}$ and ${\int }_0^{4}xdx=\frac{x^2}{2}{|}_0^4=8$.
5. Step 5: Substitute: $V=\pi \left[10\cdot \frac{16}{3}-8\right]=\pi \left[\frac{160}{3}-8\right]=\pi \left[\frac{136}{3}\right]=\frac{136\pi }{3}{\text{m}}^3$.
6. Step 6: Conclude the required volume is $\frac{136\pi }{3}{\text{m}}^3$.

Concept tested: Volume by washer method with shifted axis

Published 1 October 2026 · How this solution was prepared and checked