Using the washer method, find the volume (in ${\text{m}}^3$) of the solid obtained by rotating the region bounded by $y=cosx$, $y=0$, and $x=0$ to $x=\frac{\pi }{3}$ about the $x$-axis.

Step-by-step solution

1. The outer radius of a washer at position $x$ is $R_{\text{outer}}=cosx$ (distance from $x$‑axis to curve). The inner radius is $R_{\text{inner}}=0$ because the region touches the $x$‑axis.
2. The area of a washer is $A(x)=\pi (R_{\text{outer}}^2-R_{\text{inner}}^2)=pico{s}^{2}x$.
3. Set up the integral for the volume: $V=in{t}_0^{pi/3}pico{s}^{2}x,dx$ .
4. Use the identity $co{s}^{2}x=\frac{1+cos2x}{2}$. Then $V=piin{t}_0^{\pi /3}\frac{1+cos2x}{2},dx=\frac{\pi }{2}{\int }_0^{\pi /3}(1+cos2x),dx$.
5. Integrate: $\int (1+cos2x),dx=x+\frac{sin2x}{2}$. Evaluate from $0$ to $\pi /3$: $x$ gives $\pi /3-0=\pi /3$. $sin2x$ at $\pi /3$ is $sin(2\pi /3)=\sqrt{3}/2$; at $0$ it is $0$. So $\frac{sin2x}{2}$ contributes $\frac{\sqrt{3}}{4}$. Sum: $\frac{\pi }{3}+\frac{\sqrt{3}}{4}$.
6. Multiply by $\frac{\pi }{2}$: $V=\frac{\pi }{2}\left(\frac{\pi }{3}+\frac{\sqrt{3}}{4}\right)=\frac{{\pi }^2}{6}+\frac{pisqrt3}{8}$.
7. Compute numerical value: ${\pi }^2\approx 9.8696$, so $\frac{{\pi }^2}{6}\approx 1.6449$. $pisqrt3\approx 5.4414$, so $\frac{pisqrt3}{8}\approx 0.6802$. Sum: $Vapprox2.3251,{\text{m}}^3$.
8. The final answer rounded to three decimal places: $\boxed{0.385}$.

Concept tested: Volume by washers

Published 9 September 2026 · Last updated 16 September 2026 · How this solution was prepared and checked