Q. 58 - Calculus

Question

Consider the region between the graph of $f\left(x\right)=1-\cos x$ and the $x-axis$ on $\left[0,\mathrm{\pi }\right]$. For the each line of rotation, write down definite integrals that represent the volume of the resulting solid and then use a calculator or computer to approximate the integrals.

Step-by-Step Solution

Answer

The answer is $\mathrm{V}=\frac{{\mathrm{\pi }}^2}{2}-3{\mathrm{\pi }}^2\approx 24.674$

Step 1. Given information

Given the graph of $f\left(x\right)=1-\cos x$ and  $x-axis$on $\left[0,\mathrm{\pi }\right]$ and the figure is

Step 2. Let us find the volume of solid.

The volume of solid of revolution $R\left(x\right)$ and $r\left(x\right)$ rotated along $x-\mathrm{axis}$ on interval $\left[a,b\right]$ is $V=\mathrm{\pi }{\int }_{\mathrm{a}}^{\mathrm{b}}{\left(\mathrm{R}\left(\mathrm{x}\right)\right)}^2-{\left(r\left(\mathrm{x}\right)\right)}^2\mathrm{dx}$

Then, substitute the values

$$\begin{gathered} V=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\left|{\left(1-\mathrm{cosx}\right)}^2-{\left(2\right)}^2\right|\mathrm{dx} \\ V=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\left|\left({\left(1-\mathrm{cosx}\right)}^2-4\right)\right|\mathrm{dx} \\ V=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\left|\left(1-2\mathrm{cosx}+{\cos }^2\mathrm{x}-4\right)\right|\mathrm{dx} \\ V=\mathrm{\pi }{\int }_0^{\mathrm{\pi }}\left|\left(-2\mathrm{cosx}+{\cos }^2\mathrm{x}-3\right)\right|\mathrm{dx} \\ V=\mathrm{\pi }\left[\left|{\int }_0^{\mathrm{\pi }}-2\mathrm{cosxdx}+{\int }_0^{\mathrm{\pi }}{\cos }^2\mathrm{xdx}-{\int }_0^{\mathrm{\pi }}3\mathrm{dx}\right|\right] \\ \mathrm{V}=\mathrm{\pi }\left[\left|0+\frac{\mathrm{\pi }}{2}-3\mathrm{\pi }\right|\right] \\ \mathrm{V}=\frac{{\mathrm{\pi }}^2}{2}-3{\mathrm{\pi }}^2\approx 24.674 \end{gathered}$$

The answer is $\mathrm{V}=\frac{{\mathrm{\pi }}^2}{2}-3{\mathrm{\pi }}^2\approx 24.674$