If the variable $x$ is used for slicing, then slices are parallel to the axis of rotation, which indicates the shell method should be used. The radius of a shell is $x - \frac {1}{10}$. The height of a shell is exactly $\frac {1}{2} \ln {x} + 1$. The volume of the region is therefore given by $$\int _{\frac {1}{5}}^{1} \frac {\pi }{10} \left (10 x - 1\right ) \left (\ln {x} + 2\right )\, dx.$$ To compute the integral, we can use integration by parts. A reasonable strategy is to integrate $x - \frac {1}{10}$ and differentiate $\ln {x} + 2$. This gives the equality $$\begin {aligned} \pi \int \left (x - \frac {1}{10}\right ) \left (\ln {x} + 2\right )\, dx & = \pi \left (\left (\frac {x^{2}}{2} - \frac {x}{10}\right ) \left (\ln {x} + 2\right ) - \int \frac {1}{x} \left (\frac {x^{2}}{2} - \frac {x}{10}\right )\, dx\right ) \\ & = - \pi \left (\frac {x^{2}}{4} - \frac {x}{10}\right ) + \pi \left (\frac {x^{2}}{2} - \frac {x}{10}\right ) \left (\ln {x} + 2\right ). \end {aligned}$$ Therefore $$\begin {aligned} \pi \int _{\frac {1}{5}}^{1} \left (x - \frac {1}{10}\right ) \left (\ln {x} + 2\right )\, dx & = \left . \left [- \pi \left (\frac {x^{2}}{4} - \frac {x}{10}\right ) + \pi \left (\frac {x^{2}}{2} - \frac {x}{10}\right ) \left (\ln {x} + 2\right ) \right ] \right |_{\frac {1}{5}}^{1}\\ & = \left (\frac {13}{20} \pi \right ) - \left (\frac {\pi }{100} \right ) = \frac {16}{25} \pi . \end {aligned}$$

The mass $M$ will be given by the integral $$\int _{2 \pi }^{\frac {5}{2} \pi } \sin {x}\, dx$$ One can check that $$\begin {aligned} \int _{2 \pi }^{\frac {5}{2} \pi } \sin {x}\, dx & = 1. \end {aligned}$$ To compute the $x$-coordinate of the centroid, we also need to compute the integral $$\int _{2 \pi }^{\frac {5}{2} \pi } x \sin {x}\, dx$$ To compute the integral, we can use integration by parts. A reasonable strategy is to integrate $\sin {x}$ and differentiate $x$. This gives the equality $$\begin {aligned} \int x \sin {x}\, dx & = - x \cos {x} - \int \left (- \cos {x}\right )\, dx \\ & = - x \cos {x} + \sin {x}. \end {aligned}$$ Therefore $$\begin {aligned} \int _{2 \pi }^{\frac {5}{2} \pi } x \sin {x}\, dx & = \left . \left [- x \cos {x} + \sin {x} \right ] \right |_{2 \pi }^{\frac {5}{2} \pi }\\ & = 1 - \left (- 2 \pi \right ) = 1 + 2 \pi . \end {aligned}$$ The corret answer is the ratio of the integrals, i.e., $$\begin {aligned} \overline {x} & = \frac {1 + 2 \pi }{1} = 1 + 2 \pi . \end {aligned}$$

This integral can be computed via integration by parts. If we integrate $x$ and differentiate $\ln (x^2+1)$, we get $$\begin {aligned} \int _1^2 x \ln (x^2+1) dx & = \left . \frac {x^2}{2} \ln (x^2+1) \right |_{1}^2 - \int _1^2 \frac {x^2}{2} \frac {2x}{x^2+1} dx \\ & = 2 \ln 5 - \frac {1}{2} \ln 2 - \int _1^2 \frac {x^3}{x^2+1} dx. \end {aligned}$$ The latter integral can be simplified using polynomial long division: $\displaystyle \frac {x^3}{x^2+1} = x - \frac {x}{x^2+1}$. Therefore $$\begin {aligned} \int _1^2 x \ln (x^2+1) dx & = 2 \ln 5 - \frac {1}{2} \ln 2 - \int _1^2 x dx + \int _1^2 \frac {x}{x^2+1} dx \\ & = 2 \ln 5 - \frac {1}{2} \ln 2 - \left . \frac {x^2}{2} \right |_1^2 + \left . \frac {1}{2} \ln (x^2+1) \right |_1^2 \\ & = 2 \ln 5 - \frac {1}{2} \ln 2 - 2 + \frac {1}{2} + \frac {1}{2} \ln 5 - \frac {1}{2} \ln 2 \\ & = \frac {5}{2} \ln 5 - \frac {2}{2} \ln 4 - \frac {3}{2} - \frac {3}{2} = \ln \left ( \frac {5^5}{4} \right ) - \frac {3}{2}. \end {aligned}$$