Percentage Increase

Let $A_1$ be the surface area of the log without the hemisphere and $A_2$ be the surface area of the log and the hemisphere.

$A_1=2\left(\dfrac{\pi}{4}\right)(d^2)+\pi d(4d)=\dfrac{1}{2}\pi d^2+4 \pi d^2=\dfrac{9}{2} \pi d^2$

$A_2=\dfrac{\pi}{4} d^2+\dfrac{1}{2}(4)\left(\dfrac{\pi}{4}\right)(d^2)+\pi d(4d)=\dfrac{1}{4}\pi d^2+\dfrac{1}{2}\pi d^2+4\pi d^2=\dfrac{19}{4}\pi d^2$

$\dfrac{A_2}{A_1}=\dfrac{\dfrac{19}{4}}{\dfrac{9}{2}}=\dfrac{19}{4} \times \dfrac{2}{9}=\dfrac{19}{18}$

$A_2=\dfrac{19}{18}A_1$

$\%~increased=\left(\dfrac{19}{18}-1\right)(100)=$ $\boxed{\dfrac{50}{9}\%}$

Let $r$ be the radius of the base of the log, then the length is $8r$ . The surface area of the log is $2\pi r(8r)+2(\pi r^2)=16\pi r^2+2\pi r^2=18 \pi r^2$

The surface area of the log and the mounted hemispherical piece of wood is $2\pi r(8r)+\pi r^2 + \dfrac{1}{2}(4\pi r^2)=16 \pi r^2 + \pi r^2 + 2\pi r^2 = 19\pi r^2$

$\%~increased~in~surface~area=\dfrac{19\pi r^2-18 \pi r^2}{18\pi r^2} \times 100\%=\boxed{\dfrac{50}{9}\%}$