2011SP-II-17

$\begin{array}{rcl}
a & = & 2\pi r \times \dfrac{r}{2} \\
& = & \pi r^2
\end{array}$

對於該半球體，

$\begin{array}{rcl}
b & = & \dfrac{1}{2} \times 4\pi r^2 \\
& = & 2\pi r^2
\end{array}$

對於該直立圓錐體，

$\begin{array}{rcl}
c & = & \pi r \times \sqrt{(2r)^2+r^2} \\
& = & \sqrt{5}\pi r^2
\end{array}$

所以，$a < b < c$。