The base of a cone with radius $21 \, cm$ and height $20 \, cm$ is hemispherical. Find its curved surface area and volume.

(N/A) $1$. Slant height $(l)$ of the cone: $l = \sqrt{r^2 + h^2} = \sqrt{21^2 + 20^2} = \sqrt{441 + 400} = \sqrt{841} = 29 \, cm$.
$2$. Curved Surface Area $(CSA)$ of the cone: $CSA_{cone} = \pi rl = \frac{22}{7} \times 21 \times 29 = 22 \times 3 \times 29 = 1914 \, cm^2$.
$3$. Curved Surface Area $(CSA)$ of the hemisphere: $CSA_{hemi} = 2\pi r^2 = 2 \times \frac{22}{7} \times 21 \times 21 = 2 \times 22 \times 3 \times 21 = 2772 \, cm^2$.
$4$. Total Curved Surface Area: $1914 + 2772 = 4686 \, cm^2$.
$5$. Volume of the cone $(V_{cone})$: $V_{cone} = \frac{1}{3} \pi r^2 h = \frac{1}{3} \times \frac{22}{7} \times 21 \times 21 \times 20 = 22 \times 21 \times 20 = 9240 \, cm^3$.
$6$. Volume of the hemisphere $(V_{hemi})$: $V_{hemi} = \frac{2}{3} \pi r^3 = \frac{2}{3} \times \frac{22}{7} \times 21 \times 21 \times 21 = 2 \times 22 \times 21 \times 21 = 19404 \, cm^3$.
$7$. Total Volume: $9240 + 19404 = 28644 \, cm^3$.