$A$ solid toy is in the form of a hemisphere surmounted by a right circular cone. The height of the cone is $4\, cm$ and the diameter of the base is $8\, cm$. Determine the volume of the toy. If a cube circumscribes the toy,then find the difference of the volumes of cube and the toy. Also,find the total surface area of the toy.

(N/A) Let $r$ be the radius of the hemisphere and the cone,and $h$ be the height of the cone. Given $h = 4\, cm$ and diameter $d = 8\, cm$,so $r = 4\, cm$.
Volume of the toy $=$ Volume of the hemisphere $+$ Volume of the cone
$= \frac{2}{3} \pi r^3 + \frac{1}{3} \pi r^2 h = \frac{1}{3} \pi r^2 (2r + h)$
$= \frac{1}{3} \times \frac{22}{7} \times 4^2 \times (2 \times 4 + 4) = \frac{1}{3} \times \frac{22}{7} \times 16 \times 12 = \frac{1408}{7} \approx 201.14\, cm^3$.
$A$ cube circumscribes the toy. The height of the toy is $h + r = 4 + 4 = 8\, cm$ and the width is $8\, cm$. Thus,the edge of the cube is $a = 8\, cm$.
Volume of the cube $= a^3 = 8^3 = 512\, cm^3$.
Difference in volumes $= 512 - 201.14 = 310.86\, cm^3$.
Total surface area of the toy $=$ Curved surface area of cone $+$ Curved surface area of hemisphere
$= \pi r l + 2 \pi r^2$,where slant height $l = \sqrt{h^2 + r^2} = \sqrt{4^2 + 4^2} = 4\sqrt{2}\, cm$.
$= \pi r (l + 2r) = \frac{22}{7} \times 4 \times (4\sqrt{2} + 8) = \frac{88}{7} \times 4(\sqrt{2} + 2) = \frac{352}{7} (1.414 + 2) \approx 171.68\, cm^2$.