$A$ rocket is in the form of a right circular cylinder closed at the lower end and surmounted by a cone with the same radius as that of the cylinder. The diameter and height of the cylinder are $6 \, cm$ and $12 \, cm$,respectively. If the slant height of the conical portion is $5 \, cm$,find the total surface area and volume of the rocket. [Use $\pi = 3.14$]

(N/A) The rocket is a combination of a right circular cylinder and a cone.
Given: Diameter of the cylinder $= 6 \, cm$.
Radius of the cylinder $(r) = \frac{6}{2} = 3 \, cm$.
Height of the cylinder $(H) = 12 \, cm$.
Volume of the cylinder $= \pi r^2 H = 3.14 \times (3)^2 \times 12 = 3.14 \times 9 \times 12 = 339.12 \, cm^3$.
Curved surface area of the cylinder $= 2 \pi r H = 2 \times 3.14 \times 3 \times 12 = 226.08 \, cm^2$.
For the cone,slant height $(l) = 5 \, cm$ and radius $(r) = 3 \, cm$.
Height of the cone $(h) = \sqrt{l^2 - r^2} = \sqrt{5^2 - 3^2} = \sqrt{25 - 9} = \sqrt{16} = 4 \, cm$.
Volume of the cone $= \frac{1}{3} \pi r^2 h = \frac{1}{3} \times 3.14 \times (3)^2 \times 4 = 3.14 \times 3 \times 4 = 37.68 \, cm^3$.
Curved surface area of the cone $= \pi r l = 3.14 \times 3 \times 5 = 47.1 \, cm^2$.
Total volume of the rocket $= \text{Volume of cylinder} + \text{Volume of cone} = 339.12 + 37.68 = 376.8 \, cm^3$.
Total surface area of the rocket $= \text{CSA of cone} + \text{CSA of cylinder} + \text{Area of base of cylinder} = 47.1 + 226.08 + (\pi r^2) = 47.1 + 226.08 + (3.14 \times 3^2) = 47.1 + 226.08 + 28.26 = 301.44 \, cm^2$.