Let us denote the inner radius by rcm, outer radius by R cm and length of the metal pipe by h cm. Then,

`r =2, R =2.2 and h = 77`

(i) Inner curved surface area

`= (2pi rh) cm^(2)`

`= (2 xx (22)/(7) xx 2 xx 77) cm^(2) = 968 cm^(2)`

(ii) Outer curved surface area

`= (2pi Rh) cm^(2)`

`= (2 xx (22)/(7) xx 2.2 xx 77) cm^(2) = 1064.8 cm^(2)`

(iii) Total surface area

= (inner curved surface area) + (outer curved surface area) + (area of two bases)

`= {968+ 1064.8 + 2(pi R^(2) - pi r^(2))} cm^(2)`

`= {2032.8 + 2pi (R^(2) - r^(2))} cm^(2)`

`= {2032,8 + 2 xx (22)/(7) xx (R + r) (R -r) } cm^(2)`

`= {2032.8 + (44)/(7) xx (2.2 + 2) xx (2.2 -2)} cm^(2)`

`= {2032.8 + (44)/(7) xx 4.2 xx 0.2} cm^(2)`

` (2032.8 + (44)/(7) xx (42)/(10) xx (2)/(10)) cm^(2)`

`= (2032.8 + (132)/(25)) cm^(2) = (2032.8 + 5.28) cm^(2) = 2038.08 cm^(2)`