Question

Find the curved surface area of a bowl. Where Radius is R, Thickness is R/8.

Answer 1

If you think that the bowl as Hollow Semi-Sphere you can answer this question with ease.

Surface Area of a Sphere = 4πr^2

Surface Area of a Semi-sphere = 1/2 x (4πr^2)

Surface Area of the bowl (inner surface) = 2πR^2

radius of the outer surface is R+R/8 ---> 9R/8

Surface Area of the bowl (outer surface) = 2π(9R/8)^2

                                                                    = 162πR^2/64

Total surface area of two curved surfaces = 2πR^2 + 162πR^2/64

                                                                     = 290πR^2/64

       

                 

Answer 2

shape of bowl seems as semi-sphere .
so, curve surface area of semi-sphere is given by $A=2\pi R^2$
but thickness of bowl is given h ,
then, $A=2\pi[(R+h)^2+R^2]$

here, radius is R and thickness is R/8
so, curved surface area , A = 2π[(R + R/8)²+ R²]
= 2π[81R²/64 + R²]
= 2π[(81 + 64)R²/64]
= 2π × (145R²/64)
= 290πR²/64

hence, curved surface area of bowl is 145πR²/32