Plz help..The internal and external diameters of a hollow hemispherical vessel are 14 cm and 21 cm respectively. The cost of silver plating of 1...

The hemispherical vessal has two surfaces. Internal and external.

Let R and r be the eadius of the internal and external radius. Also radius is half the diameter.

The outer surface area of the hemisphere = (1/2)( 4*pi*R^2 ) =2 pi*(21/2)^2 = 2770.88472/4 sq cm = 692.72118

The area of the internal hemisphere = (1/2)*(4pir^2) = 2pi*r^2 = 2pi*(14/2)^2 = 1231.50432/4 sq cm = 307.8760801 sqcm

The ring or edge area = pi *R^2-Pir^2 = pi*(R^2-r^2) = pi*(21^2-14^2)/4 =769.6903001/4 = 192.422575 .

Thererefore the total area = 692.72118+307.87608+192.422575  = 1193.019835 sq cm

Therefore the plating cost @ Rs 1.60 for the above total area = 1193.019835 sq cm * Rs1.6sqcm = Rs 1908.83