A hemispherical bowl is made of …

I am sorry that I posted the half question.

After taking out the voulme of the bowl with outer radius,

we will  calculate the volume of the bowl with inner radius and subtract the both.

Inner radius = 4cm

Volume = \( {2 \pi R^3\over 3}\)

=( 2 X 22 X (4)³)/21

= 134 cm³

So, volume of steel used = 160.842-134

= 26.842 cm³

Ans. Inner Radius ( r) = 4 cm

Thickness of Bowl (t) = 0.25 cm 

Outer Radius (R) = r + t = 4 + 0.25 = 4.25 cm

Volume of Steel Used = Outer Volume of Bowl - Inner Volume of Bowl

=> \({2\over 3} \pi R^3 - {2\over3} \pi r^3 = {2\over 3}\pi [{R^3-r^3}]\)

=> \({2\over 3} \times {22\over 7} \times [{77-64}] = {2\over 3} \times {22\over 7} \times 13\)

=> 27.23 cm³

Thickness of hemispherical bowl = 0.25 cm 

Inner radius = 4cm

Outer radius = 4+0.25=4.25 cm 

Volume of steel used in making the bowl=

\(V = {2 \pi r^3 \over 3}\)

= (2*22*4.25*4.25*4.25)/3*7

= 160.842 cm³

=