Question

A hollow hemisphere bowl of thickness 1 cm has an inner radius of 6 cm. Find the volume of metal required to make the bowl.

Inner Radius r = 6 cm

Thickness, t = 1 cm

∴ Outer Radius (R) = 6 + 1 = 7 cm

∴ Volume of steel required = `2/3 πr^3 - 2/3 πr^3`

= `2/3 xx 22/7 xx square^3 - 2/3 xx 22/7 xx square^3`

= `44/21 (square - 6^3)`

= `44/21 xx (square - square)`

= `44/21 xx square`

= `5588/21` cm²

Answer 1

Inner Radius r = 6 cm

Thickness, t = 1 cm

∴ Outer Radius (R) = 6 + 1 = 7 cm

∴ Volume of steel required = `2/3 πr^3 - 2/3 πr^3`

= \[\frac{2}{3} \times \frac{22}{7} \times \boxed{7}^3 - \frac{2}{3} \times \frac{22}{7} \times \boxed{6}^3\]

= \[\frac{44}{21} (\boxed{7}^3 - 6^3)\]

= \[\frac{44}{21} \times (\boxed{343} - \boxed{216})\]

= \[\frac{44}{21} \times \boxed{127}\]

= `5588/21` cm²