Question

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

Answer 1

Inner radius, $r=6cm$

Thickness, $t=1cm$

Outer radius, $R=6+1=7cm$

Therefore,

Volume of steel required = $\frac{2}{3}\pi R^3-\frac{2}{3}\pi r^3$

$=\frac{2}{3}\times \frac{22}{7}\times 7^3-\frac{2}{3}\times \frac{22}{7}\times 6^3$

$=\frac{5588}{21}$ $c{m}^3$

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