A hollow hemispherical basin is crafted from metal with a thickness of 3 cm. Given that the interior radius of the basin is 6 cm, find the volume of metal used to construct the basin.

The correct answer is 342π cm³.

Given Data:

Interior radius of the hemispherical basin ($r$) = 6 cm
Thickness of the metal ($t$) = 3 cm

Step 1: Find the exterior radius of the basin

The exterior radius ($R$) is calculated by adding the thickness of the metal to the interior radius:

$R=r+t$

$R=6+3=9\text{cm}$

Step 2: Formula for the volume of metal used

The volume of metal used to construct a hollow hemispherical basin is the difference between the outer hemispherical volume and the inner hemispherical volume:

$V=\frac{2}{3}\pi (R^3-r^3)$

Step 3: Calculate the volume step-by-step

Substitute $R=9\text{cm}$ and $r=6\text{cm}$ into the formula:

$V=\frac{2}{3}\pi (9^3-6^3)$

Calculate the cube of each radius:

$9^3=729$

$6^3=216$

Subtract the inner radius cubed from the outer radius cubed:

$729-216=513$

Now calculate the final volume:

$V=\frac{2}{3}\pi \times 513$

$V=2\times 171\pi$

$V=342\pi {\text{cm}}^3$

Thus, the volume of metal used to construct the basin is 342π cm³.