A reservoir is shaped like a right circular cone's frustum. It measures $8 \mathrm{m}$ across the top and $4 \mathrm{m}$ across the bottom. If it is $6 \mathrm{m}$ deep its volume is ?

The concept of frustum volume is used here.

We need to determine the capacity of the frustum-shaped reservoir.

Volume of a frustum of a cone $=\frac{1}{3}\pi h\left({\mathrm{R}}^2+{\mathrm{r}}^2+\mathrm{Rr}\right)$

Where h is the height and R and r are the radii of the upper and lower ends of a frustum of a cone.

Height $h=6 \mathrm{m}$

Since, the reservoir is $8 \mathrm{m}$ across the top, upper radius $R=4 \mathrm{m}$

Similarly, lower radius $\mathrm{r}=2 \mathrm{m}$

So, volume of the reservoir $=\frac{1}{3}\times \frac{22}{7}\times 6\left(4^2+2^2+4\times 2\right){\mathrm{m}}^3$

$=\frac{22}{7}\times 2(16+4+8){\text{m}}^3$

$=\frac{22}{7}\times 2\times 28 {\text{m}}^3=176 {\text{m}}^3$

Hence, the volume of the reservoir is $176{ \mathrm{m}}^3$