Q. 60 - Calculus

Question

Two of Dr. Geek’s friends got married this year, one in June and one in July. His traditional gift for such occasions is a fancy crystal bowl, whose volume corresponds to how much he likes the future spouse of the recipient. 

a)For the June wedding, Dr. Geek gave a fancy hexagonal crystal bowl with base hexagon of radius 4 inches, top hexagon of radius 8 inches, and height of 8 inches. The area of a hexagon with radius r is $A=\frac{3\sqrt{3}}{2}{r}^2$. What is the volume of this bowl, in cubic inches?

(b) For the July wedding, Dr. Geek gave a fancy octagonal crystal bowl with base radius 4, top radius 8, and height 6 inches. The area of an octagon with radius r is $A=2\sqrt{2}{r}^2$. What is the volume of this bowl, in cubic inches? 

Step-by-Step Solution

Answer

a) The volume of hexagonal bow is $776$cubic inches.

b)The volume of octagonal bow is $633.57$ cubic inches.

Step 1. Given information

a) The hexagonal bowl with base radius $4$ inches. and top radius $8$ inches and height is $8$ inches

The area of hexagonal is $A=\frac{3\sqrt{3}}{2}{r}^2$, where $r$ is the radius.

b) The octagonal bowl with base radius  $4$ inches. and top radius $8$ inches, height  is $6$ inches.

The area of octagonal is $A=2\sqrt{2}{r}^2$, where $r$  is the radius.

Part (a) Step 1. To find the volume of hexagonal bowl:

Draw an approximated  2d diagram of the hexagonal bowl;

Here $∆AOD~∆BOC$

So,

$$\begin{gathered} \frac{AO}{BO}=\frac{AD}{BC} \\ \frac{8+x}{x}=\frac{8}{4} \\ \frac{8+x}{x}=2 \\ 8+x=2x \\ 2x-x=8 \\ x=8 \end{gathered}$$

Now the volume of the bowl is the difference between volume of the large pyramid whose radius: $8$inches ,height: $16$inches and small pyramid whose radius is $4$ inches and height is $8$ inches.

So the volume of the bowl is:

$$\begin{gathered} V=\frac{1}{3}\times \left(\frac{3\sqrt{3}}{2}\times 8^2\right)\times 16-\frac{1}{3}\times \left(\frac{3\sqrt{3}}{2}\times 4^2\right)\times 8 \\ =\frac{\sqrt{3}}{2}\left(8^2\times 16-4^2\times 8\right) \\ =\frac{\sqrt{3}}{2}\left(1024-128\right) \\ =\frac{896\sqrt{3}}{2} \\ \approx 776 \end{gathered}$$

So volume of the hexgonal bowl is $776$ cubic inches.

Part (b) Step 1. To find the volume of the octagonal bowl:

Draw an approximated  2d diagram of the hexagonal bowl;

here $∆AOD~∆BOC$

So, $$\begin{gathered} \frac{AO}{BO}=\frac{AD}{BC} \\ \frac{x+6}{x}=\frac{8}{4} \\ \frac{x+6}{x}=2 \\ 2x=x+6 \\ x=6 \end{gathered}$$

So, 

Now the volume of the bowl is the difference between the volume of the large pyramid whose radius: $8$ inches ,height: $12$inches and the small pyramid whose radius is $4$ inches and height is $6$ inches.

So the volume of the bowl is:

$$\begin{gathered} V=\frac{1}{3}\left(2\sqrt{2}\times 8^2\right)\times 12-\frac{1}{3}\left(2\sqrt{2}\times 4^2\right)\times 6 \\ =\frac{2\sqrt{2}}{3}\left(8^2\times 12-4^2\times 6\right) \\ =\frac{2\sqrt{2}}{3}\left(768-96\right) \\ =\frac{2\sqrt{2}}{3}\times 672 \\ =633.57 \end{gathered}$$

So, the volume of the octagonal bowl is $633.57$ cubic inches.