# solid geometry

Powder grain manufactured for the use of the large guns are small cylinders (composed of a nitrocellulose compound) with a number of longitudinal perforations. if each grain is 3/4 diameter, 1 3/4 in. long, and contains 7 cylindrical perforations 1/8 in. in diameter, find the amount of material in single grain.

## To answer the problem above,

To answer the problem above, we need a figure.....

On how one piece of grain would look like, I believe it looks like this:

The problem asks for the amount of material of a single grain, which means we need to get its volume of a single grain.

$$Volume \space of \space the \space material \space in \space a \space grain \space = \space Volume \space of \space small \space cylinder \space - \space Volume \space of \space seven \space cylindrical \space perforations$$

Volume of a cylinder is $$V = \pi r^2h$$

where $r$ is the radius of the base and $h$ is the height of the cylinder. Looking at the preceding figure:

$$Volume \space of \space the \space material \space in \space a \space grain \space = \pi r_1^2h_1^2 - 7(\pi r_2^2h_1^2)$$ $$Volume \space of \space the \space material \space in \space a \space grain \space = \pi \left(\frac{0.75}{2} \space in.\right)^2(1.75 \space in.)^2 - 7\left(\pi \left(\frac{0.125 \space in.}{2} \right)^2(1.75 \space in.)^2\right)$$ $$\color{green}{Volume \space of \space the \space material \space in \space a \space grain \space = 0.62 \space in^3}$$

The amount of the material in a single grain is $$\color{green}{0.62 \space in^3}$$

Or grain is cylindrical in shape...You're not mistaken....heheheheh

Alternate solutions are encouraged....