Volume of a frustrum

For derivation of the volume formula, the reader is referred to this wiki

The volume is given by

$V = \dfrac{\pi h}{3} (r_1^2 + r_1 r_2 + r_2^2 )$

$= \dfrac{7 \pi }{3} ( 10^2 + (10)(4) + 4^2 )$

$= \dfrac{7 \pi }{3} ( 156 ) = 364 \pi$

therefore, $n = 364$

Cavalieri's Principle apllies to this problem. The obliquity can be ignored.

The bases of the full cone, $r=10$ , and of the removed cone, $r=4$ , are specified in the problem. The height of the full cone is needed.

Solving $0=b+10 m\land 7=b+4 m$ gives $b\to \frac{35}{3},m\to -\frac{7}{6}$ . The $y$ intercept, $b$ , gives the height of the full cone.

The formula for the volume of a right (vertical) cone is well-known: $\frac{1}{3}\,\pi\,h\,r^2$ .

$\text{Volume}\left[\text{Cone}\left[\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 0 & \frac{35}{3} \\ \end{array} \right),10\right]\right]-\text{Volume}\left[\text{Cone}\left[\left( \begin{array}{ccc} 0 & 0 & 7 \\ 0 & 0 & \frac{35}{3} \\ \end{array} \right),4\right]\right] \Rightarrow 364\pi$

In the Cone function, the top row of the matrix is the center of the base circle and the bottom row is the location of the tip of the cone.

Full cone volume is $\frac{1}{3} \pi h r^2\text{ substituting }\left\{r\to 10,h\to \frac{35}{3}\right\} \Rightarrow \frac{3500 \pi }{9}$ .

Removed cone volume is $\frac{1}{3} \pi h r^2\text{ substituting }\left\{r\to 4,h\to \frac{35}{3}-7\right\} \Rightarrow \frac{224 \pi }{9}$ .

$\frac{3500 \pi }{9}-\frac{224 \pi }{9} \Rightarrow \frac{3500-224}{9}\pi \Rightarrow \frac{3276}{9}\pi \Rightarrow 364 \pi$