Chapter 8.3, Problem 43E

To determine

To explain : Why the given solid has the same volume as a right circular cone with base radius $3$ and height $12$ .

Explanation of Solution

Given information : A solid lies between planes perpendicular to the X-axis at $x=0$ and $x=12$ .

The cross-sections by planes perpendicular to the X-axis are circular disc whose diameters run from the line $y=\frac{x}{2}$ to the line $y=x$ .

  

Concept used :

Cavalieri’s volume Theorem:

Cavalieri’s volume Theorem says that solids with equal altitudes and identical cross-section areas at each height have same volume.

The solid lies between planes perpendicular to the X-axis at $x=0$ and $x=12$ .

So, the altitude (Spread along X-axis) of the solid is $\left(12-0\right)=12$ unit.

Now, the cross-section by the planes perpendicular to the X-axis are circular disk whose diameter run from the line $y=\frac{x}{2}$ to the line $y=x$ .

So, the diameter of the cross-sectional disc at $x=12$

  $=\left(12-\frac{12}{2}\right)\text{ unit }=12-6\text{ unit =}6\text{ unit}\text{.}$

The radius of the cross-sectional disk is $\frac{6}{2}=3\text{ unit}\text{.}$

Now, if we compare the solid with cone, we infer that the solid and cone have same altitude (spread along X-axis as $\text{12}$ units.)

At $x=12$ , the radius of the cross-sections of the solid and cone are same and is equal to $3\text{ unit}$

From the above figure it is evident that the solid and cone have thye identical cross-section at each height.

Hence, keeping in mind the cavalieri’s volume theorem, we can say that the given solid and cone have the same volume.