Volumes by Slicing Help

Introduction to Volumes by Slicing

When we learned the theory of the integral, we found that the basic idea was that one can calculate the area of an irregularly shaped region by subdividing the region into “rectangles.” We put the word “rectangle” here in quotation marks because the region is not literally broken up into rectangles; the union of the rectangles differs from the actual region under consideration by some small errors (see Fig. 8.1). But the contribution made by these errors vanishes as the mesh of the rectangles become finer and finer.

Fig. 8.1

We will now implement this same philosophy to calculate certain volumes. Some of these will be volumes that you have heard about (e.g., the sphere or cone), but have never known why the volume had the value that it had. Others will be entirely new (e.g., the paraboloid of revolution). We will again use the method of slicing.

The Basic Strategy 

Imagine a solid object situated as in Fig. 8.2. Observe the axes in the diagram, and imagine that we slice the figure with slices that are vertical (i.e., that rise out of the x-y plane) and that are perpendicular to the x -axis (and parallel to the y -axis). Look at Fig. 8.3. Notice, in the figure, that the figure extends from x = a to x = b .

Fig. 8.2

Fig. 8.3

If we can express the area of the slice at position x as a function A (x) of x , then (see Fig. 8.4) the volume of a slice of thickness Δ x at position x will be about A ( x ) · Δ x. If P = { x ₀ , x ₁ ,..., x _k } is a partition of the interval [ a, b ] then the volume of the original solid object will be about

Fig. 8.4

As the mesh of the partition becomes finer and finer, this (Riemann) sum will tend to the integral

We declare the value of this integral to be the volume V of the solid object . 

Volume of a Right Circular Cone with a Disc Cross-Section Example

Calculate the volume of the right circular cone with base a disc of radius 3 and height 6.

Solution

Examine Fig. 8.5.

Fig. 8.5

We have laid the cone on its side, so that it extends from x = 0 to x = 6. The upper edge of the figure is the line y = 3 − x /2. At position x, the height of the upper edge is 3 − x /2, and that number is also the radius of the circular slice at position x (Fig. 8.6). Thus the area of that slice is

Fig. 8.6

We find then that the volume we seek is

You Try It: Any book of tables (see [CRC]) will tell you that the volume of a right circular cone of base radius r and height h is . This formula is consistent with the result that we obtained in the last example for r = 3 and h = 6. Use the technique of the example above to verify this more general formula.

Volume of a Solid with an Equilateral Triangle Cross-Section Example

A solid has base the unit disk in the x-y plane. The vertical cross section at position x is an equilateral triangle. Calculate the volume.

Solution

Examine Fig. 8.7.

Fig. 8.7

The unit circle has equation x ² + y ² = 1. For our purposes, this is more conveniently written as

Thus the endpoints of the base of the equilateral triangle at position x are the points . In other words, the base of this triangle is

Examine Fig. 8.8.

Fig. 8.8

We see that an equilateral triangle of side b has height . Thus the area of the triangle is . In our case then, the equilateral triangular slice at position x has area

Finally, we may conclude that the volume we seek is