Question

Find equations for the bounding surfaces, set up a volume integral, and evaluate the integral to obtain a volume formula for each region. Assume that \(a, b, c, r, R,\) and h are positive constants. Find the volume of a truncated cone of height \(h\) whose ends have radii \(r\) and \(R\)

Short answer

Question: Given a truncated cone with height h and radii r and R, find the volume of the truncated cone. Answer: The volume of the truncated cone is given by the formula \(V = \frac{\pi h}{3}(R^2 + rR + r^2)\), where h is the height and r and R are the radii of the two bases.

Step-by-step solution

Visualization

Draw a larger cone of height H and base radius R, and inside it, draw a smaller cone of height h and base radius r. The larger cone will be the union of the smaller cone and the desired truncated cone.

Variable Declaration

Let's define \(V_1\) as the volume of the smaller cone, \(V_2\) as the volume of the larger cone, and V as the volume of the truncated cone. It's evident that the volume of the truncated cone will be the difference between the volumes of the two cones (i.e., \(V=V_2-V_1\)).

Find the Volume of the smaller cone (V1)

To find the volume of the smaller cone, we can use the volume formula for a cone: \(V_1 = \frac{1}{3}\pi r^2 h\)

Find the Volume of the larger cone (V2)

The total height of the larger cone is H. We'll use a similar volume formula for a cone: \(V_2 = \frac{1}{3}\pi R^2 H\)

Determine H in terms of h

To express H in terms of h, we can use similar triangles. Because the two cones are similar, their dimensions are proportional. From this proportionality, we conclude: \(\frac{H-h}{h} = \frac{R-r}{r}\) Isolate H by solving for it: \(H = \frac{(R-r)h}{r} + h\)

Substitute H back into the larger cone formula (V2)

Substitute the expression for H back into the formula for \(V_2\): \(V_2 = \frac{1}{3}\pi R^2 \left(\frac{(R-r)h}{r} + h\right)\)

Calculate the volume of the truncated cone (V)

Now, we can find the volume of the truncated cone by subtracting the volume of the smaller cone from the volume of the larger cone: \(V = V_2 - V_1\) \(V = \left[\frac{1}{3}\pi R^2 \left(\frac{(R-r)h}{r} + h\right)\right] - \left[\frac{1}{3}\pi r^2 h\right]\) Distribute \(\pi h/3\) to simplify the expression: \(V = \frac{\pi h}{3}(R^2 + rR + r^2)\) This is the formula for the volume of a truncated cone with height h and radii r and R.

Key concepts

Cone Volume Formula

The formula for the volume of a cone is an essential part of geometry. A cone can be easily visualized as a circular base tapering to a point, called the apex. For a cone with radius \( r \) and height \( h \), the volume \( V \) is given by:\[V = \frac{1}{3} \pi r^2 h\]This formula is derived by considering how a cone is related to a cylinder with the same base and height. A cone fits inside such a cylinder, taking up precisely one-third of its volume. Understanding this foundational concept can aid in visualizing more complex structures, like truncated cones.For a truncated cone, or a frustum, we modify the formula. Here, we have a cone with the top cut off parallel to its base. It has two radii — the larger base radius \( R \) and the smaller top radius \( r \), as well as a vertical height \( h \). The formula for the volume of a truncated cone is then:\[V = \frac{\pi h}{3}(R^2 + rR + r^2)\]This formula is found by subtracting the volume of the smaller cone (which is missing) from the volume of the larger cone that would include the entire shape if it were complete.

Integral Calculus

Integral calculus provides a powerful framework for finding the volume of more complex shapes, like truncated cones, through methods involving integration. When we don't have a simple formula readily available, integration allows us to consider small slices of the shape and sum them up to find the total volume.For a truncated cone, as described in our problem, the shape can be seen as a collection of many thin discs stacked along its height. These discs can vary in radius from \( r \) to \( R \). By integrating these discs — considering their varying radii as we move up the cone — we calculate the total volume.To set up the integral, we need to express the radius of each disc as a function of its height in the truncated cone. The relationship between the two can be defined using the geometry of similar triangles, giving us an equation for the radius as a function of height. We then integrate over the height of the cone, taking these varying radii into account. This approach highlights how calculus can bridge gaps where geometric intuition is not enough.

Geometry of Cones

The geometry of a cone involves understanding its basic elements and properties. A cone is a 3-dimensional shape with a circular base that converges smoothly to a point called the apex. The line from the center of the base to the apex is the perpendicular height, while any line running from the base's circumference to the apex is known as the slant height.A truncated cone, or frustum, results when a cone is sliced parallel to its base, cutting off the top and leaving two circular ends of different sizes. The geometry of the truncated cone is characterized by:

• The larger base with radius \( R \).
• The smaller top with radius \( r \).
• The perpendicular height \( h \) between these bases.

Understanding the proportionality of elements in similar cones is key. If two cones have the same shape but different sizes, they are similar, meaning their corresponding dimensions (heights, radii) are proportionally related. This concept of similarity is fundamental in deducing the relationships necessary to calculate the volumes of both complete and truncated cones.