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Chapter 5: Problem 22

Prove that the volume of a right-circular cone of height \(h\) and radius \(r\) is \(V=\frac{1}{3} \pi r^{2} h .\) (Hint: Rotate a line starting at the origin and ending at the point \((h, r)\) about the \(x\) -axis.)

Short Answer

Expert verified
The volume of a cone is \( V = \frac{1}{3} \pi r^{2} h \).

Step by step solution

01

Understand the Solid of Revolution

To find the volume of the cone, visualize the line from the origin (0, 0) to the point (h, r) on the xy-plane. By rotating this line about the x-axis, a right-circular cone is formed.
02

Set Up the Integral

When the line is rotated, the resulting cone's height lies along the x-axis, from 0 to h. The radius at any point along the x-axis can be described as a function of x. Since the line is straight from (0,0) to (h,r), the function describing the radius is linear: \( r(x) = \frac{r}{h} x \).
03

Use the Disk Method Formula

The volume of the cone can be computed by integrating the cross-sectional area of the disks formed by the rotating line segment. The formula for this is: \[ V = \pi \int_{0}^{h} [r(x)]^{2} \, dx \].
04

Substitute and Simplify

Substitute the function \( r(x) = \frac{r}{h} x \) into the integral: \[ V = \pi \int_{0}^{h} \left(\frac{r}{h} x\right)^{2} \ dx = \pi \int_{0}^{h} \frac{r^{2}}{h^{2}} x^{2} \ dx \].
05

Integrate

Now integrate the function: \[ V = \pi \frac{r^{2}}{h^{2}} \int_{0}^{h} x^{2} \, dx \]. The integral of \( x^2 \) is \( \frac{x^3}{3} \), so we get: \[ V = \pi \frac{r^{2}}{h^{2}} \left. \frac{x^{3}}{3} \right|_{0}^{h} \].
06

Evaluate the Integral

Evaluate the integral at the boundaries: \[ V = \pi \frac{r^{2}}{h^{2}} \left( \frac{h^{3}}{3} - 0 \right) = \pi \frac{r^{2}}{h^{2}} \frac{h^{3}}{3} \].
07

Simplify the Final Expression

Simplify the expression: \[ V = \pi \frac{r^{2}}{h^{2}} \frac{h^{3}}{3} = \pi \frac{r^{2} h}{3} \]. This confirms that the volume of the cone is indeed \[ V = \frac{1}{3} \pi r^{2} h \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Solid of Revolution
When studying the volume of shapes, the 'Solid of Revolution' is an essential concept. It involves rotating a 2D plane curve around an axis to create a 3D object. Imagine taking a thin strip of paper and spinning it around a stick. The shape formed is a solid of revolution. In the case of this problem, we rotate a straight line from the origin (0, 0) to a point (h, r) around the x-axis. This rotation produces a 3D cone.
Disk Method
The 'Disk Method' helps in finding the volume of a solid of revolution by slicing the solid into thin, disk-shaped pieces. To visualize this, think about stacking a pile of coins (disks), where each disk's volume is easy to calculate. The volume of each disk is given by \( \pi [r(x)]^2 dx \). Here, \[ r(x) \] stands for the radius of the disk at a particular x-value, and \[ dx \] is its thickness. By summing (integrating) all these disk volumes from 0 to h, we get the total volume of the cone.
Definite Integral
The 'Definite Integral' is a powerful tool in calculus used to compute the accumulation of quantities, such as area and volume. For our cone problem, we use the definite integral to sum up the volumes of all the disks described earlier. The integral \[ \pi \int_{0}^{h} \left(\frac{r}{h} x\right)^{2} \, dx \] accumulates the volume of disks from the base (x=0) to the top (x=h) of the cone. By solving this integral, we capture the total volume, ensuring no part of the cone is left out.
Volume of Right-Circular Cone
To find the volume of a right-circular cone with height \ h \ and radius \ r \, we start from a geometric perspective. By demonstrating the steps taken, here is a summary:
  • Visualize the cone as a solid of revolution by rotating the line from (0,0) to (h,r) around the x-axis.
  • Employ the disk method and set up the integral: \[ \pi \int_{0}^{h} \left( \frac{r}{h} x \right)^{2} dx \].
  • Simplify the integral to: \ \pi \frac{r^{2}}{h^{2}} \int_{0}^{h} x^{2} dx \.
  • Integrate \[ x^{2} \]: \ \frac{x^{3}}{3} \ and apply the boundaries from 0 to h.
  • The final volume becomes: \[ V = \frac{1}{3} \pi r^{2} h \].

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