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Chapter 5: Problem 40

Use the disk method to verify that the volume of a right circular cone is \(\frac{1}{3} \pi r^{2} h,\) where \(r\) is the radius of the base and \(h\) is the height.

Short Answer

Expert verified
The disk method confirms the formula for the volume of a right circular cone, which is \(\frac{1}{3} \pi r^{2} h\), derived by using integral calculus.

Step by step solution

01

Set up the disk method

Consider a cone as a collection of small disks stacked upon each other. The radius of each disk can be expressed as a function \(y(x) = r\frac{x}{h}\), where \(x\) is the distance from the vertex of the cone, \(r\) is the radius of the base, and \(h\) is the height of the cone. The small change in \(x\) is presented as \(dx\). The small volume \(dV\) of the tiny disk is therefore \(dV = \pi [y(x)]^2 dx = \pi [r\frac{x}{h}]^2 dx\).
02

Integrate to find the volume

Next, the integral of this expression from \(x = 0\) to \(x = h\) is taken to sum up all the small volumes \(dV\). This gives us the total volume \(V\) of the cone: \(V = \int_0^h \pi [r\frac{x}{h}]^2 dx\).
03

Evaluate the Integral

Evaluate the above integral. Upon doing so, we get: \(V = \pi \int_0^h (\frac{r^2x^2}{h^2}) dx\). Which simplifies to \(V = \frac{\pi r^2}{h^2} \int_0^h x^2 dx\). Evaluating the integral on the rhs, we get \(V = \frac{\pi r^2}{h^2} [\frac{x^3}{3}]_0^h = \frac{\pi r^2}{h^2}[\frac{h^3}{3}-0] = \frac{1}{3} \pi r^{2} h\). This matches with the formula for the volume of a cone.

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Key Concepts

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

Volume of a Cone
The volume of a cone is a fundamental geometric concept, crucial in various fields. Specifically, a right circular cone is a three-dimensional shape with a circular base and a pointed top, or apex. Its volume formula, \(\frac{1}{3} \pi r^{2} h\), where \(r\) is the radius of the base and \(h\) is the height, can be derived using calculus.

The formula essentially tells us the space occupied inside the cone. Here's a helpful way to visualize it:
  • Imagine filling up a pyramid-like shape (the cone) with tiny stacked flat circles (disks).
  • The base's radius \(r\) and the specific height \(h\) determine the cone's dimensions.
  • The factor \(\frac{1}{3}\) indicates that the cone's volume is one-third that of a cylinder with the same base and height.
Understanding this formula helps in solving problems related to geometry and physical volumes.
Definite Integral
A definite integral is a fundamental concept in calculus, representing the area under a curve on a graph within a specified interval. It is symbolized by \(\int_{a}^{b} f(x) \, dx\), where \(f(x)\) is the function, and \(a\) and \(b\) are the limits of integration.

In geometric terms, this integral helps calculate areas, volumes, and other quantities. For the volume of a cone, it allows us to sum infinitely small slices, or disks, from the base to the apex.
  • The integral finds accumulated sums over the interval \(x = 0\) to \(x = h\), where each slice's contribution to the volume is calculated.
  • In the cone's context, it calculates the total space occupied by these 'disks'.
  • The problem employs \([r \frac{x}{h}]^2\) as the function, representing each disk's radius dependent on \(x\).
This approach is powerful, not only for cones but also for various shapes and systems in physics and engineering.
Geometric Applications of Calculus
Calculus opens numerous doors in understanding geometric structures and their properties. The disk method, specifically, is a part of calculus with practical implications for finding volumes of solids of revolution, such as cones.

Using calculus, we can turn abstract math equations into practical solutions for real-world shapes:
  • For a cone, the disk method involves dividing it into an infinite number of thin slices, or disks.
  • Each disk's volume is calculated and summed, hence arriving at the total volume.
  • Besides cones, this application can be generalized to other shapes, making calculus a universal tool in geometry.
Thanks to these mathematical techniques, designing and analyzing various architectural and mechanical systems becomes feasible. From architects to engineers, understanding these applications is crucial for those dealing with spatial dimensions and structural integrity.

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Most popular questions from this chapter

Sketch the region bounded by the graphs of the algebraic functions and find the area of the region. $$ f(x)=\sqrt[3]{x-1}, g(x)=x-1 $$

(a) use a graphing utility to graph the region bounded by the graphs of the equations, (b) find the area of the region, and (c) use the integration capabilities of the graphing utility to verify your results. $$ f(x)=\frac{1}{x^{2}} e^{1 / x}, \quad y=0, \quad 1 \leq x \leq 3 $$

Think About It Consider the equation \(\frac{x^{2}}{9}+\frac{y^{2}}{4}=1 .\) (a) Use a graphing utility to graph the equation. (b) Set up the definite integral for finding the first quadrant arc length of the graph in part (a). (c) Compare the interval of integration in part (b) and the domain of the integrand. Is it possible to evaluate the definite integral? Is it possible to use Simpson's Rule to evaluate the definite integral? Explain. (You will learn how to evaluate this type of integral in Section \(6.7 .)\)

In Exercises 61 and \(62,\) use the Second Theorem of Pappus, which is stated as follows. If a segment of a plane curve \(C\) is revolved about an axis that does not intersect the curve (except possibly at its endpoints), the area \(S\) of the resulting surface of revolution is given by the product of the length of \(C\) times the distance \(d\) traveled by the centroid of \(C\). A sphere is formed by revolving the graph of \(y=\sqrt{r^{2}-x^{2}}\) about the \(x\) -axis. Use the formula for surface area, \(S=4 \pi r^{2},\) to find the centroid of the semicircle \(y=\sqrt{r^{2}-x^{2}}\)

Let \(V\) be the region in the cartesian plane consisting of all points \((x, y)\) satisfying the simultaneous conditions \(|x| \leq y \leq|x|+3\) and \(y \leq 4\) Find the centroid \((\bar{x}, \bar{y})\) of \(V\).
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