Table of contents

0. Functions7h 52m
1. Limits and Continuity2h 2m
2. Intro to Derivatives1h 33m
3. Techniques of Differentiation3h 18m
4. Applications of Derivatives2h 38m
5. Graphical Applications of Derivatives6h 2m
6. Derivatives of Inverse, Exponential, & Logarithmic Functions2h 37m
7. Antiderivatives & Indefinite Integrals1h 26m
8. Definite Integrals4h 44m
9. Graphical Applications of Integrals2h 27m
- Area Between Curves
  1h 18m
- Introduction to Volume & Disk Method
  1h 8m
10. Physics Applications of Integrals 3h 16m
- Kinematics
  1h 13m
- Work
  2h 3m
11. Integrals of Inverse, Exponential, & Logarithmic Functions2h 34m
12. Techniques of Integration7h 39m
13. Intro to Differential Equations2h 55m
14. Sequences & Series5h 36m
15. Power Series2h 19m
- Introduction to Power Series
  1h 9m
- Taylor Series & Taylor Polynomials
  1h 10m
16. Parametric Equations & Polar Coordinates7h 58m

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

Calculus

9. Graphical Applications of Integrals

Introduction to Volume & Disk Method

Previous problemNext problem

2:33 minutes

Problem 6.3.4

Textbook Question

Why is the disk method a special case of the general slicing method?

Verified step by step guidance

Understand that the general slicing method involves finding the volume of a solid by slicing it into thin cross-sectional pieces perpendicular to an axis, then summing the volumes of these slices using integration.

Recognize that each slice in the general slicing method has an area $A(x)$ that depends on the position $x$, and the volume is given by the integral $V = \int_a^b A(x) \, dx$.

Note that the disk method is a specific case where each cross-sectional slice is a disk (a circle), so the area of each slice is $A(x) = \pi [R(x)]^2$, where $R(x)$ is the radius of the disk at position $x$.

Therefore, the disk method applies the general slicing formula with the particular area function of a disk, leading to the volume formula $V = \int_a^b \pi [R(x)]^2 \, dx$.

In summary, the disk method is a special case of the general slicing method because it uses circular cross sections, making the area function a specific formula for disks within the broader slicing framework.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

General Slicing Method

The general slicing method involves finding the volume of a solid by dividing it into thin cross-sectional slices perpendicular to an axis. Each slice's area is calculated, and integrating these areas over the interval gives the total volume. This method applies to solids with varying cross-sectional shapes.

Disk Method

The disk method is a specific application of the slicing method where each cross-section is a disk (circle). It is used when a solid is formed by revolving a region around an axis, and the radius of each disk corresponds to the distance from the axis to the function curve.

Relationship Between Disk and Slicing Methods

The disk method is a special case of the slicing method because it uses circular cross-sections, simplifying the area calculation to π times the radius squared. Thus, the disk method fits within the broader slicing framework by specifying the shape of the slices as disks.

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Textbook Question

9–20. Arc length calculations Find the arc length of the following curves on the given interval.

y = (x²+2)^3/2 / 3 on [0, 1]

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Textbook Question

Suppose a cut is made through a solid object perpendicular to the x-axis at a particular point x. Explain the meaning of A(x).

Textbook Question

A solid has a circular base; cross sections perpendicular to the base are squares. What method should be used to find the volume of the solid?

Textbook Question

Consider a solid whose base is the region in the first quadrant bounded by the curve y=√3−x and the line x=2, and whose cross sections through the solid perpendicular to the x-axis are squares.

a. Find an expression for the area A(x) of a cross section of the solid at a point x in [0, 2].

Textbook Question

Let R be the region bounded by the curve y=√cos x and the x-axis on [0, π/2]. A solid of revolution is obtained by revolving R about the x-axis (see figures).

c. Write an integral for the volume of the solid.

Textbook Question

Let R be the region bounded by the curve y=cos^−1x and the x-axis on [0, 1]. A solid of revolution is obtained by revolving R about the y-axis (see figures).

b. Find an expression for the area A(y) of a cross section of the solid at a point y in [0,π/2].

Textbook Question

Let R be the region bounded by the following curves. Find the volume of the solid generated when R is revolved about the given axis.

y=4−x^2,x=2, and y=4; about the y-axis

Textbook Question

Find the area of the surface generated when the given curve is revolved about the given axis.

y=x^3/2−x^1/2 / 3, for 1≤x≤2; about the x-axis

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