Skip to main content
Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 6, Problema 6.R.56c

Comparing volumes Let R be the region bounded by y=1/x^p and the x-axis on the interval [1, a], where p>0 and a>1 (see figure). Let Vₓ and Vᵧ be the volumes of the solids generated when R is revolved about the x- and y-axes, respectively.


c. Find a general expression for Vₓ in terms of a and p. Note that p=1/2 is a special case. What is Vₓ when p=1/2?

Guida verificata passo dopo passo
1
Identify the region R bounded by the curve \(y = \frac{1}{x^p}\), the x-axis, and the vertical lines \(x=1\) and \(x=a\), where \(p > 0\) and \(a > 1\).
To find the volume \(V_x\) generated by revolving the region R about the x-axis, use the disk method. The formula for the volume is: \[V_x = \pi \int_1^a \left( y \right)^2 \, dx\] Since \(y = \frac{1}{x^p}\), substitute this into the integral.
Rewrite the integral explicitly: \[V_x = \pi \int_1^a \left( \frac{1}{x^p} \right)^2 \, dx = \pi \int_1^a x^{-2p} \, dx\]
Evaluate the integral \(\int_1^a x^{-2p} \, dx\). For \(p \neq \frac{1}{2}\), use the power rule for integration: \[\int x^m \, dx = \frac{x^{m+1}}{m+1} + C\] where \(m = -2p\). So, \[\int_1^a x^{-2p} \, dx = \left[ \frac{x^{-2p+1}}{-2p+1} \right]_1^a\]
For the special case \(p = \frac{1}{2}\), the exponent becomes \(-2p = -1\), so the integral becomes: \[\int_1^a x^{-1} \, dx = \left[ \ln|x| \right]_1^a = \ln a\] Thus, the volume \(V_x\) for \(p = \frac{1}{2}\) is: \[V_x = \pi \ln a\]

Risposta video verificata per un problema simile:

Questa soluzione video è stata consigliata dai nostri tutor come utile per risolvere questo problema.
Durata del video:
2m

Concetti chiave

Ecco i concetti essenziali che devi comprendere per rispondere correttamente alla domanda.

Volume of Solids of Revolution (Disk/Washer Method)

This method calculates the volume of a solid formed by revolving a region around an axis. When revolving around the x-axis, the volume is found by integrating π times the square of the function (radius) with respect to x over the given interval. It is essential to set up the integral correctly using the function y = 1/x^p.
Video consigliato:
04:48
Finding Volume Using Disks

Handling Parameters in Integrals

The problem involves parameters p and a, which affect the shape and bounds of the region. Understanding how to integrate functions with variable exponents and parameters is crucial. This includes recognizing special cases, such as p = 1/2, where the integral may require different techniques or simplifications.
Video consigliato:
05:59
Eliminating the Parameter

Improper Integrals and Convergence

Since the function y = 1/x^p involves powers and the interval starts at 1, it is important to consider the behavior of the integral for different values of p. Some values of p may lead to convergent or divergent integrals, affecting the existence and finiteness of the volume. Recognizing these conditions ensures correct evaluation.
Video consigliato:
Percorso guidato
11:11
Improper Integrals: Infinite Intervals
Pratica correlata
Domanda del libro di testo

{Use of Tech} Decreasing velocity A projectile is fired upward, and its velocity in m/s is given by v(t) = 200e^−t/10, for t≥0.

a. Graph the velocity function, for t≥0.

230
views
Domanda del libro di testo

14–25. {Use of Tech} Areas of regions Determine the area of the given region.


The region in the first quadrant bounded by y = x/6 and y = 1−|x/2−1|

38
views
Domanda del libro di testo

27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3.

Find the area of each of the regions R₁,R₂, and R₃.

58
views
Domanda del libro di testo

Force on a dam Find the total force on the face of a semicircular dam with a radius of 20 m when its reservoir is full of water. The diameter of the semicircle is the top of the dam.

63
views
Domanda del libro di testo

27–33. Multiple regions The regions R₁,R₂, and R₃ (see figure) are formed by the graphs of y = 2√x,y = 3−x,and x=3.


Use the shell method to find an integral, or sum of integrals, that equals the volume of the solid obtained by revolving region R₃ about the line x=3. Do not evaluate the integral.

48
views
Domanda del libro di testo

43–55. Volumes of solids Choose the general slicing method, the disk/washer method, or the shell method to answer the following questions.


The region bounded by the graph of y = 4−x² and the x-axis on the interval [−2,2] is revolved about the line x = −2. What is the volume of the solid that is generated?

80
views