/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 Scaling surface area Let \(f\) b... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 41

Scaling surface area Let \(f\) be a nonnegative function with a continuous first derivative on \([a, b]\) and suppose \(g(x)=c f(x)\) and \(h(x)=f(c x),\) where \(c>0 .\) When the curve \(y=f(x)\) on \([a, b]\) is revolved about the \(x\) -axis, the area of the resulting surface is \(A\). Evaluate the following integrals in terms of \(A\) and \(c\) a. \(\int_{a}^{b} 2 \pi g(x) \sqrt{c^{2}+g^{\prime}(x)^{2}} d x\) b. \(\int_{a / c}^{b / c} 2 \pi h(x) \sqrt{c^{2}+h^{\prime}(x)^{2}} d x\)

Short Answer

Expert verified
Answer: a. \(\int_{a}^{b} 2 \pi g(x) \sqrt{c^{2}+g^{\prime}(x)^{2}} d x = \frac{A}{2\pi}\int_{a}^{b}2\pi \cdot c \sqrt{1 + c^{2} (f'(x))^2} dx\) b. \(\int_{a / c}^{b / c} 2 \pi h(x) \sqrt{c^{2}+h^{\prime}(x)^{2}} d x = \frac{A}{2\pi c}\int_{a}^{b}2\pi \cdot \frac{1}{c} \sqrt{1 + c^{2} (f'(u))^2} du\)

Step by step solution

01

Surface Area of Revolution for f(x)

The surface area of revolution for a curve \(y=f(x)\) revolved about the x-axis on the interval \([a, b]\) is given by: $$ A = \int_{a}^{b} 2\pi f(x)\sqrt{1 + (f'(x))^2} dx. $$
02

Scaling the function f(x)

We have two functions, \(g(x) = cf(x)\) and \(h(x) = f(cx)\), which are scaled versions of \(f(x)\). We also need to find the derivatives of these two functions: 1. \(g(x) = cf(x)\), thus $$ g'(x) = cf'(x). $$ 2. \(h(x) = f(cx)\), thus $$ h'(x) = f'(cx) \cdot c. $$
03

Surface Area of Revolution for g(x)

Using the formula for the surface area, we have: $$ \int_{a}^{b} 2\pi g(x) \sqrt{1 + (g'(x))^2} dx = \int_{a}^{b} 2\pi (cf(x))\sqrt{1 + (cf'(x))^2} dx. $$ Factor out constants from the expression of the surface area: $$ c\int_{a}^{b} 2\pi f(x) \sqrt{1 + c^{2} (f'(x))^2} dx = \frac{A}{2\pi}\int_{a}^{b}2\pi \cdot c \sqrt{1 + c^{2} (f'(x))^2} dx. $$
04

Surface Area of Revolution for h(x)

Similarly, the surface area of revolution for the curve \(h(x)\) is given by: $$ \int_{a / c}^{b / c} 2\pi h(x) \sqrt{1 + (h'(x))^2} dx = \int_{a / c}^{b / c} 2\pi f(cx) \sqrt{1 + (cf'(cx))^2} dx. $$ To solve this integral, we apply the substitution method. Let \(u = cx\), then \(x = u/c\), and \(\frac{du}{dx} = c\). The new integral becomes: $$ \int_{a}^{b} 2\pi f(u) \sqrt{1 + (cf'(u))^2} \frac{1}{c} du. $$ Factor out constants from the expression of the surface area: $$ \frac{1}{c}\int_{a}^{b} 2\pi f(u) \sqrt{1 + c^{2} (f'(u))^2} du = \frac{A}{2\pi c}\int_{a}^{b}2\pi \cdot \frac{1}{c} \sqrt{1 + c^{2} (f'(u))^2} du. $$ So, the expressions for the integrals of g(x) and h(x) in terms of A and c are: a. \(\int_{a}^{b} 2 \pi g(x) \sqrt{c^{2}+g^{\prime}(x)^{2}} d x = \frac{A}{2\pi}\int_{a}^{b}2\pi \cdot c \sqrt{1 + c^{2} (f'(x))^2} dx\) b. \(\int_{a / c}^{b / c} 2 \pi h(x) \sqrt{c^{2}+h^{\prime}(x)^{2}} d x = \frac{A}{2\pi c}\int_{a}^{b}2\pi \cdot \frac{1}{c} \sqrt{1 + c^{2} (f'(u))^2} du\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Integral Calculus
Integral calculus is a cornerstone of calculus, dealing with the concept of integration. This is a process to find the accumulated quantity, like area, volume, and other quantities. It's the reverse operation of differentiation and is often used to calculate areas under curves and the total change given a rate of change.

In the context of this problem, we are interested in using integrals to calculate the surface area of a revolution. When we have a curve, like the function \( f(x) \), and we revolve it around the x-axis from \( a \) to \( b \), this creates a 3D surface. The surface area, denoted as \( A \), can be calculated using the integral \( \int_{a}^{b} 2 \pi f(x) \sqrt{1 + (f'(x))^2} \, dx \). This integral formula considers both the length of the function along the x-axis and the rate at which it changes, which is represented by its derivative \( f'(x) \).

This formula's strength lies in its capability to account for both straight and curved paths of the function, giving a precise calculation of the surface area.
Scaling Functions
Scaling a function involves modifying it by stretching or compressing it. This often means multiplying the function by a constant, \( c \), or adjusting its input.

For example, when scaling the function \( f(x) \), we have two scaled versions in this problem:
  • \( g(x) = cf(x) \) means each output value of the function is multiplied by the constant \( c \). This represents a vertical stretch or compression depending on \( c \).
  • \( h(x) = f(cx) \) means we adjust the input. This represents a horizontal scaling of the function.
Understanding how these transformations affect the shape and position of the graph is crucial in calculus. In terms of our exercise, these transformations impact the surface area integral, as they modify the curve and consequently its revolution around the axis.
Derivative of Functions
A derivative represents the rate at which a function's value changes with respect to changes in its input. For a function \( f(x) \), its derivative \( f'(x) \) indicates how quickly \( f(x) \) is changing at any point \( x \).

Derivatives play a crucial role in determining the slope of the curve at a given point, and they are vital for calculating the surface area of revolution. They help in understanding the shape and curvature of the graph, as the integral formula for the surface area includes the term \( \sqrt{1 + (f'(x))^2} \).

In our scaled functions, calculating derivatives might be slightly different due to scaling factors:
  • For \( g(x) = cf(x) \), the derivative is \( g'(x) = c f'(x) \).
  • For \( h(x) = f(cx) \), the derivative comes out as \( h'(x) = c f'(cx) \).
These derivatives are crucial for adjusting the surface area calculations as they alter the length and shape of the curve during the revolution process.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Winding a chain A 30 -m-long chain hangs vertically from a cylinder attached to a winch. Assume there is no friction in the system and the chain has a density of \(5 \mathrm{kg} / \mathrm{m}\) a. How much work is required to wind the entire chain onto the cylinder using the winch? b. How much work is required to wind the chain onto the cylinder if a \(50-\mathrm{kg}\) block is attached to the end of the chain?

Emptying a conical tank An inverted cone (base above the vertex) is \(2 \mathrm{m}\) high and has a base radius of \(\frac{1}{2} \mathrm{m} .\) If the tank is full, how much work is required to pump the water to a level \(1 \mathrm{m}\) above the top of the tank?

Surface area using technology Consider the following curves on the given intervals. a. Write the integral that gives the area of the surface generated when the curve is revolved about the given axis. b. Use a calculator or software to approximate the surface area. \(y=\ln x^{2},\) for \(1 \leq x \leq \sqrt{e} ;\) about the \(x\) -axis

Equal integrals Without evaluating integrals, explain the following equalities. (Hint: Draw pictures.) a. \(\pi \int_{0}^{4}(8-2 x)^{2} d x=2 \pi \int_{0}^{8} y\left(4-\frac{y}{2}\right) d y\) b. \(\int_{0}^{2}\left(25-\left(x^{2}+1\right)^{2}\right) d x=2 \int_{1}^{5} y \sqrt{y-1} d y\)

Painting surfaces \(A\) 1.5-mm layer of paint is applied to one side of the following surfaces. Find the approximate volume of paint needed. Assume \(x\) and \(y\) are measured in meters. The spherical zone generated when the curve \(y=\sqrt{8 x-x^{2}}\) on the interval \(1 \leq x \leq 7\) is revolved about the \(x\) -axis.
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Geometry

Read Explanation

Discrete Mathematics

Read Explanation

Pure Maths

Read Explanation

Mechanics Maths

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.