/*! 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 6 Find the area of the surface gen... [FREE SOLUTION] | 91Ó°ÊÓ

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Find the area of the surface generated when the given curve is revolved about the \(x\) -axis. $$y=12-3 x \text { on }[1,3]$$

Short Answer

Expert verified
Question: Calculate the surface area generated when the curve \(y = 12 - 3x\) is revolved about the x-axis in the interval \([1, 3]\). Answer: The surface area generated when the curve is revolved about the x-axis in the given interval is \(S = 24\pi \sqrt{10}\).

Step by step solution

01

Set up the Surface Area Formula

The formula for the surface area of revolution about the x-axis for a given curve \(y = f(x)\) in the interval \([a, b]\) is $$ S = 2 \pi \int_{a}^{b} y \sqrt{1 + (f'(x))^2} dx $$ where \(y = f(x)\) is the function we want to revolve, and \(f'(x)\) is its derivative.
02

Find the Derivative of the Function

For the given function \(y = 12 - 3x\), find its derivative with respect to x: $$ \frac{dy}{dx} = -3 $$ Since the derivative is constant, the square of the derivative is also constant: $$ \left(\frac{dy}{dx}\right)^2 = 9 $$
03

Apply the Surface Area Formula

Now, we apply the surface area formula using the given function \(y = 12 - 3x\), its derivative squared \((f'(x))^2 = 9\), and the interval [1, 3]: $$ S = 2\pi \int_{1}^{3} (12 - 3x) \sqrt{1 + 9} dx $$ Simplify the square root term: $$ S = 2\pi \int_{1}^{3} (12 - 3x) \sqrt{10} dx $$
04

Integrate with Respect to x

Now, integrate the function with respect to x in the interval [1, 3]: $$ S = 2\pi \left[ \sqrt{10} \int_{1}^{3} (12 - 3x) dx \right] $$ Integrate: $$ S = 2\pi \sqrt{10} \left[ \int_{1}^{3} 12 dx - 3\int_{1}^{3} x dx \right] $$ $$ S = 2\pi \sqrt{10} \left[ \left(12x\right) \bigg|_{1}^{3} - \left(\frac{3x^2}{2}\right) \bigg|_{1}^{3} \right] $$
05

Evaluate the Integrals

Evaluate the definite integrals by plugging in the limits of integration: $$ S = 2\pi \sqrt{10} \left[ (12\cdot3 - 12\cdot1) - \left(\frac{3\cdot3^2}{2} - \frac{3\cdot1^2}{2}\right) \right] $$ Simplify and calculate: $$ S = 2\pi \sqrt{10} \left[ 24 - \left( \frac{27}{2} - \frac{3}{2} \right) \right] $$ $$ S = 2\pi \sqrt{10} \left[ 24 - 12 \right] $$ $$ S = 2\pi \sqrt{10} \cdot 12 $$ The surface area of the surface generated when the curve \(y = 12 - 3x\) is revolved about the x-axis in the interval \([1, 3]\) is: $$ S = 24\pi \sqrt{10} $$

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Key Concepts

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

Integration
Integration is a fundamental operation in calculus that enables us to find the whole when we know the parts or the rate of change. It's like piecing together the segments to reveal the total area, volume, or, in this case, the surface area of a solid figure. Imagining we are filling a shape with tiny blocks and adding them all up, we find that integration can give us that total with perfect accuracy.

To integrate a function means to compute the integral, which in calculus represents the accumulation of quantities. When we are calculating areas, we're essentially integrating over the distance or length of an object's boundary. With surface area of revolution, the integration process takes on a spatial aspect, expanding into three dimensions.
Definite Integrals
Definite integrals specifically provide a computation of quantity where there is a clear start and end point. Think of it as a journey with a defined beginning and destination—the definite integral takes every part of that journey into account to give you the total distance traveled. In the context of the surface area problem we have, the 'journey' is the path traveled by the curve from the start of the interval \begin{document} \([a, b]\) \f{enddocument} as it is revolved around the axis.

The beauty of definite integrals lies in their versatility—they not only calculate area but also deal with various physical concepts like mass, force, and even energy, when the function under consideration represents a related rate or density. They help us transition from the world of infinitesimals to something that has tangible meaning in the real world.
Derivatives
Derivatives stand as the cornerstone of differential calculus, capturing the essence of change. When we take the derivative of a function, we're essentially asking, 'At any given moment, at any infinitesimal interval, how fast is this function changing?' It’s like looking at the speedometer of your car that tells you your instant speed, not the distance covered or the time it will take for your trip. This rate of change is crucial when dealing with the geometry of curves.

In our surface area problem, the derivative comes into play as we're considering not just the static curve, but the curve's behavior as it sweeps and creates a surface. Without understanding derivatives, we wouldn't be able to accurately calculate the new shape's characteristics because it's the derivative that precisely describes how the curve's slope changes at each point as it rotates.
Calculus Applications
Calculus isn't just about abstract symbols and theoretical concepts; it's a powerful tool with a myriad of applications in science, engineering, economics, and beyond. In the world around us, calculus allows us to solve real-life problems involving rates of change and cumulative values—such as predicting population growth, optimizing resources, and calculating the amount of material needed for construction. The process of solving our surface area problem is an excellent demonstration of calculus at work, as it combines the notions of integration and differentiation to help us understand and measure the intricate shapes that emerge from simple motions in space.

Surface area of revolution is particularly relevant in engineering and physics, where it often comes up in the design of objects like gears, propellers, and even telescopes. Understanding the relationship between a curve and the surface it generates when revolved about an axis is indispensable for creating objects with specific properties, such as minimal air resistance or maximal strength.

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Most popular questions from this chapter

Without evaluating integrals, prove that $$ \int_{0}^{2} \frac{d}{d x}\left(12 \sin \pi x^{2}\right) d x=\int_{0}^{2} \frac{d}{d x}\left(x^{10}(2-x)^{3}\right) d x $$

The burning of fossil fuels releases greenhouse gases (roughly \(60 \% \text { carbon dioxide })\) into the atmosphere. In 2010 , the United States released approximately 5.8 billion metric tons of carbon dioxide (Environmental Protection Agency estimate), while China released approximately 8.2 billion metric tons (U.S. Department of Energy estimate). Reasonable estimates of the growth rate in carbon dioxide emissions are \(4 \%\) per year for the United States and \(9 \%\) per year for China. In 2010 , the U.S. population was 309 million, growing at a rate of \(0.7 \%\) per year, and the population of China was 1.3 billion, growing at a rate of \(0.5 \%\) per year. a. Find exponential growth functions for the amount of carbon dioxide released by the United States and China. Let \(t=0\) correspond to 2010 . b. According to the models in part (a), when will Chinese emissions double those of the United States? c. What was the amount of carbon dioxide released by the United States and China per capita in \(2010 ?\) d. Find exponential growth functions for the per capita amount of carbon dioxide released by the United States and China. Let \(t=0\) correspond to 2010. e. Use the models of part (d) to determine the year in which per capita emissions in the two countries are equal.

a. Confirm that the linear approximation to \(f(x)=\tanh x\) at \(a=0\) is \(L(x)=x.\) b. Recall that the velocity of a surface wave on the ocean is \(v=\sqrt{\frac{g \lambda}{2 \pi} \tanh \frac{2 \pi d}{\lambda}} .\) In fluid dynamics, shallow water refers to water where the depth-to-wavelength ratio \(d / \lambda<0.05 .\) Use your answer to part (a) to explain why the shallow water velocity equation is \(v=\sqrt{g d}.\) c. Use the shallow-water velocity equation to explain why waves tend to slow down as they approach the shore.

Explain why l'Hôpital's Rule fails when applied to the limit \(\lim _{x \rightarrow \infty} \frac{\sinh x}{\cosh x}\) and then find the limit another way.

Miscellaneous integrals Evaluate the following integrals. \(\int_{0}^{1} \frac{16^{x}}{4^{2 x}} d x\)

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