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Chapter 6: Problem 10

Use a CAS to find the exact area of the surface generated by revolving the curve about the stated axis. $$y=\frac{1}{3} x^{3}+\frac{1}{4} x^{-1}, 1 \leq x \leq 2 ; \text { x -axis }$$

Short Answer

Expert verified
Use a CAS to compute the integral for the exact surface area.

Step by step solution

01

Understand the Problem

We need to find the exact area of the surface generated by revolving the curve \( y = \frac{1}{3} x^3 + \frac{1}{4} x^{-1} \) around the x-axis for \( 1 \leq x \leq 2 \).
02

Formula for Surface Area of Revolution

The formula for the surface area \( S \) of a curve \( y = f(x) \) revolved around the x-axis from \( x = a \) to \( x = b \) is \[ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(f'(x)\right)^2} \, dx \] where \( f'(x) \) is the derivative of \( f(x) \).
03

Differentiate the Function

Differentiate \( y = \frac{1}{3}x^3 + \frac{1}{4}x^{-1} \) to find \( f'(x) \). The derivative is: \[ f'(x) = x^2 - \frac{1}{4}x^{-2} \].
04

Set Up the Integral for Surface Area

Substitute \( f(x) \) and \( f'(x) \) into the surface area formula: \[ S = 2\pi \int_{1}^{2} \left( \frac{1}{3}x^3 + \frac{1}{4}x^{-1} \right) \sqrt{1 + \left(x^2 - \frac{1}{4}x^{-2}\right)^2} \, dx \].
05

Simplify the Expression Inside the Integral

Expand and simplify the expression \( 1 + \left(x^2 - \frac{1}{4}x^{-2}\right)^2 \) to make the integration possible: \[ 1 + (x^4 - \frac{1}{2} + \frac{1}{16x^4}) \].
06

Use a CAS for the Integral

Use a Computer Algebra System (CAS) to compute the simplified integral. Input the integral into the CAS and get the exact value for the surface area.

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Key Concepts

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

Calculus
Calculus is a branch of mathematics that deals with change and motion. It is essential for understanding how things change, whether they grow or diminish.
  • Calculus mainly focuses on two major concepts: differentiation and integration.
  • Differentiation deals with the rate of change, while integration is about accumulation and area under a curve.
Calculus helps us solve problems related to real-world phenomena involving motion, growth, decay, and other such processes.
Definite Integral
Definite integrals help calculate areas under curves. They are used extensively in calculus to find accumulated values like areas and volumes.
  • They are expressed as \( \int_{a}^{b} f(x) \, dx \), where \( a \) and \( b \) are boundaries of integration.
  • This notation represents the area under the curve \( f(x) \) from \( x = a \) to \( x = b \).
Evaluating definite integrals gives us exact numbers, often representing areas, total change, or accumulated quantities.
Differentiation
Differentiation is a fundamental concept of calculus used to compute the derivative of a function.
  • A derivative represents the rate of change of a function with respect to a variable.
  • In the context of finding surface areas, derivatives are used to determine how a curve's slope changes.
Differentiation helps solve various applied problems, from optimizing shapes to predicting future values based on current rates of change.
Computer Algebra System (CAS)
A Computer Algebra System (CAS) is a powerful tool that automates symbolic mathematics.
  • These systems can solve equations, perform algebraic manipulations, and integrate functions symbolically.
  • Using a CAS for complex calculations, like integrating complicated functions, saves time and minimizes human error.
To find a surface area of revolution, a CAS simplifies the integral calculation, turning complex integrations into manageable tasks.
Curve Revolution around Axis
When we revolve a curve around an axis, we create a 3D shape. Calculating the surface area of this shape requires calculus, specifically the concept of surface area of revolution.
  • For rotation around the x-axis, we use the function itself and its derivative in the surface area formula: \[ S = 2\pi \int_{a}^{b} f(x) \sqrt{1 + \left(f'(x)\right)^2} \, dx \]
  • This formula accounts for the radial distance around the axis and the length along the curve's path.
The result is the total surface area of the object formed by the curve's revolution, useful in engineering, physics, and many fields of science.

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Most popular questions from this chapter

Use the first and second derivatives to show that the graph of \(y=\tanh ^{-1} x\) is always increasing and has an inflection point at the origin.

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Suppose that a hollow tube rotates with a constant angular velocity of \(\omega\) rad/s about a horizontal axis at one end of the tube, as shown in the accompanying figure. Assume that an object is free to slide without friction in the tube while the tube is rotating. Let \(r\) be the distance from the object to the pivot point at time \(t \geq 0,\) and assume that the object is at rest and \(r=0\) when \(t=0 .\) It can be shown that if the tube is horizontal at time \(t=0\) and rotating as shown in the figure, then $$r=\frac{g}{2 \omega^{2}}[\sinh (\omega t)-\sin (\omega t)]$$ during the period that the object is in the tube. Assume that \(t\) is in seconds and \(r\) is in meters, and use \(g=9.8 \mathrm{m} / \mathrm{s}^{2}\) and \(\omega=2 \mathrm{rad} / \mathrm{s}\) (a) Graph \(r\) versus \(t\) for \(0 \leq t \leq 1\) (b) Assuming that the tube has a length of \(1 \mathrm{m},\) approximately how long does it take for the object to reach the end of the tube? (c) Use the result of part (b) to approximate \(d r / d t\) at the instant that the object reaches the end of the tube. (GRAPH CAN'T COPY).
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