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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. $$ y=\frac{1}{3} x^{3}+\frac{1}{4} x^{-1}, 1 \leq x \leq 2 ; x \text { -axis } $$

Short Answer

Expert verified
The exact surface area is approximately 52.817.

Step by step solution

01

Understand the Problem

We need to find the area of the surface generated by revolving the curve \( y = \frac{1}{3}x^3 + \frac{1}{4}x^{-1} \) about the x-axis over the interval \( 1 \leq x \leq 2 \).
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 \( a \) to \( b \) is: \[ S = \int_a^b 2\pi y \sqrt{1+(f'(x))^2} \, dx \]where \( f'(x) \) is the derivative of \( f(x) \) with respect to \( x \).
03

Find the Derivative

Differentiate \( y = \frac{1}{3}x^3 + \frac{1}{4}x^{-1} \) with respect to \( x \). \[ f'(x) = \frac{d}{dx} \left( \frac{1}{3}x^3 + \frac{1}{4}x^{-1} \right) = x^2 - \frac{1}{4}x^{-2} \]
04

Plug into the Surface Area Formula

Substitute \( y \) and \( f'(x) \) into the surface area formula:\[ S = \int_1^2 2\pi \left( \frac{1}{3}x^3 + \frac{1}{4}x^{-1} \right) \sqrt{1 + \left( x^2 - \frac{1}{4}x^{-2} \right)^2} \, dx \]
05

Simplify and Integrate Using CAS

Use a Computer Algebra System (CAS) to evaluate the integral:\[ S = \int_1^2 2\pi \left( \frac{1}{3}x^3 + \frac{1}{4}x^{-1} \right) \sqrt{1 + \left( x^2 - \frac{1}{4}x^{-2} \right)^2} \, dx \]The CAS evaluates this integral to find the exact surface area.

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Key Concepts

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

Integral Calculus
Integral calculus is one of the main branches of calculus and deals with the computation of integrals.It helps us calculate quantities like areas, volumes, and other values derived from the accumulation of quantities.
In this exercise, we focus on finding the surface area of a revolution by calculating the integral of a function that's revolved around an axis, specifically the x-axis.
The surface area of revolution, like many integral calculus problems, requires us to set up an integral expression. For surfaces of revolution, the integral expression incorporates both the function and its derivative, lending a complexity that makes these calculations engaging as well as challenging.
  • Integrals: The integral used in this exercise is a definite integral over a specified interval \(1 \leq x \leq 2\).
  • Surface of Revolution Principals: The formula accounts for rotation, and incorporates the curve's path from start to end points along the x-axis.
Differentiation
Differentiation is the process of finding the derivative of a function.A derivative indicates the rate at which one quantity changes with respect to another.
In this problem, differentiating the original function \(y = \frac{1}{3}x^3 + \frac{1}{4}x^{-1}\) is necessary because the surface area formula requires the function's derivative, \(f'(x)\).
This is the step where we determine \(f'(x)\) by applying basic differentiation rules:
  • Power Rule: Allows us to differentiate terms like \(x^3\).
  • Negative Power Rule: Helps differentiate terms like \(x^{-1}\).
After differentiating, we obtain \(f'(x) = x^2 - \frac{1}{4}x^{-2}\), which must then be used in the surface area integral formula.Every derivative tells us how the original function changes, crucial in finding the shape's surface when it's revolved.
Computer Algebra Systems (CAS)
Computer Algebra Systems (CAS) are powerful software tools designed to help with algebraic calculations like differentials and integrals. CAS can provide precise results and handle complex calculations that may be difficult to perform by hand.
In the context of this exercise, a CAS is used in the final step to perform the actual integration. Given the complexity of the function and its derivative when plugged into the surface area formula, manually computing the integral would be cumbersome and error-prone.
  • Automation: CAS automates the integration process, ensuring accuracy.
  • Efficiency: It quickly handles the lengthy calculations, providing exact values for advanced expressions.
Using CAS allows you to focus on understanding the setup and process rather than grappling with tedious arithmetic, making learning calculus concepts more approachable.

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Most popular questions from this chapter

Use cylindrical shells to find the volume of the cone generated when the triangle with vertices \((0,0),(0, r),(h, 0),\) where \(r>0\) and \(h>0,\) is revolved about the \(x\) -axis.

Find the area enclosed by \(y=\sinh 2 x, y=0,\) and \(x=\ln 3\)

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 \(\omega=2 \mathrm{rad} / \mathrm{s}\) $$ \begin{array}{l}{\text { (a) Graph } r \text { versus } t \text { for } 0 \leq t \leq 1} \\ {\text { (b) Assuming that the tube has a length of } 1 \mathrm{m} \text { , approxi- }} \\ {\text { mately how long does it take for the object to reach the }} \\ {\text { end of the tube? }} \\ {\text { (c) Use the result of part (b) to approximate } d r / d t \text { at the }} \\\ {\text { instant that the object reaches the end of the tube. }}\end{array} $$

Use a CAS to estimate the volume of the solid that results when the region enclosed by the curves is revolved about the stated axis. $$ y=\pi^{2} \sin x \cos ^{3} x, \quad y=4 x^{2}, \quad x=0, \quad x=\pi / 4 ; \quad x \text { -axis } $$

Prove the identities. $$ \begin{array}{l}{\text { (a) } 1-\tanh ^{2} x=\operatorname{sech}^{2} x} \\\ {\text { (b) } \tanh (x+y)=\frac{\tanh x+\tanh y}{1+\tanh x \tanh y}} \\\ {\text { (c) } \tanh 2 x=\frac{2 \tanh x}{1+\tanh ^{2} x}}\end{array} $$
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