Finding Volume with Integration

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SUMMARY

The discussion focuses on determining the appropriate formula for calculating the volume of solids of revolution, specifically when revolving the area between the curve y=(cos x)/x and the x-axis from π/6 to π/2 about the y-axis. Two methods are presented: the shell method, V = ∫(2πx f(x) dx), and the disk method, V = π ∫(y² dx). The shell method yields a volume of π, while the disk method involves a more complex integral that does not produce the same result. The consensus is to choose the method that simplifies calculations, emphasizing understanding over memorization of formulas.

PREREQUISITES
  • Understanding of calculus concepts such as integration and volume of solids of revolution.
  • Familiarity with the shell method and disk method for volume calculation.
  • Knowledge of trigonometric functions, specifically cosine and their properties.
  • Experience with definite integrals and their applications in volume problems.
NEXT STEPS
  • Study the shell method for calculating volumes of solids of revolution in detail.
  • Learn the disk method and its applications in various volume problems.
  • Explore the use of Wolfram Alpha for verifying integral calculations and results.
  • Practice solving volume problems involving trigonometric functions and integration techniques.
USEFUL FOR

Students and educators in calculus, particularly those focusing on volume calculations in solid geometry, as well as anyone looking to deepen their understanding of integration techniques in mathematical analysis.

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Homework Statement


How do you determine what formula to use in a given volume question? I always confused about it. Like, when to apply shell, when should use washer, etc.

For example:
Find the volume of the solid generated by revolving the area between the curve y=(cos x)/x and the x-axis for π/6≤x≤π/2 about the y-axis.

Homework Equations


Should I use this one
[itex]V = \int\limits_{a}^{b}2\pi xf(x)dx[/itex]
Or this one
[itex]V=π\int\limits_{a}^{b}y^2dx[/itex]

The Attempt at a Solution


Using first equation, I will get
[itex]V = \int\limits_{\pi/6}^{\pi/2} 2\pi x \frac{cosx}{x}dx = 2\pi \int\limits_{\pi/6}^{\pi/2}cosxdx = 2\pi [sinx]_{\pi/6}^{\pi/2} = 2 \pi (1 - 1/2)= \pi[/itex]

I don't know how to calculate the second one but someone write the answer as follow
[itex]\pi \int\limits_{\pi/6}^{\pi/2}(\cos x/x)^2dx= \pi (-x) \frac{(-x Si(2x)+\cos^2(x))}{(x)}|_\frac{\pi}{6}^\frac{\pi}{2}=Si(\frac{\pi}{3})-Si(\pi)+\frac{9}{2\pi}[/itex]Those answers are not equal. I have checked with wolframalpha.
 
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You choose the approach that makes the math simplest - you get to figure this out from experience doing lots of problems.

Usually the symmetry of the problem is the most important clue.
If you are in doubt, just brute-force the calculation as a triple integral.

In your example, it looks to be set up for the solid of rotation method, but either that or the disk method should produce the same result.
In general - don't try to memorize formulas - try to understand them instead.

eg. divide the volume into disks along the x-axis ... the disk at position x has thickness dx and radius R=f(x) ... so it's volume is dV: the volume of a disk is the area of the surface times the thickness. Thus: dV=πR2dx = πf2(x)dx and the total volume will be the volumes of all these disks added up thus:[tex]V=\pi \int_a^b f^2(x)dx[/tex]
 
Last edited:

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