Finding Volume with Integration

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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.
 

Answers and Replies

  • #2
Simon Bridge
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Homework Helper
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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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