The problem involves calculating the volume of a solid formed by rotating the function y = (1/x) around the x-axis, specifically between the limits x = 1 and x = 2.
Some participants have provided guidance on evaluating the integral and correcting assumptions about the shell radius in the cylindrical shells method. Multiple interpretations of the problem are being explored, particularly regarding the formulas used for volume calculation.
There are references to specific calculus textbooks and formulas, indicating that participants are drawing from established methods while also questioning their application in this context. Some participants express uncertainty about their previous calculations and seek clarification on the correct approach.
jakobs said:Homework Statement
y=(1/x) is rotating around the x-axis.
find the volume for the figure between x=1 and x=2
Homework Equations
The Attempt at a Solution
V=pi y2 = pi*(1/x)2
V= 1∫2 pi(1/x)2dx which gives V= 1∫2 pi(1/x2)dx
A primitive function for (1/x2) is -(1/x)
V=1∫2 pi((-(1/2))-(-(1/1)))
How do I go from here?
The OP correctly identified the volume of revolution to be V = \pi\int_{1}^{2} \frac{1}{x^2} dx.Zondrina said:Easily stated you want :
2π\int_{1}^{2}\frac{1}{x^2}dx
You could have also used :
π\int_{1}^{2}1dx
CAF123 said:The OP correctly identified the volume of revolution to be V = \pi\int_{1}^{2} \frac{1}{x^2} dx.What mathematical reasoning gives this?
I have come across the method of cylindrical shells, where for some rotation about y axis, V = 2\pi \int_{a}^{b} x f(x) dxZondrina said:There are two formulas I remember from Stewart calc 4th edition which highlight the formulas to use when calculating volumes of solids.
\pi\int_{a}^{b}xf(x)dx
2\pi\int_{a}^{b}f^2(x)dx
Zondrina said:Easily stated you want :
2π\int_{1}^{2}\frac{1}{x^2}dx
You could have also used :
π\int_{1}^{2}1dx
You've remembered incorrectly.Zondrina said:There are two formulas I remember from Stewart calc 4th edition which highlight the formulas to use when calculating volumes of solids.
\pi\int_{a}^{b}xf(x)dx
2\pi\int_{a}^{b}f^2(x)dx
Show us what you tried. You should have had to split it up into two integrals.CAF123 said:I have come across the method of cylindrical shells, where for some rotation about y axis, V = 2\pi \int_{a}^{b} x f(x) dx
I was wondering if someone could attempt the problem in the OP using cylindrical shells, because I am always getting either π or π/4, not the required π/2.
Perfect, I must have thought that I was rotating about the line y = 1/2 at my first attempt, but I see why it is simply y: y is the distance from the axis of rotation, which is the x axis. I now have the correct answer.vela said:The shell radius in the first integral should simply be y.
Zondrina said:You could have also used :
π\int_{1}^{2}1dx