Shell Method Example: Finding Volume Using Shell Method

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SUMMARY

The discussion centers on using the shell method to find the volume of a solid formed by rotating the area between the y-axis and the line y = x + 1 from y = 2 to y = 4 about the x-axis. The correct volume, calculated using the formula V = 2π ∫ (radius)(height of shell) dy, is determined to be (76/3)π. The confusion arises from the problem statement asking for area instead of volume, leading to a discussion on the components of the solid and their respective areas.

PREREQUISITES
  • Understanding of the shell method in calculus
  • Familiarity with integral calculus and volume calculations
  • Knowledge of the relationship between area and volume in solids of revolution
  • Ability to manipulate equations and perform definite integrals
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  • Study the shell method in detail, focusing on its application in finding volumes of solids of revolution
  • Learn about the disk and washer methods for volume calculation
  • Explore the concept of surface area of solids formed by rotation
  • Practice problems involving the integration of functions to find volumes and areas
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Students and educators in calculus, particularly those focusing on solid geometry and volume calculations using the shell method.

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Homework Statement



The area between the y-axis and the line y = x + 1 from y = 2 to
y = 4 is rotated about the x-axis. Use the shell method to find the
area of the resulting solid.



Homework Equations



V = 2pi integral (radius)(height of shell) dx or dy


The Attempt at a Solution



Well I am kind of confused with this problem, it says to find the AREA but isn't the shell method suppose to find the VOLUME?

Here is my attempt:

y= x+1 -----> x=y-1

V = 2pi integral from 2 to 4 y(y-1) dy

I then evaluated it to get = (76/3)pi as the volume.

What did I do wrong?
 
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I'll agree with what you've done so far. The volume is 76*pi/3. If they really do want area I'd add up the area of the parts of the boundary. You've got two cylinders, a disk with a hole in it and a part of a cone. It seems like it shouldn't be too hard, but I don't know what that has to do with the 'shell method'.
 

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