Integrating Volume via Shells: What Went Wrong with the Bounds?

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SUMMARY

The discussion focuses on calculating the volume of a region bounded by the curve \( \frac{1}{x^4} \), the line \( y = 0 \), and vertical lines \( x = 2 \) and \( x = 6 \) about the axis \( y = -4 \). The initial approach using cylindrical shells is identified as incorrect; the problem is better suited for the disk method. The correct height of the shell should be defined as \( x_{right} - x_{left} \), necessitating the division of the integral into two parts to accommodate the discontinuity at \( x = 6 \).

PREREQUISITES
  • Understanding of volume calculation methods, specifically the disk and shell methods.
  • Familiarity with integration techniques in calculus.
  • Knowledge of the properties of functions and their graphs, particularly \( y = \frac{1}{x^4} \).
  • Ability to manipulate and interpret mathematical expressions involving square roots and integrals.
NEXT STEPS
  • Study the disk method for volume calculations in calculus.
  • Learn how to set up and evaluate integrals for piecewise functions.
  • Explore the concept of cylindrical shells and their applications in volume problems.
  • Review the properties of the function \( y = \frac{1}{x^4} \) and its behavior over specified intervals.
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations, as well as educators seeking to clarify common misconceptions in integration techniques.

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Homework Statement


Volume of the region bounded by 1/x^4, y = 0, x = 2, and x = 6 about the axis y = -4


Homework Equations





The Attempt at a Solution



XFHCwOW.png


The height for any shell is x(y) - 2, or [itex]\sqrt[4]{\frac{1}{y}} - 2[/itex]

The radius of any shell is y + 4

So the circumference is 2*pi*(y + 4), and the surface area is then [2*pi*(y + 4)] * [itex][\sqrt[4]{\frac{1}{y}} - 2][/itex]

So I integrate this from 0 to 1/(2^4) because y(2) = 1/(2^4) which is the upper boundary on the y-axis and the lower boundary is 0.

Where did I go wrong?
 
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PhizKid said:

Homework Statement


Volume of the region bounded by 1/x^4, y = 0, x = 2, and x = 6 about the axis y = -4


Homework Equations





The Attempt at a Solution



XFHCwOW.png


The height for any shell is x(y) - 2, or [itex]\sqrt[4]{\frac{1}{y}} - 2[/itex]

The radius of any shell is y + 4

So the circumference is 2*pi*(y + 4), and the surface area is then [2*pi*(y + 4)] * [itex][\sqrt[4]{\frac{1}{y}} - 2][/itex]

So I integrate this from 0 to 1/(2^4) because y(2) = 1/(2^4) which is the upper boundary on the y-axis and the lower boundary is 0.

Where did I go wrong?

The first place you went wrong was choosing to use shells in the first place. The problem is more naturally done with disks. Anyway, given that you are using shells, remember that the height of the shell in this example is ##x_{right} - x_{left}##. The problem is that the right hand curve is not a single piece. Part of it is the line ##x=6##. You have to break the integral into two parts to account for that.
 

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