Volume of Revolution Problem: Finding the Volume of a Solid of Revolution

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SUMMARY

The volume of the solid of revolution generated by revolving the curve defined by the equation y=x^2 + x - 2 around the x-axis is calculated using the integral V = -π ∫_{-1}^1 (x^2 + x - 2)^2 dx. The correct volume is determined to be 81π/10, while the initial calculation yielded 18π/5, indicating an integration error. The bounds of integration are confirmed to be -1 and 1, where the curve intersects the x-axis.

PREREQUISITES
  • Understanding of calculus, specifically integration techniques
  • Familiarity with the concept of solids of revolution
  • Knowledge of the disk method for volume calculation
  • Ability to manipulate polynomial functions
NEXT STEPS
  • Review the disk method for calculating volumes of solids of revolution
  • Practice integration of polynomial functions using definite integrals
  • Explore common mistakes in integration and how to avoid them
  • Learn about the relationship between curves and their intersections with axes
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Students and educators in calculus, mathematicians focusing on volume calculations, and anyone interested in mastering integration techniques related to solids of revolution.

G01
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Find the volume created by revolving:
[tex]y=x^2 +x - 2[/tex]

and y=0 about the x axis.

[tex]y=x^2 +x - 2[/tex] intersects the x-axis at -1 and 1 so those are the bounds of integration.

the radius of the figure = -([tex]y=x^2 +x - 2[/tex])

so:

V = [tex]- \pi \int_{-1}^1 (x^2 + x - 2)^2 dx[/tex]

After integrating that I get 18[tex]\pi[/tex]/5. The actual answer is 81[tex]\pi[/tex]/10. What I want to know is if i set the integral up right. If I did then this is only a stupid integration mistake. If not then I hope someone finds my mistake. Thanks for the help.
 
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G01 said:
[tex]y=x^2 +x - 2[/tex] intersects the x-axis at -1 and 1 so those are the bounds of integration.
You may want to check that again :smile:
Fix this, and you'll find the given answer.
 

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