Finding Volume Using Integration: Revolving Functions Around a Line

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SUMMARY

The discussion focuses on calculating the volume of a solid formed by revolving the functions y=x² and y=1 around the line y=6. The correct approach involves determining the outer radius as the distance from y=6 to the lower function (y=x²) and the inner radius as the distance from y=6 to the upper function (y=1). The limits of integration are set from -1 to 1, leading to the correct volume calculation using the formula ∏∫(ro² - ri²)dx. The final solution confirms that the initial negative volume result was due to incorrect radius calculations.

PREREQUISITES
  • Understanding of integral calculus, specifically volume of revolution
  • Familiarity with the disk method for calculating volumes
  • Knowledge of functions and their graphs, particularly parabolas
  • Proficiency in using definite integrals for area and volume calculations
NEXT STEPS
  • Study the disk method for volume calculations in greater detail
  • Learn about the washer method for solids of revolution
  • Explore applications of integration in real-world volume problems
  • Practice more problems involving revolving functions around different axes
USEFUL FOR

Students studying calculus, particularly those focusing on volume calculations and integration techniques, as well as educators seeking to clarify concepts related to solids of revolution.

EngnrMatt
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Homework Statement



Find the volume of the solid formed by revolving the functions: y=x2 and y=1 about the line y=6

Homework Equations



∏∫(ro2-ri2)dx

The Attempt at a Solution



I found the outer radius to be (1-x2)-6 and my inner radius to be -6. Also, the limits of integration were -1 and 1. Plugging it all in got -224/15. This is obviously wrong because it is negative, and the positive value didn't work either. Does anyone have a clue on how to do this?
 
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EngnrMatt said:

Homework Statement



Find the volume of the solid formed by revolving the functions: y=x2 and y=1 about the line y=6

Homework Equations



∏∫(ro2-ri2)dx

The Attempt at a Solution



I found the outer radius to be (1-x2)-6 and my inner radius to be -6. Also, the limits of integration were -1 and 1. Plugging it all in got -224/15. This is obviously wrong because it is negative, and the positive value didn't work either. Does anyone have a clue on how to do this?

Isn't the inner radius the distance from y=1 to y=6? Isn't that 5? What's the outer radius?
 
Yes, you are right. I see now, the outer radius will be the distance between y=6 and the lower function and the inner will be the distance between y-6 and the upper function. It worked by the way, thanks.
 

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