What Methods Calculate the Volume of a Solid Revolved Around Non-Standard Axes?

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SUMMARY

The volume of the solid generated by revolving the region bounded by the equations y = 4x - x² and y = 2x about the y-axis can be calculated using the cylinder method. To express x in terms of y, one must manipulate the original equations, potentially using the completing the square technique. When the axis of revolution is not y = 0, it is effective to change variables by replacing x with x' = x - 2. The final volume is computed using the integral 2π ∫₀² (2x² - x³) dx.

PREREQUISITES
  • Understanding of solid geometry and volume calculation methods
  • Familiarity with the cylinder method and disk method for volume integration
  • Knowledge of manipulating equations to express variables in terms of others
  • Proficiency in performing definite integrals
NEXT STEPS
  • Study the cylinder method for calculating volumes of solids of revolution
  • Learn about the completing the square technique in algebra
  • Explore variable substitution methods in integral calculus
  • Practice calculating volumes using different axes of revolution
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Mathematics students, educators, and professionals involved in geometry, calculus, and engineering applications requiring volume calculations of solids of revolution.

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Find the volume of the solid generated by revolving the region bounded by y = 4x-x^2 and y = 2x about the y-axis. About the line x = 2.

Since this is rotated about the y-axis, I know I have to manipulate the equation so I can get to x = something. The problem is I cannot change the equation y = 4x-x^2.

Questions:
How would you find the volume that is rotated about the y-axis while the equations are y = form instead of x =? If it is not possible, then what do I have to do in this case?
How would you find the volume if the axis of revolution is not y = 0?
 
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You need to express x in terms of y. Use the completing the square technique.
How would you find the volume if the axis of revolution is not y = 0?
Change variables. Replace ##x## by ##x'=x-2##.
 
You can use the "cylinder method" rather that the "disk method" to integrate with respect to x. Imagine a line extending upward from each value of x. That line cuts the "region bounded by y = 4x-x^2 and y = 2x", for x between 0 and 2, in a line segment of length (4x- x^2)- 2x= 2x-x^2. Rotating around the y-axis that forms a cylinder of radius x and height 2x- x^2 so area 2\pi x(2x- x^2)= 2\pi (2x^2- x^3). Taking the "thickness" of each cylinder to be "dx", the whole volume is given by 2\pi \int_0^2 2x^2- x^3 dx.
 

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