Area of revolution of hyperbola

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SUMMARY

The volume of the solid obtained by rotating the region in the first quadrant bounded by the hyperbola defined by the equation y² - x² = 4 and the vertical lines x = 3 and x = 5 about the y-axis can be calculated using the integral ∫₃⁵ 2xπ√(4 + x²) dx. The discussion confirms that it is unnecessary to incorporate the "hole" from x = 0 to x = 3, as the volume is derived solely from the cylindrical shells between x = 3 and x = 5.

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  • Understanding of hyperbolic equations and their properties
  • Knowledge of volume calculation using cylindrical shells
  • Familiarity with integral calculus and definite integrals
  • Proficiency in using mathematical notation and symbols
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  • Study the method of cylindrical shells for volume calculations
  • Explore hyperbolic functions and their applications in calculus
  • Practice solving definite integrals involving square roots
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~Sam~
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Homework Statement


Find the volume of the solid obtained by rotating the region in the first quadrant bounded by the hyperbola y2−x2=4 and the lines y=0, x=3 and x=5 about the y− axis.


Homework Equations



Nothing specific...general equations

The Attempt at a Solution



So I would write it as
I was wondering if it was just 2x*pi*dx*(√4+x2) from x=3 and x=5? Or would I have to incorporate the "hole" from x=0 to x=3.
 
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~Sam~ said:
I was wondering if it was just 2x*pi*dx*(√4+x2) from x=3 and x=5? Or would I have to incorporate the "hole" from x=0 to x=3.

Hi ~Sam~! :smile:

(have a pi: π :wink:)

Yes, just ∫35 2x*π*dx*√(4+x2) …

you're "adding" the volumes of the cylindrical shells that make up the whole shape, so you only need the shells with radii from 3 to 5. :smile:
 

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