Work Required to Empty a Hemispherical Tank

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SUMMARY

The work required to empty a hemispherical tank of radius R meters through an outlet at the top is calculated using the integral formula w = ∫_0^R (p)(g)(π)(R²)(R - x)dx. However, the area of the slice must be correctly identified as π(R² - x²) rather than πR², as the latter applies to cylindrical shapes. The correct formulation involves measuring x from the top of the tank to simplify calculations. The final expression for the area of the slice is π(R² - x²).

PREREQUISITES
  • Understanding of calculus, specifically integration techniques.
  • Familiarity with the concepts of work and force in physics.
  • Knowledge of the geometry of hemispherical shapes.
  • Basic understanding of fluid density and gravitational acceleration.
NEXT STEPS
  • Review calculus integration techniques for volume and work calculations.
  • Study the geometry of hemispheres and their cross-sectional areas.
  • Explore physics principles related to work done by forces in fluid dynamics.
  • Practice solving similar problems involving different tank shapes and fluid densities.
USEFUL FOR

Students studying physics and calculus, particularly those focusing on fluid mechanics and work-energy principles. This discussion is also beneficial for educators seeking to clarify concepts related to integration and geometry in real-world applications.

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Homework Statement
Find the work required to empty a tank in the shape of a hemisphere of radius R meters through an outlet at the top of the tank. The density of water is p kg/m^{3}; the acceleration of a free falling body is g. (Ignore the length of the outlet at the top.)


The attempt at a solution

w = \int_a^b (density)(gravity)(Area-of-slice)(distance)dx<br />

w = \int_0^R (p)(g)(\pi)(R^{2})(R - x)dx<br />

Is this correct/complete?
 
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Hi Precursor! :smile:

(have a pi: π and a rho: ρ and an integral: ∫ and try using the X2 tag just above the Reply box :wink:)
Precursor said:
Find the work required to empty a tank in the shape of a hemisphere of radius R meters through an outlet at the top of the tank. The density of water is p kg/m^{3}; the acceleration of a free falling body is g. (Ignore the length of the outlet at the top.)

w = \int_0^R (p)(g)(\pi)(R^{2})(R - x)dx<br />

Is this correct/complete?

No, that's the correct formulal for a cylinder (Area-of-slice = πr2).

Try again! :smile:
 
So is the area of the slice actually π(1 - x²)?
 
Precursor said:
So is the area of the slice actually π(1 - x²)?

Nooo (btw, it might be easier if you measured x from the top instead of from the bottom :wink:)
 
Is it π√(R² - x²)?
 
Precursor said:
Is it π√(R² - x²)?

With x is measured from the top, yes :smile: except …

lose the square-root! :wink:
(and I'm going to bed :zzz: g'night!)​
 

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