Work Required to Empty a Hemispherical Tank

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Homework Help Overview

The problem involves calculating the work required to empty a hemispherical tank of radius R through an outlet at the top. The context includes considerations of fluid density and gravitational acceleration.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the formulation of the work integral, questioning the area of the cross-sectional slice and whether it should be based on the geometry of a hemisphere.

Discussion Status

There is ongoing exploration of the correct expression for the area of the slice, with participants suggesting different approaches and measurements. Some guidance has been offered regarding the measurement reference point for x.

Contextual Notes

Participants are navigating potential misconceptions about the geometry involved and the implications of measuring from different reference points. There is a focus on ensuring the correct area formula is used for the hemispherical shape.

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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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