Work for pumping out a halfsphere

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To calculate the work needed to pump water from a halfsphere with an 8-meter diameter, the changing radius is determined using the Pythagorean theorem. The volume of the halfsphere is calculated with the formula 2/3*r^3*π, and the mass is derived from the volume with a density of 1 kg/dm³. The discussion highlights the integration process, specifically the integral ∫(16-(4-x)²)^(3/2)dx, which yields 48π, although the reasoning behind the π is questioned. Clarifications are sought on how to utilize the volume in the context of the work calculation, particularly regarding the height through which the water is pumped. The conversation emphasizes the need for a clear understanding of the integration and application of these concepts in the homework problem.
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I have to calculate the power needed to pump out water of a halfsphere with the diameter of 8 meters only with integrating. When you pump out the water of the sphere the waterlevel is going to lower so the radius is also going do become smaller as futher the waterlevel drops.
For the changing radius you use the pythagorean theorem and get rx2=r2-(r-x)2
rx is the changing radius and x is the changing waterlevel.
Then you use the formula for the volume of a halfspher 2/3*r3*\pi. Now you have the volume so you can calculate the mass with δ=1kg/dm3.

And then a great miracle occurs when you intergrate ∫(16-(4-x)2)3/2*dx between 0 and 4 you will get 48\pi. Nobody knew why there came a \pi. Besides that our teacher said we that there is one tiny mistake in it.

It's homework because no one in class was able to solve this and i don't want to have a cliffhanger for the rest of the weak ;) I really hope you guy can help me with this one.
 
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StandUpGuy said:
I have to calculate the power needed to pump out water of a halfsphere
I assume you mean work, not power. Power will depend on the rate at which it is pumped.
Then you use the formula for the volume of a halfspher 2/3*r3*\pi. Now you have the volume so you can calculate the mass with δ=1kg/dm3.
How are you going to make use of the volume here? Don't you want to multiply the volume of an element of thickness dx by the height through which it will be pumped (x, I'm guessing)?
And then a great miracle occurs when you integrate ∫(16-(4-x)2)3/2*dx
How do you arrive at that integral?
 

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