Work for pumping out a halfsphere

[StandUpGuy]

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 r_x²=r²-(r-x)²
r_x 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.

 

[haruspex]

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?