Calculating Work for Pumping Oil from a Spherical Tank

[icesalmon]

Homework Statement

A spherical tank of radius 8 feet is half full of oil that weighs 50 pounds per cubic foot. Find the work required to pump oil out through a hole in the top of the tank.

The Attempt at a Solution

this is an example problem in my book and they start off by subdividing the region into disks of thickness Δy and radius x and by saying as a result of the increment of force for each of these disks being given by weight we have ΔF = weight → (50pounds/ft³)*volume → 50(∏x²Δy)pounds. This is where I get stuck, I have no idea where that expression, ∏x²Δy, came from. Are they using ∏r²h here for the volume of each cylindrical shell?

 

[iRaid]

Well if you're cutting the sphere in many pieces horizontally, you get circles, right? So the area of one of these circles is πx². The thickness of the circle is Δy. So with all that you can plug it into the integral for work.
Edit: Forgot to add, the radius is changing at certain spots right? So you have to take that into account.

 

[HallsofIvy]

As iRaid said, for each different level, y, we have a disk of oil with radius x, so area \pi x^2. Further, the thickness of each disk is \Delta y so its volume is \pi x^2\Delta y. Finally, you multiply by the density to get the weight, 50\pi x^2\Delta y, to get the weight that will be lifted to the top of the tank. That weight, times the height lifted, gives the work done in lifting that disk of oil to the top of the tank. Suming over all "disks" gives a Riemann sum approximating the work done lifting all of the oil. Convert that Riemann sum to the integral that gives the exact value.

 

[icesalmon]

alright thanks a lot everyone. Got it now.