A sphere of radius 16 cm is sitting on a table. We cut the sphere with a plane parallel to the table at a distance 24 cm from the table. Find the volume V of the part of the sphere above the cutting plane.

V= 2680.82 cm^3


Now suppose the portion of the sphere below the plane is filled with a liquid of density ρ=1.3 g/cm^3.

Find the amount of work W needed to pump the liquid over the top of the plane.

W= ?? (kgm^2)/sec^2


Thanks everyone

 Oct 6, 2022
edited by Guest  Oct 6, 2022

I let the central crossection of the sphere be  \(x^2+y^2=16^2\)


So the crossection of the top half is    \(y=\sqrt{256-x^2}\)


The table is at y=-16


So the cut is at y=8


Using Pythagoras you will find that the positive end of the cut is at  \(x=8\sqrt3\)



The volume of the sphere section from x=0 to 8sqrt3  will be

\(V=\displaystyle \pi\int_0^{8\sqrt3}\;256-x^2\;dx\\ V=\pi \left[256x-\frac{x^3}{3}\right]_0^{8\sqrt3}\\ V=\pi \left[256*8\sqrt3 -\frac{8^3*3*\sqrt3}{3}\right]\\ V=\pi \left[4*8^3\sqrt3 - 8^3\sqrt3\right]\\ V=\pi \left[3*8^3\sqrt3 \right]\\ \)

The volume of the rectangular p[rism section from  x=0 to x=8sqrt3 will be


\(V=\pi \displaystyle \int_o^{8\sqrt3} 8^2dx\\ V=\pi \displaystyle \left[64x\right ]_o^{8\sqrt3} \\ V=64*8\sqrt3\;\pi\\ V=8^3\sqrt3\;\pi\)



Difference in volume is   

 \(V=2*8^3\sqrt3\;\pi\\ V=1024\sqrt3\;\pi\)


This is top and bottom so you can divide by 2 but it is only half of the needed region so you have to multiply by 2

So that is the  volume of the top bit.


You should be able to do at least a bit more by yourself.   indecision  laugh

 Oct 6, 2022

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