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
+118609 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 \(x=8\sqrt3\)
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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
\(V=\pi \displaystyle \int_o^{8\sqrt3} 8^2dx\\ V=\pi \displaystyle \left[64x\right ]_o^{8\sqrt3} \\ V=64*8\sqrt3\;\pi\\ V=8^3\sqrt3\;\pi\)
Subtract
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. 
