What if the volume of a cylinder as a function of it's height/radius? Full question in the description box below.

Question

7958 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-01-04T19:36:00

#V_1(h) = 1/16 h(144-h^2)" cm"^3" "# for #h in [0, 12]#

#V_2(r) = 2pir^2 sqrt(36-r^2)" cm"^3" "# for #r in [0, 6]#

If we take a vertical slice through the centre of the sphere and cylinder, then it looks like this:

enter image source here

Where #r# is the radius of the cylinder, #R# the radius of the sphere and #h# the height of the cylinder.

Then the volume of the cylinder is the area of its base, multiplied by the height. That is:

#V = h pi r^2#

Note that the radius #r# varies as the height #h# varies, always satisfying Pythagoras formula:

#(2R)^2 = h^2+(2r)^2#

That is:

#4R^2 = h^2+4r^2#

Hence we can write #r# in terms of #h# or #h# in terms of #r#:

#r = 1/4 sqrt(4R^2-h^2)#

#h = sqrt(4R^2-4r^2) = 2sqrt(R^2-r^2)#

We can substitute these formulae into our prior formula for the volume of the cylinder to find:

#V_1(h) = h pi r^2 = h pi (1/4 sqrt(4R^2-h^2))^2 = 1/16 h(4R^2-h^2)#

#V_2(r) = h pi r^2 = (2sqrt(R^2-r^2)) pi r^2 = 2pir^2 sqrt(R^2-r^2)#

Finally, substituting #R=6# (cm) we get:

#V_1(h) = 1/16 h(144-h^2)" cm"^3" "# for #h in [0, 12]#

#V_2(r) = 2pir^2 sqrt(36-r^2)" cm"^3" "# for #r in [0, 6]#