Question 1168325

The volume of the water in a hemisphere ( the water in the hemispherical bowl)  is given by 

{{{V = (pi/3)h^2*(3r-h)}}}, where {{{r}}} is the radius and {{{h }}}is height.

Given that the radius of the hemisphere is {{{r=2m}}} and the volume of water is {{{V=2.05m^3}}}, we have:


{{{2.05 = (pi/3)h^2*(3*2-h)}}}

{{{2.05 = (pi/3)h^2*(6-h)}}}

{{{1.9576058000303127= 6h^2-h^3}}}

{{{6h^2-h^3-1.9576058000303127=0}}}.......factor

{{{-1 (h - 5.9446) (h - 0.60222) (h + 0.546824) = 0}}}

{{{-1 (h - 5.9446)=0}}}=>{{{h=5.9446}}}.....disregard because is greater than diameter of the sphere

{{{(h + 0.546824) = 0}}}=> {{{h=- 0.546824}}}.....disregard negative solution

{{{(h - 0.60222)=0}}}=>{{{h=0.60222}}}

answer:

=> {{{h}}}≈{{{0.602}}}