Question

A hemispherical bowl of radius 12 inches is filled to a depth of h​ inches, where 0less than or equalshless than or equals12. Find the volume of water in the bowl as a function of h.​ (Check the special cases hequals0 and hequals12​.)

Answer 1

Suppose the bowl is situated such that the rim of the bowl touches the x axis, and the semicircular cross section of the bowl lies below the x-axis (in (iii) and (iv) quadrant ). Then the equation of the cross section of the bowl would be
`x^2+y^2=144`, where y≤ 0,

⇒
`y=-√(144-x^2)`

Here, h represents the depth of water,

Thus, by using shell method,

The volume of the disk would be,

`V(h) = \pi \int_(-12)^(-12+h) x^2 dx`

`= \pi \int_(-12)^(-12+h) (144-y^2) dy`

`= \pi |144y-(y^3)/(3)|_(-12)^(-12+h)`

`=\pi [ (144(-12+h)-((-12+h)^3)/(3)-144(-12)+((-12)^3)/(3)}]`

`=\pi [ -1728 + 144h - (1)/(3)(-1728+h^3+432h-36h^2)+1728-(1728)/(3)]`

`=\pi [ 144h - (1)/(3)(h^3+432h-36h^2}{3}]`

`=\pi [ 144h - (h^3)/(3) - 144h + 12h^2]`

`=\pi ( 12h^2 - (h^3)/(3))`

Special cases :

If h = 0,

`V(0) = 0`

If h = 12,

`V(12) = \pi ( 1728 - 576) = 1152\pi`