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

Answer:

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,

⇒ $y=-\sqrt{144-x^2}$

Here, h represents the depth of water,

Thus, by using shell method,

The volume of the disk would be,

$V(h) = \pi \int_{-12}^{-12+h} x^2 dx$

$= \pi \int_{-12}^{-12+h} (144-y^2) dy$

$= \pi |144y-\frac{y^3}{3}|_{-12}^{-12+h}$

$=\pi [ (144(-12+h)-\frac{(-12+h)^3}{3}-144(-12)+\frac{(-12)^3}{3}}]$

$=\pi [ -1728 + 144h - \frac{1}{3}(-1728+h^3+432h-36h^2)+1728-\frac{1728}{3}]$

$=\pi [ 144h - \frac{1}{3}(h^3+432h-36h^2}{3}]$

$=\pi [ 144h - \frac{h^3}{3} - 144h + 12h^2]$

$=\pi ( 12h^2 - \frac{h^3}{3})$

Special cases :

If h = 0,

$V(0) = 0$

If h = 12,

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