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 [tex]x^2+y^2=144[/tex], where y≤ 0,
⇒ [tex]y=-\sqrt{144-x^2}[/tex]
Here, h represents the depth of water,
Thus, by using shell method,
The volume of the disk would be,
[tex]V(h) = \pi \int_{-12}^{-12+h} x^2 dx[/tex]
[tex]= \pi \int_{-12}^{-12+h} (144-y^2) dy[/tex]
[tex]= \pi |144y-\frac{y^3}{3}|_{-12}^{-12+h}[/tex]
[tex]=\pi [ (144(-12+h)-\frac{(-12+h)^3}{3}-144(-12)+\frac{(-12)^3}{3}}][/tex]
[tex]=\pi [ -1728 + 144h - \frac{1}{3}(-1728+h^3+432h-36h^2)+1728-\frac{1728}{3}][/tex]
[tex]=\pi [ 144h - \frac{1}{3}(h^3+432h-36h^2}{3}][/tex]
[tex]=\pi [ 144h - \frac{h^3}{3} - 144h + 12h^2][/tex]
[tex]=\pi ( 12h^2 - \frac{h^3}{3})[/tex]
Special cases :
If h = 0,
[tex]V(0) = 0[/tex]
If h = 12,
[tex]V(12) = \pi ( 1728 - 576) = 1152\pi [/tex]
