Question

A hemispherical bowl of internal diameter 12 cm, full of water, is emptied into a cylindrical can 8 cm in diameter. How deep is the water in the can?​

Answer 1

Step-by-step explanation:

The volume $V_s$ of a sphere of radius R is is given by

$V_s = \frac{4\pi}{3}R^3$

so a hemisphere will only have half the volume of a sphere so we can write its volume $V_h$ as

$V_h = \frac{1}{2}V_s = \frac{2\pi}{3}R^3$

Since the diameter D is twice the radius R, we can rewrite $V_h$ in terms of D as

$V_h = \frac{2\pi}{3}\left(\frac{1}{2}D\right)^3 = \frac{\pi}{12}D^3$

$\:\:\:\:\:\:\:=\frac{\pi}{12}(12\:\text{cm})^3 = 452.4\:\text{cm}^3$

Now represents the amount of water inside the hemisphere, which will be then transferred to a cylinder 8 cm in diameter. The volume of this cylindrical can $V_c$ must equal the volume of the hemisphere:

$V_h = V_c = \pi R^2h = \pi\left(\frac{1}{2}D\right)^2h$

$\:\:\:\:\:\:\:=\frac{\pi}{4}D^2h$

where h is the height (or depth) of the can. Solving for h, we get

$h = \dfrac{4V_h}{\pi D^2} = \dfrac{4(452.4\:\text{cm}^3)}{\pi(8\:\text{cm})^2}$

$\:\:\:\:\:= 9\:\text{cm}$