Chapter 8.3, Problem 73E

a.

To determine

To find : the volume of water in the bowl.

Answer to Problem 73E

The volume of water in the bowl is $\frac{\pi h^2\left(3a-h\right)}{3}$ .

Explanation of Solution

Given information : A hemispherical bowl of radius $a$ contains water to a depth $h$ .

Calculation :

By the washer method.

Using the region bounded by the right half of the semi circle, which has equation.

  $\begin{array}{l}x=\sqrt{a^2-{\left(y-a\right)}^2}\\ \Rightarrow \sqrt{2ay-y^2}\end{array}$

The volume of water at depth $h$ is given by:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{h}{\int }}\pi \left(2ay-y^2\right)}dy\\ \Rightarrow \frac{\pi h^2\left(3a-h\right)}{3}\end{array}$

b.

To determine

To calculate : the speed of the water rising in the bowl.

Answer to Problem 73E

The speed of the water rising in the bowl is $\frac{1}{120\pi }$ .

Explanation of Solution

Given information : Water runs into a sunken concrete hemispherical bowl of radius $5$ m at the of $0.2m^3/\sec$ and the water is $4m$ deep.

Calculation :

From the part (a).

  $\begin{array}{l}\frac{dV}{dh}=\pi \left(10h-h^2\right)\\ \frac{dV}{dh}=0.2\\ \frac{dV}{dh}=\frac{dV}{dh}\cdot \frac{dh}{dt}\end{array}$

So, for $\frac{dh}{dt}$ using the two equations above, this should give us.

  ${\frac{dh}{dt}|}_{h=4}=\frac{1}{120\pi }$