Chapter 7.3, Problem 81E

(a)

To determine

The volume of water in the bowl.

Answer to Problem 81E

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

Explanation of Solution

Given:

Radius of a hemispherical bowl is a and contains water to the depth h.

Calculation:

The equation for a hemispherical bowl can be written as:

  $x^2+y^2+z^2=a^2$

As, the radius of bowl is a and depth is h then, using spherical coordinate system the bound for volume can be written as:

  $\frac{a-h}{\cos \varphi }\le \rho \le \alpha ,0\le \varphi \le \arccos \left[\frac{a-h}{a}\right]\text{ (0}\le \theta \le \text{2}\prod \text{)}$

Hence, volume of water in hemisphere is:

  $\begin{array}{l}V={\displaystyle \underset{0}{\overset{2\prod }{\int }}{\displaystyle \underset{0}{\overset{\arccos (\frac{a-h}{a})}{\int }}{\displaystyle \underset{\frac{a-h}{\cos \varphi }}{\overset{a}{\int }}{\rho }^2\sin (\varphi )d\rho d\varphi d\theta }}}\\ V={\displaystyle \underset{0}{\overset{2\prod }{\int }}{\displaystyle \underset{0}{\overset{\arccos (\frac{a-h}{a})}{\int }}{\left[\frac{{\rho }^3}{3}\right]}_{\frac{a-h}{\cos \varphi }}^a}}\sin (\varphi )d\varphi d\theta \\ V=\frac{\prod h^2}{3}(3a-h)\end{array}$

Conclusion:

  $x.f(x)=\sin x$ in the interval $0\le x\le \prod$ is proved.

(b)

To determine

The rate at which water level rising.

Answer to Problem 81E

The rate at which water level rising is $\frac{1}{120\prod }\frac{m}{\sec }$ .

Explanation of Solution

Given:

The radius of hemispherical bowl is $5\text{ m}$ and the rate at which water runs in hemispherical bowl is $0.2\frac{m^3}{\sec }$ .

Calculation:

  $V=\frac{\prod h^2}{3}(3a-h)$

Differentiate both sides with respect to h.

  $\begin{array}{l}\frac{dV}{dt}=a\prod (2h)\frac{dh}{dt}-\frac{\prod }{3}(3h^2)\frac{dh}{dt}\\ $ 0.2=2(5)(4)\prod \frac{dh}{dt}-\prod {(4)}^2\frac{dh}{dt}$0.2=40\prod \frac{dh}{dt}-16\prod \frac{dh}{dt}\\ 0.2=24\prod \frac{dh}{dt}\\ \frac{1}{5}=24\prod \frac{dh}{dt}\\ \frac{dh}{dt}=\frac{1}{120\prod }\frac{m}{\sec }\end{array}$