Chapter 8, Problem 22RE

a.

To determine

To find : the volume of the bowl.

Answer to Problem 22RE

The volume of the bowl is $4\pi$ .

Explanation of Solution

Given information : the section of the parabola $y^2=\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ from $y=0$ to $y=2$ is revolved about the y-axis from a bowl.

Calculation :

The volume of the bowl:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{2}{\int }}\pi \left(2y\right)}dy\\ {\displaystyle \underset{0}{\overset{2}{\int }}\left(2\pi y\right)}dy\\ \Rightarrow \pi y^2\\ \Rightarrow 4\pi \end{array}$

b.

To determine

To find : the weight that bowl is holding when it is filled to a depth of $k$ unit.

Answer to Problem 22RE

The bowl is hold $\pi k^2$ .

Explanation of Solution

Given information : the section of the parabola $y^2=\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ from $y=0$ to $y=2$ is revolved about the y -axis from a bowl.

Calculation :

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{k}{\int }}\pi \left(2y\right)}dy\\ {\displaystyle \underset{0}{\overset{k}{\int }}\left(2\pi y\right)}dy\\ \Rightarrow \pi y^2\\ \Rightarrow \pi k^2\end{array}$

c.

To determine

To find : the speed of depth $k$ increasing when $k=1$ .

Answer to Problem 22RE

The liquid rising at a rate of $\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$\pi $}\right.$ .

Explanation of Solution

Given information : the section of the parabola $y^2=\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ from $y=0$ to $y=2$ is revolved about the y -axis from a bowl.

Calculation :

The derivative of the integral with respect to time:

  $\frac{dV}{dt}=\frac{d}{dt}\left[{\displaystyle \underset{0}{\overset{k}{\int }}\pi 2ydy}\right]$

Since, $k=1$ and $\frac{dV}{dt}=2$

  $\begin{array}{l}2=\pi \cdot 2\cdot \frac{dk}{dt}\\ \Rightarrow \frac{1}{\pi }\end{array}$