Chapter 8.3, Problem 50E

a.

To determine

To find : the volume of the bowl.

Answer to Problem 50E

The volume of the bowl is $25\pi$ .

Explanation of Solution

Given information : A bowl has a shape that can be generated by revolving the graph of $y=\raisebox{1ex}{$x^2$}\!\left/ \!\raisebox{-1ex}{$2$}\right.$ between $y=0$ and $y=5$ about y -axis.

Calculation :

The volume of the bowl is given by:

  $\begin{array}{l}{\displaystyle \underset{0}{\overset{5}{\int }}\pi \cdot {\left(\sqrt{2y}\right)}^{2}dy}\\ \Rightarrow 25\pi \end{array}$

b.

To determine

To find: How fast will the water level in the bowl be rising when the water is $4$ unit deep.

Answer to Problem 50E

The water level in the bowl be rising at a rate of $\frac{3}{8\pi }$ .

Explanation of Solution

Given information : If we fill the bowl with water at a constant rate of $3$ cubic units per second.

Calculation :

The derivative of integral with respect to time.

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

Since $h=4\&\frac{dV}{dt}=3$

Solving the above

  $\begin{array}{l}3=\pi \cdot 8\cdot \frac{dh}{dt}\\ \Rightarrow \frac{3}{8\pi }\end{array}$

Hence, the liquid is rising at a rate of $\frac{3}{8\pi }$