Chapter 7.3, Problem 58E

(a)

To determine

The volume of the bowl.

Answer to Problem 58E

The volume of the bowl is $25\prod$

Explanation of Solution

Given:

The shape of bowl generated by the graph of $y=\frac{x^2}{2}$ between $y=0$ and $y=5$ about the y-axis.

Calculation:

The volume of bowl can be find out as:

  $\begin{array}{l}V=2\prod {\displaystyle \underset{0}{\overset{\sqrt{10}}{\int }}(x)(5-\frac{x^2}{2})dx}\\ V=2\prod {\displaystyle \underset{0}{\overset{\sqrt{10}}{\int }}(5x-\frac{x^3}{2})}dx\\ V=2\prod {\left[\frac{5x^2}{2}-\frac{x^4}{8}\right]}_0^{\sqrt{10}}\\ V=2\prod \left[\frac{5(10)}{2}-\frac{100}{8}\right]\\ V=2\prod \left[25-\frac{25}{2}\right]\\ V=2\prod \left[\frac{25}{2}\right]\\ V=25\prod \end{array}$

Conclusion:

The volume of the bowl is $25\prod$

(b)

To determine

To find the rate at which water level in the bowl rises.

Answer to Problem 58E

The rate at which water level in the bowl rises is $0.1193\frac{ft}{\sec }$

Explanation of Solution

Given:

The rate at which bowl fills with water is $3\text{ cubic units per second}$ .

Calculation: $\frac{dV}{dt}=3$ and height is $4\text{ units}$ .

Assume height to be y and radius to be x.

Then, volume is:

  $V=A\times H$

Where;

V is the volume of bowl.

A is the area of bowl.

H is the height.

  $\begin{array}{l}V=A\times H\\ V=(\prod x^2)\times y\\ V=\prod y^2\end{array}$

To find the rate, differentiate the volume with respect to t.

  $\begin{array}{l}\frac{dV}{dt}=\prod (2y)\left(\frac{dy}{dt}\right)\\ 3=\prod (8)\left(\frac{dy}{dt}\right)\\ \left(\frac{dy}{dt}\right)=\frac{3}{8\prod }\\ \left(\frac{dy}{dt}\right)=0.1193\end{array}$

Conclusion:

The rate at which water level in the bowl rises is $0.1193\frac{ft}{\sec }$ .