Chapter 7.3, Problem 61E

To determine

To Find: The value of a.

Answer to Problem 61E

The value of a is $5$ .

Explanation of Solution

Given information: The region is enclosed in the first quadrant between graph of $y=ax-x^2$ , and x-axis.

Calculation:

Given equation is:

  $\begin{array}{l}y=ax-x^2=0\\ x(a-x)=0\\ x=0,x=a\end{array}$

This means that the parabola $y=ax-x^2$ intersects the x-axis at two points and those points are: $x=0\text{ and }x=a$ .

Consider, the solid is generated by revolving the region around x-axis. Hence, volume can be find out as:

  $\begin{array}{l}V=\prod {\displaystyle \underset{0}{\overset{a}{\int }}{(ax-x^2)}^{2}dx}\\ V=\prod \left[{\displaystyle \underset{0}{\overset{a}{\int }}{(ax)}^{2}dx+{\displaystyle \underset{0}{\overset{a}{\int }}{x}^{4}dx}}-{\displaystyle \underset{0}{\overset{a}{\int }}2a{x}^{3}dx}\right]\\ V=(\prod )\left[(a^2.{(\frac{x^3}{3})}_0^a)\right]+{\left[\frac{x^5}{5}\right]}_0^a-2a.{\left[\frac{x^4}{4}\right]}_0^a\\ V=(\prod )\left[a^2.\frac{a^3}{3}\right]+\left[\frac{a^5}{5}\right]-2a.\left[\frac{a^4}{4}\right]\\ V=(\prod )\left[\frac{a^5}{3}+\frac{a^5}{5}-\frac{a^5}{2}\right]\\ V=\frac{\prod a^5}{30}\end{array}$

Now, consider the solid is revolved around y-axis then, the volume is:

  $\begin{array}{l}V=2\prod {\displaystyle \underset{0}{\overset{a}{\int }}x(ax-x^2)dx}\\ V=2\prod \left[{\displaystyle \underset{0}{\overset{a}{\int }}a{x}^2-x^{3}dx}\right]\\ V=(2\prod )\left[(a.{(\frac{x^3}{3})}_0^a)\right]-{\left[\frac{x^4}{4}\right]}_0^a\\ V=(2\prod )\left[a.\frac{a^3}{3}\right]-\left[\frac{a^4}{4}\right]\\ V=(2\prod )\left[\frac{a^4}{3}-\frac{a^4}{4}\right]\\ V=\frac{\prod a^4}{6}\end{array}$

As, the volume is independent of axis on which it is being revolved then,

  $\begin{array}{l}\frac{\prod a^4}{6}=\frac{\prod a^5}{30}\\ a=5\end{array}$