Chapter 6, Problem 13RE

To determine

To Find: The volume of the solid that is obtained by rotating the region that is bounded by the given curves about the specified axis.

Answer to Problem 13RE

The required volume is $\pi \left[\frac{4}{3}{\left(h^2+2ah\right)}^{\frac{3}{2}}\right]$ .

Explanation of Solution

Given:

The given expression is $x^2-y^2=a^2$ .

The value of $x=a+h$

Calculation:

First find the points of the intersection of the curves $x^2-y^2=a^2$ or $x=\sqrt{a^2+y^2}$ .

Then

  $\begin{array}{l}\sqrt{a^2+y^2}={\left(a+h\right)}^2\\ y=\pm \sqrt{h\left(h+2a\right)}\end{array}$

Consider the sketch for the curve is shown in Figure 1

  

Figure 1

Rotate the shaded region about the y axis as it is rotated in this way a solid is obtained that is cut a slice from the solid horizontal as the washer is obtained.

Consider the outer radius is $a+h$ .

The cross sectional area of the washer is,

  $\begin{array}{c}A\left(y\right)=\pi \left[{\left(a+h\right)}^2-{\left(\sqrt{a^2+y^2}\right)}^3\right]\\ =\pi \left[\left(h^2+2ah\right)-y^2\right]\end{array}$

Consider that,

  $\begin{array}{c}h\left(h+2a\right)=h^2+2ah\\ =k\end{array}$

Then, the volume solid is,

  $\begin{array}{c}V={\displaystyle {\int }_{\sqrt{h\left(h+2a\right)}}^{\sqrt{h\left(h+2a\right)}}A\left(y\right)dy}\\ =\pi {\left[ky-\frac{y^3}{3}\right]}_3^{\sqrt{k}}\\ =\pi \left[\frac{4}{3}{k}^{\frac{3}{2}}\right]\\ =\pi \left[\frac{4}{3}{\left(h^2+2ah\right)}^{\frac{3}{2}}\right]\end{array}$