Science:Math Exam Resources/Courses/MATH101 C/April 2024/Question 19/Solution 1

Washer (or annulus) cross-section of a volume of revolution

Let us use the method of washers. (We could also use the cylindrical shells.) The curves intersect at $(0, 0)$ and $(1 / a, 1)$ .

To find the volume of the solid formed by rotating the region around the $x$ -axis, think of this solid as a sum of cross-sections. For each $x$ from $0$ to $1 / a$ , as shown in the figure to the right, the cross section at $x$ is a ring with outer radius $R = \sqrt{a x}$ and inner radius $r = a^{2} x^{2}$ . Thus, the area of this cross section is $π (R^{2} - r^{2}) = π (a x - a^{4} x^{4})$ . Each cross section should be thought of as having a "thickness" of $d x$ . Therefore, adding up the cross sections to find the volume, we get

$\begin{aligned} V_{x} = \int_{0}^{1 / a} π (a x - a^{4} x^{4}) d x = π {[\frac{a x^{2}}{2} - \frac{a^{4} x^{5}}{5}] |}_{0}^{1 / a} = π (\frac{a^{- 1}}{2} - \frac{a^{- 1}}{5}) = \frac{3 π}{10 a} . \end{aligned}$

For the solid formed by rotating the region around the $y$ -axis, we must first rewrite the equations of the curves so that they give $x$ in terms of $y$ , instead of $y$ in terms of $x$ . The curve $y = a^{2} x^{2}$ can also be written as $x = \sqrt{y} / a$ , and the curve $y = \sqrt{a x}$ can be written as $x = y^{2} / a$ . We omit the figure in this case but encourage you to draw your own. You will see that for each $y \in [0, 1]$ , the cross section of this new solid at $y$ is a ring with outer radius $R = \sqrt{y} / a$ and inner radius $r = y^{2} / a$ . This cross section has an area of $π (R^{2} - r^{2}) = π (y / a^{2} - y^{4} / a^{2})$ , and a "thickness" of $d y$ . By adding up all these cross sections, the volume is

$\begin{aligned} V_{y} = \int_{0}^{1} π (\frac{y}{a^{2}} - \frac{y^{4}}{a^{2}}) d y = π {[\frac{y^{2}}{2 a^{2}} - \frac{y^{5}}{5 a^{2}}] |}_{0}^{1} = \frac{π}{a^{2}} (\frac{1}{2} - \frac{1}{5}) = \frac{3 π}{10 a^{2}} . \end{aligned}$

Finally, let us find which $a$ makes it so that $V_{y} = 2 V_{x}$ . We would like $a$ to satisfy $\frac{3 π}{10 a^{2}} = 2 \cdot \frac{3 π}{10 a^{2}}$ . The solution is $a = 1 / 2$ .