Science:Math Exam Resources/Courses/MATH101 C/April 2025/Question 06/Solution 1

First, we identify the points where the two curve intersect; setting $x^n=x^{-n}$, we get the intersection point $x=1$. The region $R$ can be split into two subregions, one under $y=x^n$ over $0\le x\le 1,$ and one under $y=x^{-n}$ over $1\le x<\mathrm{\infty }.$ The two subregions are illustrated in the attached Figure. Thus, the required volume is given by integrating over the two subregions, $$\begin{array}{rl}V(n)& =\underset{0}{\overset{1}{\int }}\pi x^{2n}dx+\underset{1}{\overset{\mathrm{\infty }}{\int }}\frac{\pi }{x^{2n}}dx\\ & =\frac{\pi x^{2n+1}}{2n+1}{|}_0^1+\underset{a\to \mathrm{\infty }}{\lim }\frac{\pi }{(-2n+1)x^{2n-1}}{|}_1^a\\ & =\frac{\pi }{2n+1}+\underset{a\to \mathrm{\infty }}{\lim }(\frac{\pi }{(-2n+1)a^{2n-1}}-\frac{\pi }{-2n+1})\\ \end{array}$$

Now, looking at the limit expression on the right, we notice that only the first term depends on $a$. Since we are given $n>1$, we know the exponent in $a^{2n-1}$ is positive. Using this information we can say $\underset{a\to \mathrm{\infty }}{\lim }\frac{1}{a^{2n-1}}=0$. Hence we compute the limit as follows $\underset{a\to \mathrm{\infty }}{\lim }(\frac{\pi }{(-2n+1)a^{2n-1}}-\frac{\pi }{-2n+1})=\underset{a\to \mathrm{\infty }}{\lim }\left(\frac{\pi }{(-2n+1)a^{2n-1}}\right)-\frac{\pi }{-2n+1}=-\frac{\pi }{-2n+1}.$

Finally, we get the following simplified expression for the volume $$V(n)=\frac{\pi }{2n+1}-\frac{\pi }{-2n+1}=\frac{4n\pi }{4n^2-1}.$$