Science:Math Exam Resources/Courses/MATH101 A/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 $0\leq x\leq 1,$ and one under $y=x^{-n}$ over $1\leq x<\infty .$ The two subregions are illustrated in the attached Figure. Thus, the required volume is given by integrating over the two subregions, $${\begin{aligned}V(n)&=\int _{0}^{1}\pi x^{2n}dx+\int _{1}^{\infty }{\frac {\pi }{x^{2n}}}dx\\&={\frac {\pi x^{2n+1}}{2n+1}}{\Big |}_{0}^{1}+\lim _{a\to \infty }{\frac {\pi }{(-2n+1)x^{2n-1}}}{\Big |}_{1}^{a}\\&={\frac {\pi }{2n+1}}+\lim _{a\to \infty }\left({\frac {\pi }{(-2n+1)a^{2n-1}}}-{\frac {\pi }{-2n+1}}\right)\\\end{aligned}}$$

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 $\lim _{a\to \infty }{\frac {1}{a^{2n-1}}}=0$. Hence we compute the limit as follows $\lim _{a\to \infty }\left({\frac {\pi }{(-2n+1)a^{2n-1}}}-{\frac {\pi }{-2n+1}}\right)=\lim _{a\to \infty }\left({\frac {\pi }{(-2n+1)a^{2n-1}}}\right)-{\frac {\pi }{-2n+1}}=-{\frac {\pi }{-2n+1}}.$

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