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

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MATH101 C April 2025

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  •  April 2024  •  

Question 06
Consider the region $R$ in the first quadrant bounded by the $x$-axis and the curves $y=x^{-n}$ and $y=x^n$, where $n>1$ is a real number. Region $R$ bounded by the $x$-axis and the curves $y=x^{-n}$ and $y=x^n$. Find the volume $V(n)$ of the solid generated by revolving $R$ about the $x$-axis. Express your answer in terms of $n$.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Start by finding where the two curves $y=x^n$ and $y=x^{-n}$ intersect. Use this to split the region into two parts: one for $0\le x\le 1$ and one for $x\ge 1$. We can now think of the volume as the sum of which two integrals?

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Solution

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 $$V(n)=\frac{\pi }{2n+1}-\frac{\pi }{-2n+1}=\frac{4n\pi }{4n^2-1}.$$

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