Science:Math Exam Resources/Courses/MATH103/April 2005/Question 05 (b)/Solution 1

For similar reasons to part (a) we will have two cases because we will have to worry about the integral we are doing. There will be a unique situation if we are ever integrating 1/x which occurs when p=1/2 because

$V=\underset{1}{\overset{B}{\int }}\pi {\left[\frac{1}{x^p}\right]}^{2}dx=\underset{1}{\overset{B}{\int }}\pi \frac{1}{x^{2p}}dx.$

If $p\ne 1/2$ then when we rotate the function $y=f(x)=\frac{1}{x^p}$ about the x-axis over the interval [1,B], the volume obtained is

$\begin{array}{rl}V& =\underset{1}{\overset{B}{\int }}\pi {\left[\frac{1}{x^p}\right]}^{2}dx=\pi \underset{1}{\overset{B}{\int }}\frac{1}{x^{2p}}dx=\pi \underset{1}{\overset{B}{\int }}{x}^{-2p}dx\\ & =\pi {\left[\frac{x^{-2p+1}}{-2p+1}\right]}_1^B=\frac{\pi }{1-2p}(B^{1-2p}-1)\end{array}.$

If p=1/2 then

$\begin{array}{rl}V& =\underset{1}{\overset{B}{\int }}\pi {\left[\frac{1}{x^{1/2}}\right]}^{2}dx=\pi \underset{1}{\overset{B}{\int }}\frac{1}{x}dx\\ & =\pi {[\ln x]}_{x=1}^{x=B}=\pi \ln B\end{array}.$