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

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

If $p\neq {}1/2$ 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

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

If p=1/2 then

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