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

MATH103 April 2005

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Question 05 (b)
Consider the curve given by $\displaystyle y=f(x)={\frac {1}{x^{p}}}$ for $\displaystyle 1\leq x\leq B$ where $\displaystyle p,B$ are constants with $\displaystyle B>1$ and $\displaystyle p>0$. (b) Compute the volume obtained by revolving the same region about the x-axis. (Later we will refer to this volume as V).

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
The volume V obtained by rotating a function y=f(x) around the x-axis (for $a\leq x\leq b$) is $V=\int _{a}^{b}\pi [f(x)]^{2}dx.$

Hint 2
We have $f(x)^{2}={\frac {1}{x^{2p}}}.$ Are there any values of p that makes this integral special compared to other values?

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. 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 ${\begin{aligned}V&=\int _{1}^{B}\pi \left[{\frac {1}{x^{1/2}}}\right]^{2}dx=\pi \int _{1}^{B}{\frac {1}{x}}dx\\&=\pi \left[\ln {x}\right]_{x=1}^{x=B}=\pi \ln {B}\end{aligned}}.$

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