Science:Math Exam Resources/Courses/MATH101/April 2008/Question 02 (b)

MATH101 April 2008

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Question 02 (b)
Full-Solution Problem. Justify your answers and show all your work. Simplification of answers is not required. Let $R$ be the unbounded region that lies under the curve $y=1/x^{p}$, above the $x$-axis, and to the right of the vertical line $x=1$. For what values of the constant $p$ does the solid obtained by rotating $R$ about the $x$-axis have finite volume?

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
For a function $f(x)$, the volume of revolution about the x-axis from a to b of this function can be computed by $V=\int _{a}^{b}\pi f(x)^{2}\,dx$

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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. Let $f(x)={\tfrac {1}{x^{p}}}$. Then, we wish to know when is $V=\int _{1}^{\infty }\pi f(x)^{2}\,dx$ a finite number. To solve this, we compute directly that $V=\int _{1}^{\infty }\pi f(x)^{2}\,dx=\pi \int _{1}^{\infty }{\frac {1}{x^{2p}}}\,dx=\pi \lim _{b\to \infty }\int _{1}^{b}{\frac {dx}{x^{2p}}}$ This integral has to be treated in two cases. One when $2p=1$ and one when $2p\neq 1$. When $2p=1$ we have $V=\pi \lim _{b\to \infty }\int _{1}^{b}{\frac {dx}{x^{2p}}}=\pi \lim _{b\to \infty }\left.\ln |x|\right|_{1}^{b}=\lim _{b\to \infty }\pi \ln |b|$ and this diverges. When $2p\neq 1$, we have ${\begin{aligned}V&=\pi \lim _{b\to \infty }\int _{1}^{b}{\frac {dx}{x^{2p}}}\\&=\pi \lim _{b\to \infty }\left.{\frac {x^{1-2p}}{1-2p}}\right|_{1}^{b}\\&=\lim _{b\to \infty }{\frac {\pi b^{1-2p}}{1-2p}}-{\frac {\pi }{1-2p}}\end{aligned}}$ Now, the above limit converges when $1-2p<0$ and diverges when $1-2p>0$ (this is because the term will appear in the denominator in the first case and in the numerator in this second case). Thus, we get convergence only when ${\tfrac {1}{2}}<p$ as required.

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