Science:Math Exam Resources/Courses/MATH101/April 2018/Question 01 (iv)

MATH101 April 2018

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Question 01 (iv)
Let ${\textstyle y=f(x)={1/x^{p}}}$ on the interval ${\textstyle [1,\infty )}$ . For which values of ${\textstyle p}$ do we get a finite volume by revolving ${\textstyle f(x)}$ about the ${\textstyle x}$ -axis?

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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 graph ${\textstyle y=f(x)}$ , the formula for the volume of the solid given by rotating the portion of the curve on the interval ${\textstyle [a,b]}$ about the ${\textstyle x}$ -axis is $V=\int _{a}^{b}\pi f(x)^{2}\;dx.$

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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. Using the formula in the hint, the volume of the solid is given by $\int _{1}^{\infty }\pi {\bigg (}{\frac {1}{x^{p}}}{\bigg )}^{2}\;dx=\pi \int _{1}^{\infty }{\frac {1}{x^{2p}}}\;dx,$ so the question reduces to asking for which values of ${\textstyle p}$ does the integral ${\textstyle \int _{1}^{\infty }{\frac {1}{x^{2p}}}\;dx}$ converge? We know that this integral converges if and only if ${\textstyle 2p>1}$ so the volume is finite if and only if ${\textstyle p>1/2}$ . Answer: The correct answer is ${\textstyle {\color {blue}p>1/2}}$ .