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given the function $\frac {1}{x^p}$ revolved around the x-axis find a p value such that the volume and the surface area of the revolved solid is infinite.

I am not sure if such a function actually exist, thus far i have only figured out that functions either converge and diverge for the volumen and surface area. i would appreciate any help.

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  • $\begingroup$ Do you have integrals representing the volume and surface area for general $p$? If so please post them as part of the question... $\endgroup$ – rogerl Nov 21 '17 at 17:43
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Let $p=\dfrac{1}{2}$. Then $f(x)=\dfrac{1}{\sqrt{x}}$ and the volume of the horn generated by revolving the region under $f(x)$ around the x axis is $\pi \int_1^{\infty} \dfrac{dx}{x}=\infty$. The surface area of that solid is $2\pi \int_1^{\infty}\,\dfrac{1}{\sqrt{x}}\sqrt {1+(4\,{x}^{3})^{-1}}dx=2\pi \int_1^{\infty}\,{\frac {\sqrt {4\,{x}^{3}+1}}{2{x}^{2}}}dx\geq \pi \int_1^{\infty}\,{\frac {2\sqrt {\,{x}^{3}}}{{x}^{2}}}dx=2\pi \int_1^{\infty}\,{\frac {1}{\sqrt{x}}}dx=\infty$. It will still work for any $p$ satisfying $0<p\leq \dfrac{1}{2}.$ For $\dfrac{1}{2}<p\leq 1$ the volume will be finite but the surface area will still be infinite. For $p>1$ both will be finite. Here is the work for the general case: Volume $\pi \int_1^{\infty} \dfrac{dx}{x^{2p}}=\infty$ if $p\leq \dfrac{1}{2}.$ Surface area is $2\pi \int_1^{\infty}\,f(x)\sqrt {1+(f'(x))^{2}}dx=2\pi \int_1^{\infty}\,\dfrac{1}{{x}^{p}}\sqrt {1+\dfrac{p^2}{x^{2p+2}}}dx=2\pi \int_1^{\infty}\,\dfrac{1}{{x}^{2p+1}}\sqrt {x^{2p+2}+p^2}dx$ which is very close to $2\pi \int_1^{\infty}\,\dfrac{1}{{x}^{2p+1}}\sqrt {x^{2p+2}}dx=2\pi \int_1^{\infty}\,\dfrac{1}{{x}^{p}} dx$. Therefore the surface area is infinite when $0<p\leq 1$ and finite otherwise.

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