How can I compute the volume of $$S=\{(x,y,z)\in\mathbb R^3\, |\, x^2+y^2 \le\frac{1}{(1+z)^2}, 0\le z\le 1\}$$ by exclusively using integration? I know that I can use Cavalieri, but I don't understand the solution, a step by step explanation would be much appreciated. The result we are looking for is $\frac{\pi}{2}$.
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$\begingroup$ So you do not want to view it as a solid of revolution? I suggest that you view this as the volume that appears when rotating $x=1/(1+z)$ around the $z$-axis (and when $0\leq z\leq 1$). Then $$\text{volume}\,(S)=\int_0^1 \pi \frac{1}{(1+z)^2}\,dz=\bigl[-\pi/(1+z)\bigr]_0^1=\frac{\pi}{2}.$$ It cannot be easier, I think. $\endgroup$– mickepJun 12, 2016 at 8:50
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$\begingroup$ Yes thank you, but what is the intuition behind it? How do I know that I have to rotate $x$ around the $z$-axis? $\endgroup$– ZetaJun 12, 2016 at 8:54
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$\begingroup$ See here, for instance: phengkimving.com/calc_of_one_real_var/12_app_of_the_intgrl/… $\endgroup$– Intelligenti paucaJun 12, 2016 at 8:56
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$\begingroup$ The appearance of $x^2+y^2$ implies that the domain is rotational symmetric in those variables, i.e. that $S$ is a body obtained via rotation around the $z$-axis. $\endgroup$– mickepJun 12, 2016 at 8:56
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$\begingroup$ Perfect, thank you both. $\endgroup$– ZetaJun 12, 2016 at 8:58
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1 Answer
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Use cylindrical polar coordinates to get the volume as the following integral $$ \int_0^{2\pi}\int_{0}^1\int_{0}^{1\over 1+z}rdrdzd\theta =\pi\int_0^1{1\over(1+z)^2}dz={\pi\over2} $$
answered Jun 12, 2016 at 9:12
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$\begingroup$ how to find upper and lower limit of the integrals? $\endgroup$– ZetaJun 12, 2016 at 9:37
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$\begingroup$ Your set is $\{(r,\theta,z):r^2\le{1\over(z+1)^2},z\in[0,1]\}$. The first inequality gives the upper bound for $r$ as ${1\over1+z}$ $\endgroup$ Jun 12, 2016 at 9:40