Let $S_{n}(R)$ be a surface area of a $n+1$-dimensional sphere of radius $R$. It is easy to obtain $S_n(R)=S_n(1)R^{n}$. In particular, $S_1(R)=2\pi R$ and $S_2(R)=4\pi R^2$. Consider the sphere $S_n(1)$ centred at the origin with radius 1. A vertical hyperplane $x_1=x$ intersects the sphere and leaves a cuts, which should be a $n$-dimensional sphere $S_{n-1}(r)$. The intuition is that you will obtain a circle when cutting a sphere by a plane. So the nearby surface area of the original $n+1$ sphere can be locally approximated by $\Delta x\cdot S_{n-1}(r)$, the surface area of a slim stripe. Inspired by the calculus approach and summing up the surface areas of all the slim stripes and having $\Delta\rightarrow0$, I have\begin{equation}\label{eq:5}S_n(1)=2\int_0^1S_{n-1}(r)dx,\end{equation} with $r=\sqrt{1-x^2}$. Hence \begin{eqnarray*}S_n(1)&amp;=&amp; 2\int_0^1S_{n-1}(r)dx =2\int_0^1S_{n-1}(1)r^{n-1}dx\\ &amp;=&amp;2 S_{n-1}(1)\int_0^1r^{n-1}dx=2S_{n-1}(1)\int_0^1 (\sqrt{1-x^2})^{n-1}dx.\end{eqnarray*} Let $t=1-x^2$, so $x=\sqrt{1-t}$, and $dx=-1/2\cdot (1-t)^{-1/2}dt$. The last expression becomes \begin{eqnarray*}S_n(1)&amp;=&amp; 2S_{n-1}(1)\int_0^1 (\sqrt{1-x^2})^{n-1}dx=-2S_{n-1}(1)\int_1^0 t^{(n-1)/2}\cdot \frac{1}{2}\cdot (1-t)^{-1/2}dt\\ &amp;=&amp;S_{n-1}(1)\int_0^1 t^{(n-1)/2} (1-t)^{-1/2}dt=S_{n-1}(1)\mathrm{B}(\frac{n+1}{2},\frac{1}{2}),\end{eqnarray*}where the notation $\mathrm{B}$ stands for the Beta function. I therefore obtain a recursive relationship \begin{equation}\label{eq:4}S_{n}(1)=S_{n-1}(1)\frac{\Gamma(\frac{n+1}{2})\Gamma(\frac{1}{2})}{\Gamma(\frac{n+2}{2})}.\end{equation} Utilizing \eqref{eq:4} I conclude that \begin{eqnarray*}S_n(1)&amp;=&amp;S_1(1) \frac{\Gamma(\frac{3}{2})\left[\Gamma(\frac{1}{2})\right]^{n-1}}{\Gamma(\frac{n+2}{2})}=S_1(1) \frac{\frac{\sqrt{\pi}}{2}(\sqrt{\pi})^{n-1}}{\Gamma(\frac{n+2}{2})}\\ &amp;=&amp;2\pi \frac{\frac{1}{2}\pi^{n/2}}{\Gamma(\frac{n+2}{2})}=\frac{\pi^{\frac{n+2}{2}}}{\Gamma(\frac{n+2}{2})}.\end{eqnarray*}  However, when substituting $n=1,2$ into he last expression, I get \begin{eqnarray*}S_1(1)&amp;=&amp;\frac{\pi^{3/2}}{\Gamma(3/2)}=2\pi,\\S_2(1)&amp;=&amp;\frac{\pi^2}{\Gamma(2)}=\pi^2,\end{eqnarray*}where the second equation  is plainly wrong. I trust my calculation is correct. The only possible explanation is therefore my approach to establish the recursive relationship (the first equation) must be wrong. But I don't see where the problem is. Any suggestions or comments are highly appreciated.
I am using two books for my calculus refresher.   Thomas' Calculus  Higher Math for Beginners by Ya. B. Zeldovich   My question is : When applying Integral Calculus for calculation of volumes of solids, generated by curves revolved around an axis, we use slices of 'cylinders' to approximate the volume of the resulting solid and then integrate the sum of those infinitesimal cylinders. However, when we are using the same techniques to calculate the surface area of the surfaces generated by revolving curves around an axis, we consider the 'slope' of the differential length of curve 'dx', calculate the total length of the curve and derive the required surface area.  Are we not supposed to use the same 'slope' for calculating the volumes of the infinitesimal 'cylinders' for calculation of volumes? Shouldn't we use 'sliced portions of 'cones' as the infinitesimal volumes?? When it come to calculation of volumes of solids of revolution, why are we neglecting the slope of the curve for the differential length and simply assume that it is an infinitesimal cylinder??  Ex: Let us say we want to calculate the surface area and the volume of the solid generated when the parabola y = 10 . x^2 is revolved about the y-axis, with limits of x from 0 to 5.  In such cases, when calculating the volume of the solid, we consider infinitesimal 'cylinders', ignoring the 'slope' of the curve for the differential element 'dx', but when calculating the surface area, we consider the 'slope' of the differential element 'dx'.
Why and how is the speed of light in vacuum constant, i.e., independent of reference frame? I am looking for a step by step instruction to guide me in the ways of installing, configuring and debugging many of the issues new users have when dealing with Wine in Ubuntu.  What is the recommended way of installing Wine, some of the key issues that it has and how to solve them?  If I have a specific problem with an App in Wine, how do I go about solving it? What are the recommended steps to ensure that I can fix a problem related to an app I wish to use? Let $A$ be a UFD. Assume that $a,b \in A$ are relatively prime, $c \in A$ and $a | bc$. To prove that $a|c$, is the following approach correct (or do you have to use some type of prime factorization argument)?  By the relatively prime assumption, $\gcd(a,b)$ is a unit. Call that unit $u \in A^{\times}$. Then we can write $u = ax + by$ for some $x,y \in A$. But then, $c = c \cdot u = c(ax+by)=cax+cby$. However, it is clear that $a|cax$ and the fact that $a|bc$ implies $a|cby$. Thus, $a|cax+cby=c$. How to make shortcuts on Ubuntu 17.10? Please give me feedback on my answer to this question. Question: For all $ n\geq1:\binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n} $   is equal to $ \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1} $. Answer: False, since;  Let $a=1,  b=-1$  and $n=1$  ,$k=0$   , using Binomial expression.  $$(a+b)^n=  \sum   \binom{n}{k}  a^{k}b^{n-k}$$  Then $(1+-1)^{2(1)}=\binom{2}{0}1^{0}(-1)^{2-0}+\binom{2}{1}1^{1}(-1)^{2-1}+\binom{2}{2}1^{2}(-1)^{2-2}=\binom{2}{0}-\binom{2}{1}+\binom{2}{2}$  Thus :For all $n\geq1:  \binom{2n}{0}+\binom{2n}{2}+\binom{2n}{4}+\cdots+\binom{2n}{2k}+\cdots+\binom{2n}{2n}   \ne \binom{2n}{1}+\binom{2n}{3}+\binom{2n}{5}+\cdots+\binom{2n}{2k+1}+\cdots+\binom{2n}{2n-1}$  . Is there a Windows 10 Hotkey or Shortcut for switching to a specific display? 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