Let the side of the following mentioned figures is s: The area of square is s^2 , the 3D area(volume) of a cube is s^3 ,the 4D area/volume of 4D hypercube can be said s^4 and so on. Now if radius is r, the area of circle is 𝛑r^2 , the 3D area(volume) of a sphere is(4/3)𝛑r^3 , what will be the 4D area/volume of 4D sphere and 5D area/volume of 5D sphere?


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Question Number 3843 by Rasheed Soomro last updated on 22/Dec/15

$L e t t h e s i d e o f t h e f o l l o w i n g m e n t i o n e d$ $f i g u r e s i s s :$ $T h e a r e a o f s q u a r e i s s^{2}, t h e 3 D a r e a (v o l u m e)$ $o f a c u b e i s s^{3}, t h e 4 D a r e a / v o l u m e o f 4 D$ $h y p e r c u b e c a n b e s a i d s^{4} a n d s o o n .$ $N o w i f r a d i u s i s r, t h e a r e a o f c i r c l e i s π r^{2},$ $t h e 3 D a r e a (v o l u m e) o f a s p h e r e i s \frac{4}{3} π r^{3},$ $w h a t w i l l b e t h e 4 D a r e a / v o l u m e o f 4 D$ $s p h e r e a n d 5 D a r e a / v o l u m e o f 5 D s p h e r e ?$
Commented by prakash jain last updated on 22/Dec/15

$Volume V_{n} = \frac{π^{n / 2} r^{n}}{Γ (1 + \frac{n}{2})}$ $Surface area S_{n - 1} = \frac{2 π^{n / 2} r^{n - 1}}{Γ (\frac{n}{2})} (s u r f a c e a r e a)$
Commented by Rasheed Soomro last updated on 22/Dec/15

$Volume V_{n} = \frac{π^{n / 2} r^{n}}{Γ (1 + \frac{n}{2})}$ $V_{3} = \frac{π^{3 / 2} r^{3}}{Γ (1 + \frac{3}{2})} \overset{?}{=} \frac{4}{3} π r^{3}$ $F o r 4 D S_{4 - 1} = S_{3} = \frac{2 π^{4 / 2} r^{4 - 1}}{Γ (\frac{4}{2})} [H o w m a n y d i m e n t i o n s ?]$ $W e c o n s i d e r s u r f a c e a s 2 D$
Commented by prakash jain last updated on 22/Dec/15

$For 3 - D s u r f a c e a r e a r^{2}$ $For 4 - D s u r f a c e a r e a r^{3}$ $S_{n - 1} is surface area equivalent n d i m e n t i o n a l$ $s p h e r e . W ill be proportional to r^{n - 1} .$ $Γ (\frac{5}{2}) = \frac{3}{2} Γ (\frac{1}{2}) = \frac{3 \sqrt{π}}{4}$
Commented by Rasheed Soomro last updated on 24/Dec/15

$TH α n k S!$ $I f f o r n d i m e n t i o n s s u r f a c e i s o f n - 1$ $d i m e n t i o n s . T h i s i s c o n s i s t a n t w i t h 3 D$ $f i g u r e s . B e c a u s e i n 3 D, s u r f a c e i s 2 D .$
	Answered by Filup last updated on 22/Dec/15

	$Volume is area rotated around axis .$ $∴ 4 D a r e a / v o l u m e i s t h e v o l u m e$ $r o t a t e d a r o u n d a n a x i s . Where you rotate$ $the whole 3 D solid around the z axis$ $(I think z axis)$
	Commented by Filup last updated on 22/Dec/15

	$t h i s w a s m e a n t t o b e a c o m m e n t$
	Commented by Rasheed Soomro last updated on 22/Dec/15

	$F o r m u l a o f v o l u m e o f 4 D s p h e r e, i n t e r m s$ $o f r i s r e q u i r e d .$ $F o r m u l a s o f s q u a r e, c u b e, h y p e r c u b e e t c$ $f o r m a G P : s^{2}, s^{3}, s^{4}, . . .$ $F o r m u l a s o f c i r c l e, s p h e r e, e t c f o r m a s e q u e n c e$ $π r^{2}, \frac{4}{3} π r^{3}, . . . .$ $W h a t i s n e x t t e r m ?$ $I f w e a d d o n e t e r m a t t h e b e g i n i n g, t h e f i r s t s e q u e n c e$ $s h o u l d b e : s, s^{2}, s^{3}, . . .$ $s m a y b e c o n s i d e r e d l e n g t h o f l i n e s e g m e n t .$ $T h e s e c o n d s e q u e n c e, I t h i n k s h o u l d s t a r t w i t h 0$ $(0 c a n b e c o n s i d e r e d a s^{'} {l e n g t h}^{'} o f a p o i n t) :$ $0, π r^{2}, \frac{4}{3} π r^{3}, . . .$ $P o i n t, C i r c l e, S p h e r e, . . .$
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