V-4-3-R-3-prove-

Question Number 60817 by ANTARES VY last updated on 26/May/19

V = \frac{4}{3} π R^{3} prove

Commented by Prithwish sen last updated on 26/May/19

http://mathcentral.uregina.ca/QQ/database/QQ.09.01/rahul1.html

Commented by Prithwish sen last updated on 26/May/19

This explanation is geometry based .

Answered by ajfour last updated on 26/May/19

Commented by ajfour last updated on 26/May/19

A n i n t e r m e d i a t e d i f f e r e n t i a l

l a y e r h a s v o l u m e d V = 4 π r^{2} d r

\Rightarrow V = \int_{0}^{R} 4 π r^{2} d r = \frac{4 π R^{3}}{3} .

Commented by mr W last updated on 26/May/19

o r

t h e s u r f a c e c a n b e d i v i d e d i n t o i n f i n i t e

s m a l l p i e c e s, e a c h p i e c e i s t h e b a s e o f

a p y r a m i d w h o s e t o p i s t h e c e n t e r o f

t h e s p h e r e . t h e v o l u m e o f t h e p y r a m i d

i s

Δ V = \frac{h Δ S}{3} w i t h h = R

t h e v o l u m e o f t h e s p h e r e i s

V = Σ Δ V = \frac{h}{3} Σ Δ S

\Rightarrow V = \frac{h}{3} S = \frac{h 4 π R^{2}}{3} = \frac{4 π R^{3}}{3}

Commented by mr W last updated on 26/May/19

Answered by tanmay last updated on 26/May/19

Commented by tanmay last updated on 26/May/19

d v = (r d θ) \times (r s i n θ d ϕ) \times (d r)

V = \int_{0}^{R} r^{2} d r \int_{0}^{π} s i n θ d θ \int_{0}^{2 π} d ϕ

= \frac{R^{3}}{3} \times 2 \times 2 π

= \frac{4}{3} π R^{3}

Answered by Kunal12588 last updated on 26/May/19

t h e r e i s a b r i l l i a n t w a y I w a n t t o s h a r e

Commented by Kunal12588 last updated on 26/May/19

Commented by Kunal12588 last updated on 26/May/19

Commented by Prithwish sen last updated on 26/May/19

which book ? Sir .

Commented by Kunal12588 last updated on 26/May/19

P r e c a l c u l u s M a t h e m a t i c s

Commented by Prithwish sen last updated on 26/May/19

thank you sir

Answered by tanmay last updated on 26/May/19

e q n o f s p h e r e c e n t r e o r i g i n (0, 0, 0) a n d r a d i u s R

\frac{x^{2}}{R^{2}} + \frac{y^{2}}{R^{2}} + \frac{z^{2}}{R^{2}} = 1

u_{1} = {(\frac{x}{R})}^{2} \to x = R \times u_{1}^{\frac{1}{2}} \to d x = \frac{R}{2} \times u^{\frac{- 1}{2}} {d u}_{1} = \frac{R}{2} \times u_{1}^{\frac{1}{2} - 1} {d u}_{1}

d y = \frac{R}{2} \times u_{2}^{\frac{1}{2} - 1} {d u}_{2}

d z = \frac{R}{2} \times u_{3}^{\frac{1}{2} - 1} {d u}_{3}

v o l u m e o f s p h e r e = 8 \int_{0}^{1} \int_{0}^{1} \int_{0}^{1} {(\frac{R}{2})}^{3} u_{1}^{\frac{1}{2} - 1} u_{2}^{\frac{1}{2} - 1} u_{3}^{\frac{1}{2} - 1} {d u}_{1} {d u}_{2} {d u}_{3}

= 8 \times \frac{R^{3}}{8} \times \frac{⌈ (\frac{1}{2}) \times ⌈ (\frac{1}{2}) \times ⌈ (\frac{1}{2})}{⌈ (1 + \frac{1}{2} + \frac{1}{2} + \frac{1}{2})}

= R^{3} \times \frac{{(\sqrt{π})}^{3}}{⌈ (\frac{3}{2} + 1)} = \frac{π \sqrt{π}}{\frac{3}{4} \sqrt{π}} \times R^{3} = \frac{4 π R^{3}}{3}

u s i n g D i r c h h i l e t t h e o r e m \dots a t t a c h i n g t h e m

n o t e b e l o w

⌈ (\frac{3}{2} + 1) = \frac{3}{2} ⌈ (\frac{1}{2} + 1) = \frac{3}{2} \times \frac{1}{2} ⌈ (\frac{1}{2}) = \frac{3}{4} \times \sqrt{π}

Commented by tanmay last updated on 26/May/19

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