Question-183165

Question Number 183165 by universe last updated on 21/Dec/22

Answered by mr W last updated on 22/Dec/22

Commented by mr W last updated on 21/Dec/22

r = a \cos θ

d r = - a \sin θ d θ

d A = - 2 θ r d r = 2 a^{2} θ \cos θ \sin θ d θ = a^{2} θ \sin 2 θ d θ

z = \sqrt{a^{2} - r^{2}} = \sqrt{a^{2} (1 - \cos^{2} θ)} = a \sin θ

d V = 2 z d A = 2 a^{3} θ \sin 2 θ \sin θ d θ

V = 2 a^{3} \int_{0}^{\frac{π}{2}} θ \sin 2 θ \sin θ d θ

V = 2 a^{3} \times \frac{3 π - 4}{9}

\Rightarrow V = \frac{2 (3 π - 4) a^{3}}{9} \approx 1.2055 a^{3}

Commented by universe last updated on 22/Dec/22

t h a n k s s i r

Commented by mr W last updated on 22/Dec/22

i f t h e r a d i u s o f t h e s p h e r e i s R > a,

a n d l e t \frac{a}{R} = λ < 1,

t h e n w e h a v e

z = \sqrt{R^{2} - r^{2}} = R \sqrt{1 - λ^{2} \cos^{2} θ}

d A = λ^{2} R^{2} θ \sin 2 θ d θ

d V = 2 z d A = 2 λ^{2} R^{3} θ \sin 2 θ \sqrt{1 - λ^{2} \cos^{2} θ} d θ

V = 2 λ^{2} R^{3} \int_{0}^{\frac{π}{2}} θ \sin 2 θ \sqrt{1 - λ^{2} \cos^{2} θ} d θ

t h i s i n t e g r a l c a n n o t b e e x a c t l y s o l v e d .

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