Find-the-volume-that-remains-after-the-hole-of-radius-1-bored-through-the-center-of-a-solid-sphere-of-radius-3-a-18pi-b-28-3-pi-c-36pi-d-56pi-3-

Question Number 124666 by benjo_mathlover last updated on 05/Dec/20

F i n d t h e v o l u m e t h a t r e m a i n s

a f t e r t h e h o l e o f r a d i u s 1 b o r e d

t h r o u g h t h e c e n t e r o f a s o l i d

s p h e r e o f r a d i u s 3 .

(a) 18 π (b) \frac{28}{3} π (c) 36 π (d) \frac{56 π}{3}

Commented by ajfour last updated on 05/Dec/20

Commented by ajfour last updated on 05/Dec/20

c a p v o l u m e V .

V = \int_{h}^{R} π r^{2} d y

r = R \sin θ, y = R \cos θ

= \int_{0}^{α} π (R^{2} \sin^{2} θ) R \sin θ d θ

\sin α = \frac{1}{3}, \cos α = \frac{2 \sqrt{2}}{3}

= π R^{3} \int_{\frac{2 \sqrt{2}}{3}}^{1} (1 - t^{2}) d t; t = \cos θ

= π R^{3} (t - \frac{t^{3}}{3}) ∣_{\frac{2 \sqrt{2}}{3}}^{1}

= 27 π [\frac{2}{3} - (\frac{2 \sqrt{2}}{3} - \frac{16 \sqrt{2}}{81})]

= 27 π [\frac{2}{3} - \frac{38 \sqrt{2}}{81}]

r e m a i n i n g v o l u m e

= \frac{4}{3} π (27) - 2 π {(1)}^{2} (2 \sqrt{2}) - 2 V

= 36 π - 4 \sqrt{2} π - 36 π + \frac{76 \sqrt{2}}{3} π

V_{r e m a i n i n g} = \frac{64 \sqrt{2} π}{3} .

Commented by benjo_mathlover last updated on 05/Dec/20

n o t h i n g a v a i l a b l e c h o i c e a n s w e r

i s c o r r e c t ?

Commented by mr W last updated on 05/Dec/20

y e s . a l l a n s w e r s g i v e n a r e w r o n g .

i g o t a f o r m u l a f o r t h e r e m a i n i n g

v o l u m e w h i c h i s

V_{r e m a i n} = \frac{4 π {(R^{2} - r^{2})}^{\frac{3}{2}}}{3}

Answered by mr W last updated on 05/Dec/20

M e t h o d I

R = r a d i u s o f s p h e r e

r = r a d i u s o f h o l e

h = \sqrt{R^{2} - r^{2}} = \sqrt{3^{2} - 1^{2}} = 2 \sqrt{2}

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

V = \frac{4 π R^{3}}{3}

v o l u m e o f c y l i n d e r w i t h h e i g h t 2 h

V_{1} = 2 h π r^{2}

v o l u m e o f e a c h c a p

V_{2} = π {(Δ h)}^{2} (R - \frac{Δ h}{3}) w i t h Δ h = R - h

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

V_{R e m a i n} = V - V_{1} - 2 V_{2}

= \frac{4 π R^{3}}{3} - 2 π {h r}^{2} - 2 π {(R - h)}^{2} \frac{(2 R + h)}{3}

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

= \frac{108 - 12 \sqrt{2} - 4 \times 27 + 4 \times 19 \sqrt{2}}{3} π

= \frac{64 \sqrt{2} π}{3}

Commented by mr W last updated on 05/Dec/20

Commented by mr W last updated on 05/Dec/20

M e t h o d I I

r ⩽ x ⩽ R

h (x) = \sqrt{R^{2} - x^{2}}

d V = 2 h (x) 2 π x = 4 π x \sqrt{R^{2} - x^{2}}

V_{R e m a i n} = \int_{r}^{R} 4 π x \sqrt{R^{2} - x^{2}} d x

= - 2 π \int_{r}^{R} \sqrt{R^{2} - x^{2}} d (R^{2} - x^{2})

= \frac{4 π}{3} {[{(R^{2} - x^{2})}^{\frac{3}{2}}]}_{R}^{r}

= \frac{4 π {(R^{2} - r^{2})}^{\frac{3}{2}}}{3}

= \frac{4 π {(3^{2} - 1^{2})}^{\frac{3}{2}}}{3}

= \frac{64 \sqrt{2} π}{3}

Commented by benjo_mathlover last updated on 05/Dec/20

t y p o s i r . \frac{4 π \times 8 \sqrt{8}}{3} = \frac{64 π \sqrt{2}}{3}