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Question Number 35635 by ajfour last updated on 21/May/18

Commented by ajfour last updated on 21/May/18

$F i n d t h e v o l u m e c o m m o n t o t h e$ $t w o s p h e r e s .$
	Answered by ajfour last updated on 21/May/18
	$Let centre of lower sphere be origin and line joining the centres, the z-axis. eqn. of lower sphere is z^2 =R^2 −(x^2 +y^2 ) eqn. of upper sphere is (z−d)^2 =r^2 −(x^2 +y^2 ) equating the two we get z_c ^2 −(z_c −d)^2 = R^2 −r^2 or (2z_c −d)d=R^2 −r^2 ...(i) Required volume V_c is V_c =∫_(d−r) ^( z_c ) π[r^2 −(d−z)^2 ]dz + ∫_z_c ^( R) π(R^2 −z^2 )dz =π{[r^2 z+(((d−z)^3 )/3)]_(d−r) ^z_c +(R^2 z−(z^3 /3))_z_c ^R } =π[r^2 z_c +(((d−z_c )^3 )/3)−r^2 (d−r)−(r^3 /3)] +π[R^3 −(R^3 /3)−R^2 z_c +(z_c ^3 /3)] V_c =π[((2(R^3 +r^3 ))/3)−(R^2 −r^2 )z_c −r^2 d +(d^3 /3)−d^2 z_c +z_c ^2 d ] ...(ii) Using (i) : V_c = π[((2(R^3 +r^3 ))/3)+(d^3 /3)−r^2 d−2z_c ^2 d+z_c ^2 d] V_c =π[((2(R^3 +r^3 ))/3)+(d^3 /3)−r^2 d−(((d^2 +R^2 −r^2 )^2 )/(4d))] .$
	$L e t c e n t r e o f l o w e r s p h e r e b e$ $o r i g i n a n d l i n e j o i n i n g t h e$ $c e n t r e s, t h e z - a x i s .$ $e q n . o f l o w e r s p h e r e i s$ $z^{2} = R^{2} - (x^{2} + y^{2})$ $e q n . o f u p p e r s p h e r e i s$ ${(z - d)}^{2} = r^{2} - (x^{2} + y^{2})$ $e q u a t i n g t h e t w o w e g e t$ $z_{c}^{2} - {(z_{c} - d)}^{2} = R^{2} - r^{2}$ $o r (2 z_{c} - d) d = R^{2} - r^{2} . . . (i)$ $R e q u i r e d v o l u m e V_{c} i s$ $V_{c} = \int_{d - r}^{z_{c}} π [r^{2} - {(d - z)}^{2}] d z +$ $\int_{z_{c}}^{R} π (R^{2} - z^{2}) d z$ $= π {{[r^{2} z + \frac{{(d - z)}^{3}}{3}]}_{d - r}^{z_{c}} + {(R^{2} z - \frac{z^{3}}{3})}_{z_{c}}^{R}}$ $= π [r^{2} z_{c} + \frac{{(d - z_{c})}^{3}}{3} - r^{2} (d - r) - \frac{r^{3}}{3}]$ $+ π [R^{3} - \frac{R^{3}}{3} - R^{2} z_{c} + \frac{z_{c}^{3}}{3}]$ $V_{c} = π [\frac{2 (R^{3} + r^{3})}{3} - (R^{2} - r^{2}) z_{c} - r^{2} d$ $+ \frac{d^{3}}{3} - d^{2} z_{c} + z_{c}^{2} d] . . . (i i)$ $U s i n g (i) :$ $V_{c} = π [\frac{2 (R^{3} + r^{3})}{3} + \frac{d^{3}}{3} - r^{2} d - 2 z_{c}^{2} d + z_{c}^{2} d]$ $V_{c} = π [\frac{2 (R^{3} + r^{3})}{3} + \frac{d^{3}}{3} - r^{2} d - \frac{{(d^{2} + R^{2} - r^{2})}^{2}}{4 d}] .$
	Commented by ajfour last updated on 21/May/18

	$I f d = R + r t h e n V_{c} = 0$ $l e t s c h e c k :$ $V_{c} ∣_{d = R + r} = π [\frac{2 (R^{3} + r^{3})}{3} + \frac{{(R + r)}^{3}}{3}$ $- r^{2} (R + r) - \frac{{[{(R + r)}^{2} + R^{2} - r^{2}]}^{2}}{4 (R + r)}]$ $V_{c} = π [\frac{2}{3} (R^{3} + r^{3}) + \frac{{(R + r)}^{3}}{3}$ $- r^{2} (R + r) - \frac{{(R + r)}^{3}}{4} - \frac{(R + r) {(R - r)}^{2}}{4}$ $- \frac{{(R + r)}^{2} (R - r)}{2}]$ $= π [(\frac{2}{3} + \frac{1}{3} - \frac{1}{4} - \frac{1}{4} - \frac{1}{2}) R^{3}$ $+ (0 + 1 - \frac{3}{4} - \frac{1}{4} + \frac{1}{2}) R^{2} r$ $+ (0 + 1 - 1 - \frac{3}{4} + \frac{1}{4} + \frac{1}{2}) r^{2} R$ $+ (+ \frac{2}{3} + \frac{1}{3} - 1 - \frac{1}{4} - \frac{1}{4} + \frac{1}{2}) r^{3}]$ $= 0 (i n d e e d) ∙$
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