Question Number 35635 by ajfour last updated on 21/May/18
Commented by ajfour last updated on 21/May/18
$${Find}\:{the}\:{volume}\:{common}\:{to}\:{the} \\ $$$${two}\:{spheres}. \\ $$
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))] .](https://www.tinkutara.com/question/Q35638.png)
$${Let}\:{centre}\:{of}\:{lower}\:{sphere}\:{be}\: \\ $$$${origin}\:{and}\:{line}\:{joining}\:{the} \\ $$$${centres},\:{the}\:{z}-{axis}. \\ $$$${eqn}.\:{of}\:\:{lower}\:{sphere}\:{is} \\ $$$$\:\:\:\:\:\:\:\:{z}^{\mathrm{2}} ={R}^{\mathrm{2}} −\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)\: \\ $$$${eqn}.\:{of}\:{upper}\:{sphere}\:{is} \\ $$$$\:\:\:\:\:\:\:\left({z}−{d}\right)^{\mathrm{2}} ={r}^{\mathrm{2}} −\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right) \\ $$$${equating}\:{the}\:{two}\:{we}\:{get} \\ $$$$\:\:\:{z}_{{c}} ^{\mathrm{2}} −\left({z}_{{c}} −{d}\right)^{\mathrm{2}} \:=\:{R}^{\mathrm{2}} −{r}^{\mathrm{2}} \\ $$$${or}\:\:\:\:\:\left(\mathrm{2}{z}_{{c}} −{d}\right){d}={R}^{\mathrm{2}} −{r}^{\mathrm{2}} \:\:\:\:…\left({i}\right) \\ $$$${Required}\:{volume}\:{V}_{{c}} \:{is} \\ $$$$\:\:\:{V}_{{c}} =\int_{{d}−{r}} ^{\:\:{z}_{{c}} } \pi\left[{r}^{\mathrm{2}} −\left({d}−{z}\right)^{\mathrm{2}} \right]{dz}\:+ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\int_{{z}_{{c}} } ^{\:\:{R}} \pi\left({R}^{\mathrm{2}} −{z}^{\mathrm{2}} \right){dz} \\ $$$$\:\:=\pi\left\{\left[{r}^{\mathrm{2}} {z}+\frac{\left({d}−{z}\right)^{\mathrm{3}} }{\mathrm{3}}\right]_{{d}−{r}} ^{{z}_{{c}} } +\left({R}^{\mathrm{2}} {z}−\frac{{z}^{\mathrm{3}} }{\mathrm{3}}\right)_{{z}_{{c}} } ^{{R}} \right\} \\ $$$$\:=\pi\left[{r}^{\mathrm{2}} {z}_{{c}} +\frac{\left({d}−{z}_{{c}} \right)^{\mathrm{3}} }{\mathrm{3}}−{r}^{\mathrm{2}} \left({d}−{r}\right)−\frac{{r}^{\mathrm{3}} }{\mathrm{3}}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:+\pi\left[{R}^{\mathrm{3}} −\frac{{R}^{\mathrm{3}} }{\mathrm{3}}−{R}^{\mathrm{2}} {z}_{{c}} +\frac{{z}_{{c}} ^{\mathrm{3}} }{\mathrm{3}}\right] \\ $$$$\:{V}_{{c}} =\pi\left[\frac{\mathrm{2}\left({R}^{\mathrm{3}} +{r}^{\mathrm{3}} \right)}{\mathrm{3}}−\left({R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right){z}_{{c}} −{r}^{\mathrm{2}} {d}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\frac{{d}^{\mathrm{3}} }{\mathrm{3}}−{d}^{\mathrm{2}} {z}_{{c}} +{z}_{{c}} ^{\mathrm{2}} {d}\:\right]\:\:\:\:…\left({ii}\right) \\ $$$${Using}\:\left({i}\right)\:: \\ $$$${V}_{{c}} =\:\pi\left[\frac{\mathrm{2}\left({R}^{\mathrm{3}} +{r}^{\mathrm{3}} \right)}{\mathrm{3}}+\frac{{d}^{\mathrm{3}} }{\mathrm{3}}−{r}^{\mathrm{2}} {d}−\mathrm{2}{z}_{{c}} ^{\mathrm{2}} {d}+{z}_{{c}} ^{\mathrm{2}} {d}\right] \\ $$$$\:\:\:{V}_{{c}} \:=\pi\left[\frac{\mathrm{2}\left(\boldsymbol{{R}}^{\mathrm{3}} +\boldsymbol{{r}}^{\mathrm{3}} \right)}{\mathrm{3}}+\frac{\boldsymbol{{d}}^{\mathrm{3}} }{\mathrm{3}}−\boldsymbol{{r}}^{\mathrm{2}} \boldsymbol{{d}}−\frac{\left(\boldsymbol{{d}}^{\mathrm{2}} +\boldsymbol{{R}}^{\mathrm{2}} −\boldsymbol{{r}}^{\mathrm{2}} \right)^{\mathrm{2}} }{\mathrm{4}\boldsymbol{{d}}}\right]\:. \\ $$$$ \\ $$
Commented by ajfour last updated on 21/May/18
![If d=R+r then V_c =0 lets check: V_c ∣_(d=R+r) =π[((2(R^3 +r^3 ))/3)+(((R+r)^3 )/3) −r^2 (R+r)−(([(R+r)^2 +R^2 −r^2 ]^2 )/(4(R+r)))] V_c =π[(2/3)(R^3 +r^3 )+(((R+r)^3 )/3) −r^2 (R+r)−(((R+r)^3 )/4)−(((R+r)(R−r)^2 )/4) −(((R+r)^2 (R−r))/2) ] =π[((2/3)+(1/3)−(1/4)−(1/4)−(1/2))R^3 +(0+1−(3/4)−(1/4)+(1/2))R^2 r +(0+1−1−(3/4)+(1/4)+(1/2))r^2 R +(+(2/3)+(1/3)−1−(1/4)−(1/4)+(1/2))r^3 ] = 0 (indeed) •](https://www.tinkutara.com/question/Q35641.png)
$${If}\:{d}={R}+{r}\:\:{then}\:\:{V}_{{c}} =\mathrm{0} \\ $$$${lets}\:{check}: \\ $$$${V}_{{c}} \mid_{{d}={R}+{r}} =\pi\left[\frac{\mathrm{2}\left({R}^{\mathrm{3}} +{r}^{\mathrm{3}} \right)}{\mathrm{3}}+\frac{\left({R}+{r}\right)^{\mathrm{3}} }{\mathrm{3}}\right. \\ $$$$\left.\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:−{r}^{\mathrm{2}} \left({R}+{r}\right)−\frac{\left[\left({R}+{r}\right)^{\mathrm{2}} +{R}^{\mathrm{2}} −{r}^{\mathrm{2}} \right]^{\mathrm{2}} }{\mathrm{4}\left({R}+{r}\right)}\right] \\ $$$${V}_{{c}} \:=\pi\left[\frac{\mathrm{2}}{\mathrm{3}}\left({R}^{\mathrm{3}} +{r}^{\mathrm{3}} \right)+\frac{\left({R}+{r}\right)^{\mathrm{3}} }{\mathrm{3}}\right. \\ $$$$\:\:\:−{r}^{\mathrm{2}} \left({R}+{r}\right)−\frac{\left({R}+{r}\right)^{\mathrm{3}} }{\mathrm{4}}−\frac{\left({R}+{r}\right)\left({R}−{r}\right)^{\mathrm{2}} }{\mathrm{4}}\: \\ $$$$\left.\:\:\:\:\:\:\:\:−\frac{\left({R}+{r}\right)^{\mathrm{2}} \left({R}−{r}\right)}{\mathrm{2}}\:\right] \\ $$$$\:\:\:=\pi\left[\left(\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\right){R}^{\mathrm{3}} \right. \\ $$$$\:\:\:+\left(\mathrm{0}+\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right){R}^{\mathrm{2}} {r} \\ $$$$\:\:\:+\left(\mathrm{0}+\mathrm{1}−\mathrm{1}−\frac{\mathrm{3}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right){r}^{\mathrm{2}} {R} \\ $$$$\left.\:\:\:+\left(+\frac{\mathrm{2}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{3}}−\mathrm{1}−\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}}\right){r}^{\mathrm{3}} \:\right] \\ $$$$=\:\mathrm{0}\:\:\left({indeed}\right)\:\bullet \\ $$