Question Number 20157 by ajfour last updated on 23/Aug/17
Commented by ajfour last updated on 23/Aug/17
$${Compute}\:{the}\:{volume}\:{common}\:{to}\:{the} \\ $$$${sphere}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{4}\:\:{and}\:\:{the}\:{inside}\:{of} \\ $$$${the}\:{paraboloid}\:\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} =\mathrm{3}{z}\:. \\ $$
Answered by ajfour last updated on 23/Aug/17
Commented by ajfour last updated on 23/Aug/17
![Let x^2 +y^2 =r^2 ⇒ equation of sphere becomes r^2 +z^2 =4 equation of paraboloid is r^2 =3z intersection is a circle of radius r_0 . r_0 ^2 +z_0 ^2 =4 & r_0 ^2 =3z_0 ⇒ z_0 ^2 +3z_0 =4 ⇒ z_0 =1 , r_0 =(√3) . If enclosed volume is V, V=∫_0 ^( r_0 ) (2πr)dr(Δz) =2π∫_0 ^( (√3)) ((√(4−r^2 ))−(r^2 /3))rdr =2π[−(1/3)(4−r^2 )^(3/2) −(r^4 /(12))]∣_0 ^(√3) =2π[−(1/3)+(8/3)−(3/4)] =2π(((28−9)/(12))) V =((19π)/6) .](Q20178.png)
$${Let}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\Rightarrow\:{equation}\:{of}\:{sphere}\:{becomes} \\ $$$$\:\:\:\:\:\:{r}^{\mathrm{2}} +{z}^{\mathrm{2}} =\mathrm{4} \\ $$$${equation}\:{of}\:{paraboloid}\:{is} \\ $$$$\:\:\:\:\:\:{r}^{\mathrm{2}} =\mathrm{3}{z} \\ $$$${intersection}\:{is}\:{a}\:{circle}\:{of}\: \\ $$$${radius}\:{r}_{\mathrm{0}} .\:\:\:\:\:\:\: \\ $$$$\:\:\:\:{r}_{\mathrm{0}} ^{\mathrm{2}} +{z}_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{4}\:\:\:\:\&\:\:\:{r}_{\mathrm{0}} ^{\mathrm{2}} =\mathrm{3}{z}_{\mathrm{0}} \\ $$$$\Rightarrow\:\:\:\:{z}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{3}{z}_{\mathrm{0}} =\mathrm{4}\:\:\Rightarrow\:\:\:{z}_{\mathrm{0}} =\mathrm{1}\:,\:{r}_{\mathrm{0}} =\sqrt{\mathrm{3}}\:. \\ $$$${If}\:{enclosed}\:{volume}\:{is}\:{V}, \\ $$$$\:\:{V}=\int_{\mathrm{0}} ^{\:\:{r}_{\mathrm{0}} } \left(\mathrm{2}\pi{r}\right){dr}\left(\Delta{z}\right) \\ $$$$\:\:\:\:=\mathrm{2}\pi\int_{\mathrm{0}} ^{\:\:\sqrt{\mathrm{3}}} \left(\sqrt{\mathrm{4}−{r}^{\mathrm{2}} }−\frac{{r}^{\mathrm{2}} }{\mathrm{3}}\right){rdr} \\ $$$$\:\:=\mathrm{2}\pi\left[−\frac{\mathrm{1}}{\mathrm{3}}\left(\mathrm{4}−{r}^{\mathrm{2}} \right)^{\mathrm{3}/\mathrm{2}} −\frac{{r}^{\mathrm{4}} }{\mathrm{12}}\right]\mid_{\mathrm{0}} ^{\sqrt{\mathrm{3}}} \: \\ $$$$\:=\mathrm{2}\pi\left[−\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{8}}{\mathrm{3}}−\frac{\mathrm{3}}{\mathrm{4}}\right]\:=\mathrm{2}\pi\left(\frac{\mathrm{28}−\mathrm{9}}{\mathrm{12}}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{V}\:=\frac{\mathrm{19}\pi}{\mathrm{6}}\:. \\ $$