Question Number 20162 by ajfour last updated on 23/Aug/17
$${Compute}\:{the}\:{volume}\:{of}\:{a}\:{solid} \\ $$$${bounded}\:{by}\:{a}\:{surface}\:{with}\:{equation} \\ $$$$\:\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \right)^{\mathrm{2}} ={a}^{\mathrm{3}} {x}\:. \\ $$
Answered by ajfour last updated on 23/Aug/17
![surface is symmetrical about x axis and ranges from x=0 to x=a . Let x^2 +y^2 =r^2 , Then x^2 +r^2 =a(√a)((√x)) Volume V=∫_0 ^( a) πr^2 dx =π∫_0 ^( a) [a(√a)((√x))−x^2 ]dx =π[((2a(√a)(x(√x)))/3)−(x^3 /3)]∣_0 ^a V =(π/3)a^3 .](Q20183.png)
$${surface}\:\:{is}\:{symmetrical}\:{about} \\ $$$${x}\:{axis}\:{and}\:{ranges}\:{from}\:{x}=\mathrm{0}\:{to} \\ $$$${x}={a}\:. \\ $$$${Let}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} ={r}^{\mathrm{2}} \:,\:{Then}\: \\ $$$$\:\:{x}^{\mathrm{2}} +{r}^{\mathrm{2}} ={a}\sqrt{{a}}\left(\sqrt{{x}}\right) \\ $$$$\:\:{Volume}\:{V}=\int_{\mathrm{0}} ^{\:\:{a}} \pi{r}^{\mathrm{2}} {dx} \\ $$$$\:\:\:\:=\pi\int_{\mathrm{0}} ^{\:\:{a}} \left[{a}\sqrt{{a}}\left(\sqrt{{x}}\right)−{x}^{\mathrm{2}} \right]{dx} \\ $$$$\:\:\:\:=\pi\left[\frac{\mathrm{2}{a}\sqrt{{a}}\left({x}\sqrt{{x}}\right)}{\mathrm{3}}−\frac{{x}^{\mathrm{3}} }{\mathrm{3}}\right]\mid_{\mathrm{0}} ^{{a}} \\ $$$$\:\:\:{V}\:=\frac{\pi}{\mathrm{3}}{a}^{\mathrm{3}} . \\ $$