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Compute-the-volume-of-a-solid-bounded-by-a-surface-with-equation-x-2-y-2-z-2-2-a-3-x-




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

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