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Question Number 211749 by universe last updated on 19/Sep/24

volume bounded by the curve z = (√(3x^2 +3y^2 )) and x^2 +y^2 +z^2 = 6^2

$${volume}\:{bounded}\:{by}\:{the}\:{curve} \\ $$$$\:{z}\:=\:\sqrt{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{3}{y}^{\mathrm{2}} }\:\:\:{and}\:{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} =\:\mathrm{6}^{\mathrm{2}} \\ $$

Answered by mr W last updated on 20/Sep/24

Commented by mr W last updated on 20/Sep/24

z=(√(3(x^2 +y^2 )))=(√3)r h=(√3)r ((√3)r)^2 +r^2 =R^2 =6^2 ⇒r=3 ⇒h=3(√3) V=((2πR^2 (R−h))/3)=((2π×6^2 ×3(2−(√3)))/3) =72(2−(√3))π

$${z}=\sqrt{\mathrm{3}\left({x}^{\mathrm{2}} +{y}^{\mathrm{2}} \right)}=\sqrt{\mathrm{3}}{r} \\ $$$${h}=\sqrt{\mathrm{3}}{r} \\ $$$$\left(\sqrt{\mathrm{3}}{r}\right)^{\mathrm{2}} +{r}^{\mathrm{2}} ={R}^{\mathrm{2}} =\mathrm{6}^{\mathrm{2}} \\ $$$$\Rightarrow{r}=\mathrm{3}\:\Rightarrow{h}=\mathrm{3}\sqrt{\mathrm{3}} \\ $$$${V}=\frac{\mathrm{2}\pi{R}^{\mathrm{2}} \left({R}−{h}\right)}{\mathrm{3}}=\frac{\mathrm{2}\pi×\mathrm{6}^{\mathrm{2}} ×\mathrm{3}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)}{\mathrm{3}} \\ $$$$\:\:\:=\mathrm{72}\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\pi \\ $$

Commented by universe last updated on 20/Sep/24

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by universe last updated on 20/Sep/24

can u explain V = ((2πR^2 (R−r))/3)

$${can}\:{u}\:{explain}\:{V}\:=\:\frac{\mathrm{2}\pi{R}^{\mathrm{2}} \left({R}−{r}\right)}{\mathrm{3}} \\ $$

Commented by mr W last updated on 20/Sep/24

Commented by universe last updated on 20/Sep/24

thanks sir

$${thanks}\:{sir} \\ $$

Answered by BHOOPENDRA last updated on 20/Sep/24

Answered by BHOOPENDRA last updated on 20/Sep/24

V=∫∫∫ ρ^2 sinφ dρ dθ dφ =∫_0 ^(π/6) ∫_0 ^(2π) ∫_0 ^6 ρ^2 sinφ dρ dθ dφ =∫_0 ^(π/6) ∫_0 ^(2π) (((6^3 −0^3 )/3)) sinφ dθ dφ =∫_0 ^(π/6) 72(2π−0)sinφ dφ =−144π cosφ∣_0 ^(π/6) = 144π (((2 −(√3))/2)) =72π (2−(√3))

$${V}=\int\int\int\:\rho^{\mathrm{2}} {sin}\phi\:{d}\rho\:{d}\theta\:{d}\phi \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \int_{\mathrm{0}} ^{\mathrm{2}\pi} \:\int_{\mathrm{0}} ^{\mathrm{6}} \:\rho^{\mathrm{2}} \:{sin}\phi\:{d}\rho\:{d}\theta\:{d}\phi \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \int_{\mathrm{0}} ^{\mathrm{2}\pi} \left(\frac{\mathrm{6}^{\mathrm{3}} −\mathrm{0}^{\mathrm{3}} }{\mathrm{3}}\right)\:{sin}\phi\:{d}\theta\:{d}\phi \\ $$$$\:\:\:=\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \mathrm{72}\left(\mathrm{2}\pi−\mathrm{0}\right){sin}\phi\:{d}\phi \\ $$$$\:\:\:=−\mathrm{144}\pi\:{cos}\phi\mid_{\mathrm{0}} ^{\frac{\pi}{\mathrm{6}}} \\ $$$$\:\:=\:\mathrm{144}\pi\:\left(\frac{\mathrm{2}\:−\sqrt{\mathrm{3}}}{\mathrm{2}}\right) \\ $$$$\:\:\:\:=\mathrm{72}\pi\:\left(\mathrm{2}−\sqrt{\mathrm{3}}\right) \\ $$

Commented by universe last updated on 20/Sep/24

thank you sir

$${thank}\:{you}\:{sir} \\ $$

Commented by mr W last updated on 20/Sep/24

V=72π(2−(√3)) >0

$${V}=\mathrm{72}\pi\left(\mathrm{2}−\sqrt{\mathrm{3}}\right)\:>\mathrm{0} \\ $$

Commented by BHOOPENDRA last updated on 20/Sep/24

Thanks Mr.W it was typo

$${Thanks}\:{Mr}.{W}\:{it}\:{was}\:{typo} \\ $$

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