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Question Number 182962 by universe last updated on 17/Dec/22

Answered by mr W last updated on 17/Dec/22

Commented by mr W last updated on 17/Dec/22

V=∫_0 ^2 xzdy =∫_0 ^2 (4−y^2 )(2−y)dy =∫_0 ^2 (8−2y^2 −4y+y^3 )dy =[8y−((2y^3 )/3)−2y^2 +(y^4 /4)]_0 ^2 =8×2−((2×2^3 )/3)−2×2^2 +(2^4 /4) =((20)/3)

$${V}=\int_{\mathrm{0}} ^{\mathrm{2}} {xzdy} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{4}−{y}^{\mathrm{2}} \right)\left(\mathrm{2}−{y}\right){dy} \\ $$$$\:\:=\int_{\mathrm{0}} ^{\mathrm{2}} \left(\mathrm{8}−\mathrm{2}{y}^{\mathrm{2}} −\mathrm{4}{y}+{y}^{\mathrm{3}} \right){dy} \\ $$$$\:\:=\left[\mathrm{8}{y}−\frac{\mathrm{2}{y}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}{y}^{\mathrm{2}} +\frac{{y}^{\mathrm{4}} }{\mathrm{4}}\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\:=\mathrm{8}×\mathrm{2}−\frac{\mathrm{2}×\mathrm{2}^{\mathrm{3}} }{\mathrm{3}}−\mathrm{2}×\mathrm{2}^{\mathrm{2}} +\frac{\mathrm{2}^{\mathrm{4}} }{\mathrm{4}} \\ $$$$\:\:=\frac{\mathrm{20}}{\mathrm{3}} \\ $$

Commented by universe last updated on 18/Dec/22

thanks sir

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

Answered by manxsol last updated on 18/Dec/22

D={(x,y,z)/0≪x≪4; 0≪y≪(√(4−x)) ; 0≪z≪2−y x is variable free ∫_0 ^( 4) ∫_0 ^(√(4−x)) ∫_(.0) ^(2−y) dzdydx ∫_0 ^(√(4−x)) (2−y) dy=2y−(y^2 /2) ∣_0 ^(√(4−x)) ∫_0 ^4 2(√(4−x))−(((4−x))/2) 2×(((−2)/3))((√(4−x)))^3 −2x+(x^2 /4) ∣_0 ^4 0−2×4+(4^2 /4)−(2×(((−2)/3))(√(4−0))^3 ) 6.666667=((20)/3)

$$\mathbb{D}=\left\{\left(\boldsymbol{\mathrm{x}},{y},{z}\right)/\mathrm{0}\ll{x}\ll\mathrm{4};\right. \\ $$$$\mathrm{0}\ll{y}\ll\sqrt{\mathrm{4}−{x}}\:\:; \\ $$$$\mathrm{0}\ll{z}\ll\mathrm{2}−{y} \\ $$$${x}\:{is}\:{variable}\:{free} \\ $$$$\int_{\mathrm{0}} ^{\:\mathrm{4}} \int_{\mathrm{0}} ^{\sqrt{\mathrm{4}−{x}}} \int_{.\mathrm{0}} ^{\mathrm{2}−{y}} {dzdydx} \\ $$$$\int_{\mathrm{0}} ^{\sqrt{\mathrm{4}−{x}}} \left(\mathrm{2}−{y}\right)\:{dy}=\mathrm{2}{y}−\frac{{y}^{\mathrm{2}} }{\mathrm{2}}\:\mid_{\mathrm{0}} ^{\sqrt{\mathrm{4}−{x}}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{4}} \mathrm{2}\sqrt{\mathrm{4}−{x}}−\frac{\left(\mathrm{4}−{x}\right)}{\mathrm{2}} \\ $$$$\mathrm{2}×\left(\frac{−\mathrm{2}}{\mathrm{3}}\right)\left(\sqrt{\mathrm{4}−{x}}\right)^{\mathrm{3}} −\mathrm{2}{x}+\frac{{x}^{\mathrm{2}} }{\mathrm{4}}\:\mid_{\mathrm{0}} ^{\mathrm{4}} \\ $$$$\mathrm{0}−\mathrm{2}×\mathrm{4}+\frac{\mathrm{4}^{\mathrm{2}} }{\mathrm{4}}−\left(\mathrm{2}×\left(\frac{−\mathrm{2}}{\mathrm{3}}\right)\sqrt{\mathrm{4}−\mathrm{0}}\:^{\mathrm{3}} \right) \\ $$$$\mathrm{6}.\mathrm{666667}=\frac{\mathrm{20}}{\mathrm{3}} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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