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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=02xzdy=02(4y2)(2y)dy=02(82y24y+y3)dy=[8y2y332y2+y44]02=8×22×2332×22+244=203

Commented by universe last updated on 18/Dec/22

thanks sir

thankssir

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)

D={(x,y,z)/0x4;0y4x;0z2yxisvariablefree0404x.02ydzdydx04x(2y)dy=2yy2204x0424x(4x)22×(23)(4x)32x+x240402×4+424(2×(23)403)6.666667=203

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