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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

$$V={\int }_0^{2}xzdy$$$$={\int }_0^2(4-y^2)(2-y)dy$$$$={\int }_0^2(8-2y^2-4y+y^3)dy$$$$={[8y-\frac{2y^3}{3}-2y^2+\frac{y^4}{4}]}_0^2$$$$=8\times 2-\frac{2\times 2^3}{3}-2\times 2^2+\frac{2^4}{4}$$$$=\frac{20}{3}$$

Commented by universe last updated on 18/Dec/22

$$thankssir$$

Answered by manxsol last updated on 18/Dec/22

$$\mathbb{D}=\{(\mathbf{x},y,z)/0\ll x\ll 4;$$$$0\ll y\ll \sqrt{4-x};$$$$0\ll z\ll 2-y$$$$xisvariablefree$$$${\int }_0^4{\int }_0^{\sqrt{4-x}}{\int }_{.0}^{2-y}dzdydx$$$${\int }_0^{\sqrt{4-x}}(2-y)dy=2y-\frac{y^2}{2}{\mid }_0^{\sqrt{4-x}}$$$${\int }_0^{4}2\sqrt{4-x}-\frac{(4-x)}{2}$$$$2\times \left(\frac{-2}{3}\right){\left(\sqrt{4-x}\right)}^3-2x+\frac{x^2}{4}{\mid }_0^4$$$$0-2\times 4+\frac{4^2}{4}-\left(2\times \left(\frac{-2}{3}\right)\sqrt{4-0}{}^3\right)$$$$6.666667=\frac{20}{3}$$$$$$$$$$$$$$$$$$$$$$

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