VC4.30

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The limits of integration for z are from z=0 to z=8-4x-2y,for y=0 to 4-x and those for x are from x=0 to x=2.

Therefore,\iiint _{V}\phi dV=\int _{{x=0}}^{{2}}\int _{{y=0}}^{{4-2x}}\int _{{z=0}}^{{8-4x-2y}}(45x^{2}y)dxdydz\,

=\int _{{x=0}}^{{2}}\int _{{y=0}}^{{4-2x}}x^{2}y[z]_{{z=0}}^{{8-4x-2y}}dxdy\,

=45\int _{{x=0}}^{{2}}\int _{{y=0}}^{{4-2x}}x^{2}y(8-4x-2y)dxdy\, =45\int _{{x=0}}^{{2}}\int _{{y=0}}^{{4-2x}}[x^{2}(8-4x)y-2x^{2}y^{2}]dxdy\, =45\int _{{x=0}}^{{2}}[x^{2}(8-4x){\frac  {y^{2}}{2}}-2x^{2}({\frac  {y^{3}}{3}})]_{{0}}^{{4-2x}}dx\, =45\int _{{x=0}}^{{2}}[x^{2}(4-2x)^{3}-{\frac  {2}{3}}x^{2}(4-2x)^{3}]dx\, =15\int _{{x=0}}^{{2}}x^{2}(4-2x)^{3}dx=15\int _{{x=0}}^{{2}}x^{2}(64-96x+48x^{2}-8x^{3})dx\, =15[{\frac  {64}{3}}x^{3}-24x^{4}+{\frac  {48}{5}}x^{5}-{\frac  {4}{3}}x^{6}]_{0}^{2}\, =15[{\frac  {64\times 8}{3}}-(24\times 16)+{\frac  {48\times 32}{5}}-{\frac  {4\times 64}{3}}]=128\, [on simplification]

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