VC4.35

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Given cylinder is z=4-x^{2}\, --(1)

The limits of y are from 0 to 2 and those of x from 0 to 2.Also dV=dx dy dz.

Therefore,\iiint _{V}(2x+y)dV\,

=\int _{{x=0}}^{{2}}\int _{{y=0}}^{{2}}\int _{{z=0}}^{{4-x^{2}}}(2x+y)dxdydz\,

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

=\int _{{x=0}}^{{2}}\int _{{y=0}}^{{2}}(2x+y)(4-x^{2})dxdy\,

=\int _{{x=0}}^{{2}}(4-x^{2})[2xy+{\frac  {y^{2}}{2}}]_{0}^{2}dx\,

=2\int _{{x=0}}^{{2}}(4-x^{2})(4x+2)dx\,

=2\int _{{x=0}}^{{2}}(8x-2x^{3}+4-x^{2})dx=2[4x^{2}-{\frac  {x^{4}}{2}}+4x-{\frac  {x^{3}}{3}}]_{0}^{2}\,

=2[16-8+8-{\frac  {8}{3}}]={\frac  {80}{3}}\,

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