VC4.31

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Here,z=0 to z=1,y=0 to y=1,and x from 0 to 1.

Therefore,the required volume integral is

\int _{{x=0}}^{{1}}\int _{{y=0}}^{{1}}\int _{{z=0}}^{{1}}(2xzi-xj+y^{2}k)dxdydz\,

=2i\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}xzdxdydz-j\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}xdxdydz+k\int _{0}^{1}\int _{0}^{1}\int _{0}^{1}y^{2}dxdydz\,

=2i\int _{0}^{1}\int _{0}^{1}x[{\frac  {z^{2}}{2}}]_{0}^{1}dxdy-j\int _{0}^{1}\int _{0}^{1}[z]_{0}^{1}dxdy+k\int _{0}^{1}\int _{0}^{1}y^{2}[z]_{0}^{1}dxdy\,

=i\int _{0}^{1}\int _{0}^{1}xdxdy-j\int _{0}^{1}\int _{0}^{1}xdxdy+k\int _{0}^{1}\int _{0}^{1}y^{2}dxdy\,

=i\int _{0}^{1}x[y]_{0}^{1}dx-j\int _{0}^{1}x[y]_{0}^{1}dx+k\int _{0}^{1}[{\frac  {y^{3}}{3}}]_{0}^{1}dx\,

=i\int _{0}^{1}xdx-j\int _{0}^{1}\int _{0}^{1}xdx+{\frac  {1}{3}}\int _{0}^{1}dx\,

={\frac  {1}{2}}i-{\frac  {1}{2}}j+{\frac  {1}{3}}k\,

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