MvCalc46

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The top surface is z=4-x^{2}-y^{2}=f(x,y)\,.

The projection R of this surface on the xy plane into the square is the square itself with the given vertices. Thus

V=\iint _{R}f(x,y)dxdy=\int _{0}^{1}\int _{0}^{1}(4-x^{2}-y^{2})dxdy\,

=\int _{0}^{1}[4y-x^{2}y-{\frac  {y^{3}}{3}}]_{0}^{1}dx\,

=\int _{0}^{1}[{\frac  {11}{3}}-x^{2}]dx=[{\frac  {11}{3}}x-{\frac  {x^{3}}{3}}]_{0}^{1}={\frac  {10}{3}}\,

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