MvCalc42

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The tetrahedron is bounded below by z=0(xy plane) and bounded above by the surface z={\frac  {6-6x-3y}{2}}=f(x,y)\,

The projection of R of this surface z=f(x,y)\, onto the xy-plane is the triangular region,say OAB bounded by the lines x=0,y=0 and 6x+3y=6\,. This injected region R can be covered by varying y from 0 to 2-2x and x from 0 to 1.Thus,the required volume of the tetrahedron is given by

V=\iint _{R}f(x,y)dxdy\,

=\int _{0}^{1}\int _{{0}}^{{2-2x}}{\frac  {6-6x-3y}{2}}dydx\,

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

=3\int _{0}^{1}(1-x)^{2}dx=1\,

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