MvCalc48

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The required volume is given by V=\iint _{R}zdxdy\, where z=1-x-y\, and R is the region bounded by x=0,y=0,x+y=1\,

Hence,we have V=\int _{0}^{1}\int _{{0}}^{{1-y}}(1-x-y)dxdy\,

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

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

={\frac  {1}{2}}\int _{0}^{1}(1-y)^{2}dy={\frac  {1}{6}}\,

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