"z= 1-x; z= \\frac{2-y}{2}\\\\\n1-x = \\frac{2-y}{2}\\\\\n2-2x=2-y\\\\\ny=2x\\\\\nz= 0 \\implies y= 2\\\\\nVolume = \\int\\int_D\\int _{z=0}^{\\frac{2-y}{2}}dzdA = \\int\\int_D \\frac{2-y}{2}dA\\\\\n=\\int_0^1 \\int_{2x}^{2} \\frac{2-y}{2}dydx\\\\\n=x^2-2x+1\\\\\n=\\int _0^1\\left(x^2-2x+1\\right)dx\\\\\n=\\int _0^1x^2- \\int _0^12x+\\int _0^11dx\\\\\n=\\frac{1}{3}"
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