Answer to Question #224872 in Chemical Engineering for Lokika

Question #224872

Solve ∫y=0 ^ 1 ∫x=y^ 2 ^ 1∫z=0 ^ 1-x x dzdxdy


1
Expert's answer
2021-08-18T09:46:01-0400

"\\int_{y=0}^1 \u222b_{x=y^ 2} ^ 1\u222b_{z=0} ^ {1-x} x dzdxdy\\\\\n=x\\left(-x+1\\right)\\\\\n=\\int _0^1\\int _{y^2}^1x\\left(-x+1\\right)dxdy\\\\\n=\\int _0^1\\left(\\frac{1-y^4}{2}-\\frac{1-y^6}{3}\\right)dy\\\\\n=\\frac{4}{35}"


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