Answer to Question #224872 in Chemical Engineering for Lokika

Question

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

Answer 1

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