(873, #31) The figure shows the region of integration of the integral

Rewrite this integral as an equivalent integral in the orders ..dxdydz, ...dydzdx, and ...dzdxdy.

Solution: The controlling sets of inequalities are

0 <= z <= 1 - y
sqrt(x) <= y <= 1
0 <= x <= 1

The follwing are three equivalent ways of writing these inequalities and the associated integrals:

0 <= z <= 1
0<= x <= 1
0 <= y <= 1
0 <= y <= 1 - z
0 <= z <= 1 - y
0 <= x <= y2
0 <= x <= y2
sqrt(x) <= y <= 1
0 <= z <= 1- y