Final answer:
To rewrite the triple integral ∫10 ∫0-1 ∫y^20 dz dy dx in five different orders, we need to understand the region of integration and adjust the bounds for each variable accordingly for each order.
Step-by-step explanation:
The original order of integration is ∫10 ∫0-1 ∫y^20 dz dy dx for a certain triple integral. To re-express this integral in different orders of integration, we need to understand the volume being integrated over and the bounds in terms of the other variables.
- (dy, dz, dx): We first need to express the bounds for y and dz in terms of x and z. Since the original inner bounds are from z=0 to z=y^2 and y ranges from -1 to 0, for any fixed x, y would go from -1 to 0, and z would go from 0 to y^2. Thus, the integral in this order would be ∫10 ∫y^20 ∫0-1 dy dz dx.
- (dy, dx, dz): With 'dy' first, we maintain the bounds of y from -1 to 0. To get the 'dx' bounds, note that x is being integrated from 0 to 1 irrespective of y and z. Hence, the integral would be ∫y^20 ∫10 ∫0-1 dx dy dz.
- (dx, dy, dz): Here the bounds for x don't depend on y or z and remain from 0 to 1. The bound for y ranges from -1 to 0, and the bound for z is from 0 to y^2 given a certain y. So, the integral becomes ∫10 ∫0-1 ∫y^20 dz dy dx.
- (dx, dz, dy): Starting with 'dx', x still ranges from 0 to 1. To integrate 'dz' next, we keep the bounds from 0 to y^2, and lastly for 'dy', we maintain the bounds from -1 to 0. Therefore, the integral is ∫10 ∫y^20 ∫0-1 dz dx dy.
- (dz, dx, dy): The 'dz' integral would still be from 0 to y^2 for given y bounds from -1 to 0. The x bounds are independent, from 0 to 1. Thus, the integral can be rewritten as ∫0-1 ∫10 ∫y^20 dx dz dy.