   Chapter 15.6, Problem 33E

Chapter
Section
Textbook Problem

The figure shows the region of integration for the integral ∫ 0 1 ∫ x 1 ∫ 0 1 − y   f ( x ,   y ,   z )   d z   d y   d x Rewrite this integral as an equivalent iterated integral in the five other orders. To determine

To express: The integral 01x101yf(x,y,z)dzdydx in five different ways.

Explanation

Given:

The region of the given integral is given in the form of figure.

The region are y=x and z=1y.

Calculation:

Let D1,D2,D3 be the respective projections of E on xy, yz and zx-planes.

Since the variable D1 is the projection of E on xy-plane, set z=0.

Then, the equation becomes,

z=1y0=1yy=1

The graph of the xy-plane is shown below in Figure 1.

From Figure 1, it is observed that y varies from 0 to 1, x varies from 0 to y2 and z varies from 0 to 1y.

Therefore, 010y201yf(x,y,z)dzdxdy

The variable D2 is the projection of E on yz-plane.

The graph of the yz-plane is shown below in Figure 2.

From Figure 2, it is observed that z varies from 0 to 1, y varies from 0 to 1z and x varies from 0 to y2

Therefore, 0101z0y2f(x,y,z)dxdydz

Also, from Figure 2, it is observed that y varies from 0 to 1, z varies from 0 to 1y and x varies from 0 to y2

0101y0y2f(x,y,z)dxdzdy

The variable D3 is the projection of E on yz-plane.

The graph of yz-plane is shown below in Figure 3.

From Figure 3, it is observed that x varies from 0 to 1, z varies from 0 to 1x and y varies from x  to 1z

Therefore, 0101xx1zf(x,y,z)dydzdx

Also, from Figure 3, it is observed that z varies from 0 to 1, x varies from 0 to (1z)2 and y varies from x to 1z

Therefore, 010

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