   Chapter 15.6, Problem 34E

Chapter
Section
Textbook Problem

The figure shows the region of integration for the integral ∫ 0 1 ∫ 0 1 − x 2 ∫ 0 1 − x   f ( x ,   y ,   z )   d y   d z   d x Rewrite this integral as an equivalent iterated integral in the five other orders. To determine

To express: The integral 0101x201xf(x,y,z)dydzdx in five different ways.

Explanation

Given:

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

The region are y=1x and z=1x2.

Calculation:

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

The variable D1 is the projection of E on xz-plane.

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

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

Therefore, 0101z01xf(x,y,z)dydxdz

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

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

From Figure 2, it is observed that y varies from 0 to 1, x varies from 0 to 1y and z varies from 0 to 1x2

Therefore, 0101y01x2f(x,y,z)dzdxdy

Also, from Figure 2, observed that x varies from 0 to 1, y varies from 0 to 1x and z varies from 0 to 1x2

0101x01x2f(x,y,z)dzdydx

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

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

In this plane, the surface z=1x2 intersects the plane y=1x. Therefore equate both the planes to get z=2yy2 so, the projection on this plane is divided into two regions named as R1 and R2. In this region R1, x varies from 0 to 1y and R2, x varies from 0 to 1z. So, the integral is equal.

The integral for R1 is, 01011z01zf(x,y,z)dxdydz+0111z101yf(x,y,z)dxdydz

The integral for R1 is, 0102yy201yf(x,y,z)dxdzdy+012yy2101zf(x,y,z)dxdzdy

Hence, the iterated integral 0101x201xf(x,y,z)dydzdx is expressed in 5 different ways as follows

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