   Chapter 15.6, Problem 29E

Chapter
Section
Textbook Problem

Express the integral ∭ E f   ( x ,   y ,   z )   d V ,   as an iterated integral in six different ways, where E is the solid bounded by the given surfaces.29. y = 4 - x2 - 4z2, y = 0

To determine

To express: The integral Ef(x,y,z)dV in six different ways.

Explanation

Given:

The region E is the solid bounded by the surfaces y=4x24z2 and y=0 .

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 xy-plane. So, set z=0 . Then, the equation becomes,

y=4x24z2y=4x24(0)2y=4x2

The graph of the above function is shown below in Figure 1.

From Figure 1, it is observed that x varies from 2 to 2 and y varies from 0 to 4x2 . To get the limits of z, Solve the given equations as below.

y=4x24z24z2=4x2yz2=4x2y4z=±124x2y

Hence, E={(x,y,z)|2x2,0y4x2,124x2yz124x2y}

Therefore, 2204x2124x2y124x2yf(x,y,z)dzdydx

Also, from Figure 1, it is observed that y varies from 0 to 4, x varies from 124y to 124y and z varies from 124x2y to 124x2y .

Hence, E={(x,y,z)|0y4,124yx124y,124x2yz124x2y}

Therefore, 04124y124y124x2y124x2yf(x,y,z)dzdxdy

The variable D2 is the projection of E on yz-plane. So, set x=0 . Then , the equation becomes,

y=4x24z2y=4(0)24z2y=44z2

The graph of the above function is shown below in Figure 2.

From Figure 2, it is observed that y varies from 0 to 4 and z varies from 124y to 124y . To get the limits of x, Solve the given equations as below.

y=4x24z2x2=44z2yx=±44z2y

Hence, E={(x,y,z)|0y4,124yz124y,44z2yx44z2y}

Therefore, E=04124y124y4y4z24y4z2f(x,y,z)dxdzdy

Also, from Figure 2, it is observed that z varies from 1 to 1, y varies from 0 to 44z2 and x varies from 44z2y to 44z2y .

Hence, E={(x,y,z)|1z1,0y44z2,4y4z2x4y4z2}

Therefore, E=11044z24y4z24y4z2f(x,y,z)dxdydz

The variable D3 is the projection of E on zx-plane

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