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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550
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.

32. x = 2, y = 2, z = 0, x + y – 2z = 2

To determine

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

Explanation

Given:

The region E is the solid bounded by the surfaces,

x=2,y=2,z=0, and x+y2z=2

Calculation:

Let E be the solid and if D1,D2,D3 be the 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,

x+y2(0)=2x+y0=2x+y=2y=2x

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

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

x+y2z=22z=2xyz=2xy2z=12(x+y2)

Hence, E={(x,y,z)|0x2,2xy2,0z12(x+y2)}.

Therefore,

E=022x2012(x+y2)f(x,y,z)dzdydx

Also, from Figure 1, it is observed that y varies from 0 to 2, x varies from 2y to 2 and z varies from 0 to 12(x+y2).

Hence, E={(x,y,z)|0y2,2yx2,0z12(x+y2)}

Therefore, E=022y2012(x+y2)f(x,y,z)dzdxdy

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

x+y2z=20+y2z=2y=2+2z

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

From Figure 2, it is observed that y varies from 0 to 2 and z varies from 0 to y2. To get the limits of x, solve the given equations as below.

x+y2z=2x=2y+2z

Hence, E={(x,y,z)|0y2,0zy2,2y+2zx2}.

Therefore, E=020y22y+2z2f(x,y,z)dxdzdy.

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

Hence, E={(x,y,z)|0z1,2zy2,2y+2zx2}.

Therefore, E=012z22y+2z2f(x,y,z)dxdydz

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

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Chapter 15 Solutions

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