   Chapter 15.6, Problem 31E

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.31. y =x2, z = 0, y + 2z = 4

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=x2 , z=0 and y+2z=4 .

Calculation:

Let 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,

y+2z=4y+2(0)=4y=4

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 x2 to 4. To get the limits of z, solve the given equations as below.

y+2z=42z=4yz=12(4y)z=2y2

Hence, E={(x,y,z)|2x2,x2y4,0z2y2}

Therefore, E=22x2402y2f(x,y,z)dzdydx

Also, from Figure 1, it is observed that y varies from 0 to 4, x varies from y to y and z varies from 0 to 2y2 .

Hence, E={(x,y,z)|0y4,yxy,0z2y2}

Therefore, E=04yy02y2f(x,y,z)dzdxdy

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

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 0 to 2y2 . To get the limits of x, Solve the given equations as below.

x2=yx=±y

Hence, E={(x,y,z)|0y4,0z2y2,yxy}

Therefore, E=0402y2yyf(x,y,z)dxdzdy .

Also, from Figure 2, it is observed that z varies from 0 to 2 , y varies from 0 to 42z and x varies from y to y .

Hence, E={(x,y,z)|0z2,0y42z,yxy}

Therefore, E=02042zyyf(x,y,z)dxdydz

The variable D3 is the projection of E on z x-plane. So, set y=0 . Then , the equation becomes,

y+2z=40+2z=4z=42z=2

The graph of the above function is shown below in Figure 3.

From Figure 3, it is observed that x varies from 2 to 2, z varies from 0 to 2x22 and To get the limits of y, Solve the given equations as below

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