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Calculus: Early Transcendentals

8th Edition
James Stewart
ISBN: 9781285741550

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BuyFindarrow_forward

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.

30. y2 + z2 = 9, x = -2, x = 2

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 y2+z2=9 , x=2 and x=2 .

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,

y2+z2=9y2+(0)2=9y2=9y=±3

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 3 to 3. To get the limits of z, Solve the given equations as below.

y2+z2=9z2=9y2z=±9y2

Hence, E={(x,y,z)|2x2,3y3,9y2z129y2}

Therefore, E=22339y29y2f(x,y,z)dzdydx

Also, from Figure 1, it is observed that y varies from 3 to 3. x varies from 2 to 2 and z varies from 9y2 to 9y2 .

Hence,

E={(x,y,z)|3x3,2y2,9y2z129y2}

Therefore, E=33229y29y2f(x,y,z)dzdxdy

The variable D2 is the projection of E on yz-plane. From the given that x varies from 2 to 2

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

From Figure 2, it is observed that y varies from 3 to 3.and z varies from 9y2 to 9y2 and also from the given observe that x varies from 2 to 2.

Hence,

E={(x,y,z)|3y3,9y2z9y2,2x2}

Therefore, E=339y29y222f(x,y,z)dxdzdy

Also, from Figure 2, it is observed that z varies from 3 to 3, y varies from 9y2 to 9y2 and x varies from 2 to 2 .

Hence, E={(x,y,z)|3z3,9z2y9z2,2x2}

Therefore, E=339z29z222f(x,y,z)dxdydz

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

y2+z2=9(0)2+z2=9z2=9z=±3

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

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Chapter 15 Solutions

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