   Chapter 15.6, Problem 2E

Chapter
Section
Textbook Problem

Evaluate the integral ∭ E ( x y   +   z 2 )   d v , where E   =   { ( x ,   y ,   z ) |   0   ≤   x   2 ,   0 ≤   y   ≤   1 ,   0 ≤   z   ≤   3 } using three different orders of integration.3-8 Evaluate the iterated integral.

To determine

To evaluate: The integral of E(xy+z2)dV by arranging different orders of integration.

Explanation

Given:

The function is f(x,y,z)=xy+z2 .

The region is E={(x,y,z)|0x2,0y1,0z3} .

Calculation:

Case 1: Order of integration is x, y and z.

The given integral with the above order is, E(xy+z2)dV=030102(xy+z2)dxdydz .

Integrate the given integral with respect to x and apply the limit of it.

E(xy+z2)dV=0301[yx22+z2x]02dydz=0301[(y(2)22+z2(2))(y(0)22+z2(0))]dydz=0301[(4y2+2z2)(0+0)]dydz=20301[y+z2]dydz

Integrate the given integral with respect to y and apply the limit of it.

E(xy+z2)dV=203[y22+z2y]01dz=203[((1)22+z2(1))((0)22+z2(0))]dz=203[12+z2]dz

Integrate the given integral with respect to z and apply the limit of it.

E(xy+z2)dV=2[12z+z33]03=2[(12(3)+(3)33)(12(0)+(0)33)]=2[32+273]=633

Simplify further to obtain the value.

E(xy+z2)dV=21

Thus, the value of the integral with the order x,y and z is 21.

Case 2:

Order of integration proceed with the order y, then z and x.

The given integral with the above order is, E(xy+z2)dV=020301(xy+z2)dydzdx .

Integrate the given integral with respect to y and apply the limit of it.

E(xy+z2)dV=0203[xy22+z2y]01dzdx=0203[(x(1)22+z2(1))(x(0)22+z2(0))]dzdx=0203[12x+z2]dzdx

Integrate the given integral with respect to z and apply the limit of it

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