   Chapter 15.6, Problem 6E

Chapter
Section
Textbook Problem

Evaluate the iterated integral.6. ∫ 0 1 ∫ 0 1 ∫ 0 1   − z 2   z y   +   1   d x   d z   d y

To determine

To evaluate: The iterated integral.

Explanation

Given:

The function is f(x,y,z)=zy+1 .

The region is B={(x,y,z)|0x1z2,0y1,0z1} .

Formula used:

If g(x) is the function of x and h(y) is the function of y then,

abcdg(x)h(y)dydx=abg(x)dxcdh(y)dy (1)

Calculation:

The given integral is, 010101z2zy+1dxdzdy .

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

010101z2zy+1dxdzdy=0101[zy+1x]01z2dzdy=0101[(zy+11z2)(zy+1(0))]dzdy=0101[z1z2y+1]dzdy

By the equation (1), 010101z2zy+1dxdzdy=011y+1dy01[z1z2]dz (2)

By substitution method, let u=z2 .

Then, the derivative of u is du=2dz .

Thus, there is a change in the limit values of the integration such as,

If z=0 , then u=0 and if z=1 , then u=1

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