   Chapter 15.8, Problem 18E

Chapter
Section
Textbook Problem

Sketch the solid whose volume is given by the integral and evaluate the integral.18. ∫ 0 π / 4 ∫ 0 2 π ∫ 2 sec ϕ ρ2 sin ϕ dρ dθ dϕ

To determine

The solid whose volume is given by the iterated integral and evaluate it.

Explanation

Given:

The region D is {(ρ,ϕ,θ)|0ρsecϕ,0ϕπ4,0θ2π}.

Formula used:

If g(x) is the function of x and h(y) is the function of y and k(z) is the function of z  then, abcdefg(x)h(y)k(z)dzdydx=abg(x)dxcdh(y)dyefk(z)dz (1)

Calculation:

From the given condition,

ρ=secϕρsecϕ=1ρcosϕ=1z=1

Since θ varies between 0 and 2π and ϕ=π4, the region lies under the plane z=1 and inside the cone π4. So, the outline sketch of the given region D is shown below in the Figure 1.

Use the equation (1) and separate the given iterated integral and then integrate with respect to θ and ρ

0π402π0secϕρ2sinϕdρdθdϕ=02πdθ0π40secϕρ2sinϕdρdϕ=[θ]02π0π4[ρ3sinϕ3]0secϕdϕ=[2π0]0π4

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