   Chapter 15.8, Problem 44E

Chapter
Section
Textbook Problem

A model for the density δ of the earth’s atmosphere near its surface is δ = 619.09 - 0.000097ρ where ρ (the distance from the center of the earth) is measured in meters and δ is measured in kilograms per cubic meter. If we take the surface of the earth to be a sphere with radius 6370 km, then this model is a reasonable one for 6.370 × 106 ≤ ρ ≤ 6.375 × 106. Use this model to estimate the mass of the atmosphere between the ground and an altitude of 5 km.

To determine

To estimate: The mass of the atmosphere between the ground and an altitude of 5 km

Explanation

Formula used:

If f is a spherical region E given by aρb,αθβ,cϕd , then, Ef(x,y,z)dV=αβabcdf(ρsinϕcosθ,ρsinϕsinθ,ρcosϕ)ρ2sinϕdϕdρdθ (1)

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 (2)

The spherical coordinates (ρ,θ,ϕ) corresponding to the rectangular coordinates (x,y,z) is,

ρ=x2+y2+z2ϕ=cos1(zρ)θ=cos1(xρsinϕ)

Given:

The density model of the earth’s atmosphere is δ=619.090.000097ρ .

Earth is of sphere in shape with radius 6370 km and 6.370×106ρ6.375×106 .

Calculation:

From the given condition, it is observed that ρ varies from 6.370×106 to 6.375×106 , θ varies from 0 to 2π and ϕ varies from 0 to π . Then, by the equation (1), the mass of the given solid is,

m=02π0π6.370×1066.375×106(619.090.000097ρ)dzdydx=02π0π6.370×1066.375×106(619.090.000097ρ)(ρ2sinϕ)dρdϕdθ=02π0π6.370×1066.375×106(619

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