Question

The soot produced by a garbage incinerator spreads out in a circular pattern. The depth, H(r), in millimeters, of the soot deposited each month at a distance r kilometers from the incinerator is given by H(r)=0.116eâ2.3r.Write a definite integral (with independent variable r) giving the total volume of soot deposited within 5 kilometers of the incinerator each month.

Answer 1

V = integral of [ 0.116e^(-2.3r) * 2πr (10^3) ] dr with upper limit of 5 and lower limit of 0. Unit in m^3.

Explanation:

H(r) = 0.116e^(-2.3r)

H(r) = millimeters

r = kilometers

We'll use Riemann's sum to approximate the total area underneath the region of the integral.

Using Riemann's sum, we break the region into rings of radius r and width ∆r.

Area of rings = πr^2

Area of the rings with radius r and width ∆r, then becomes=

π(r+ ∆r)^2 - πr^2

On expanding:

Area = π[ r^2 + 2r∆r + (∆r)^2] - πr^2

= πr^2 + 2πr∆r + π(∆r)^2 - πr^2

= 2πr∆r + π(∆r)^2

Area = ∆r(2πr + π∆r)

Area/∆r = 2πr + π∆r

At lim ∆r tends to zero

Area/∆r = 2πr + 0 = 2πr

Area = 2πr∆r

Volume = Area * depth

= Area * H(r)

∆V approximately equal to:

2πr∆r * H(r)

Sum of the contribution for all the rings for the volume (total volume):

V approximately sum of [H(r) *2πr∆r]

V ≈ ∑H(r)· 2πr∆r.

Taking the limit as ∆r tends to zero,

V = integral of [ 0.116e^(-2.3r) * 2πr ] dr with upper limit of 5 and lower limit of 0.

The term H(r) and Area are not in the same unit. We would convert both to meters.

H(r) = mm = 10^(-3)m

Area = km^2 = km*km

= (10^3)m * (10^3)m = (10^6)m^2

V = integral of [(10^-3m) * 0.116e^(-2.3r) *2πr (10^6m^2) dr] with upper limit of 5 and lower limit of 0.

V = integral of [ 0.116e^(-2.3r) * 2πr (10^3) dr ] with upper limit of 5 and lower limit of 0. Unit in m^3.