When an oil well burns, sediment is carried up into the air by the flames and is eventually deposited on the ground. Less sediment is deposited further away from the well. Experimental evidence indicates that the density (in tons per square mile) at a distance r in any direction from the burning oil well is give by the formula D(r) = 7/(1+r^2).

In order to estimate how large a clean-up crew to hire, a company must estimate how many tons of sediment must be removed within 100 miles of the well.

Find and evaluate the integral that represents this total deposit. Round to nearest tenth.

consider the deposit as a series of concentric rings of radius r and width dr. The amount of sediment in each ring is just the area times the density: 2πr dr * D(r)

So, add up all the rings to get the total mass:

m(r) = ∫[0,100] 2πr * 7/(1+r^2) dr
= 7π ∫ 2r/(1+r^2) dr
= 7π ln(1+r^2) [0,100]
= 7π ln(10001)
= 202.5 tons

area of ring is pi*r^2, not 2pi*r (circumference).

To find the total amount of sediment that needs to be removed within 100 miles of the well, we need to evaluate the integral of the density function, D(r), over the range from 0 to 100 miles. The integral represents the accumulation of sediment over the entire range.

The integral can be expressed as:

∫[from 0 to 100] 7/(1+r^2) dr

To evaluate this integral, we can make a substitution by letting u = 1 + r^2. Taking the derivative of both sides, we get du = 2r dr. Solving for dr, we have dr = du/ (2r).

Now, let's rewrite the integral using the substitution:

∫[from 0 to 100] (7/u)(du/2r)

Next, we can split the fraction into two separate terms:

(7/2)∫[from 0 to 100] (1/u) (1/r) du

Since the integral is with respect to u, we can treat r as a constant. Also, the term (1/r) is also a constant, so we can move it outside the integral:

(7/2)(1/r)∫[from 0 to 100] (1/u) du

The integral of (1/u) with respect to u is simply ln|u|. Therefore, we have:

(7/2)(1/r) [ln|u|] evaluated from 0 to 100

Substituting back u = 1 + r^2 and simplifying, we get:

(7/2)(1/r)[ln(1+100^2) - ln(1+0^2)]

Simplifying further:

(7/2)(1/r)[ln(1+100^2) - ln(1)]

Since ln(1) = 0, we have:

(7/2)(1/r)ln(1+100^2)

To find the approximate value, we substitute the value of 100 for r:

(7/2)(1/100)ln(1+100^2) ≈ 0.035ln(10001) ≈ 0.035(9.210) ≈ 0.321

Therefore, the approximate total deposit of sediment within 100 miles of the well is approximately 0.3 tons per square mile.