8.Rev, #6. A cut by a plane perpendicular to the x-axis at x meets the doyghnut in an annulus with outer radius 3 + \sqrt{1 − x^{2}} and inner radius 3 − \sqrt{1 − x^{2}} for x in the range −1 ≤ x ≤ 1. The area of this annulus is π\{(3+\sqrt{1 − x^{2}})^{2} −(3−\sqrt{1 − x^{2}})^{2}\} = 12π\sqrt{1 − x^{2}}.

The volume is ∫_{−1}^{1} 12π\sqrt{1 − x^{2}} dx = 12π · π/2 = 6π^{2}.

8.Rev, #9. The arclength of y = e^{x} from x = 1 to x = 2 is given by the integral

∫_{1}^{2} \sqrt{1 + e^{2x}} dx = 4.785 by Simpson’s rule. The substitution w^{2} = 1 + e^{2x} will lead

eventually to the same result.

8.Rev, #15. An income stream of $100 per year for 20 years, assuming 10% annual interest rate compounded continuously has a present value of

P = ∫_{0}^{20} 100e^{−0.1t} dt = 100(\frac{−1}{0.1})(e^{−2} − e^{0}) = $864.66.

The future value of this amount is B = $864.66e^{0.1×20} = $6389.06. The balance will reach $5000 in y years if

5000 = ∫_{0}^{y} 100e^{0.1(y−t)} dt = 1000(e^{0.1y} − 1)

so 5 + 1 = e^{0.1y} and 0.1y = ln 6, so y = 17.92years.

8.Rev, #18. If Prob{0 ≤ t ≤ 6} = 0.10, then

0.1 = ∫_{0}^{6} ce^{−ct} dt = 1 − e^{−6c}

so e^{−6c} = 0.9, −6c = ln 0.9, and c = 0.0176. Then Prob{6 ≤ t ≤ 12} = ∫_{6}^{12} ce^{−ct} dt = e^{−6c}(1 − e^{−6c}) = 0.9(1 − 0.9) = 0.09.

8.Rev, #25. The pressure at depth y below the surface is 62.4ylb/ft^{2}. The bottom has area πft^{2} and the force on the bottom, at depth 2ft, is 62.4 × 2 × π = 392lb. The area of the cylindrical region of the side between depth y and y + Δy is 2πΔyft^{2} and the force on this region is approximately 62.4y2πΔylb. The total force on the side is

∫_{0}^{2} 62.4y2π dy = 784lb.

the total force on side and bottom is 1176 lb.

8.Rev, #26. The apple is about 3.5 inches high. Slicing horizontally we get shapes which are approximately disks, so teh volume is about π ∫_{0}^{3.5} r(y)^{2} dy where r(y) is the approximate radius. From the graph of teh apple we have

\begin{matrix}y & 0 & 0.5 & 1.0 & 1.5 & 2.0 & 2.5 & 3.0 & 3.5 \\ r(y) & 0.5 & 1.5 & 1.75 & 2.0 & 2.25 & 2.25 & 2.0 & 1.5 \end{matrix} .

Using the trapezoid rule we find

V = 0.5π\{\frac{1}{2}0.5^{2} + 1.5^{2} + 1.75^{2} + 2.0^{2} + ⋯ + 2.0^{2} + \frac{1}{2}1.5^{2}\} = 38.8in^{3}.

The weight is 38.8 × 0.03 = 1.16lb and the cost is 1.16 × 80 = 93cents.