8.3, #9. A spherical planet of radius R and density ρ has volume V = (4/3)πR^{3} and mass M = (4/3)πR^{3}ρ. The escape velocity v for an object from this planet is given in Example 5 on page 441: v = \sqrt{2GM/R}. In particular, v = \sqrt{8GπR^{2}ρ/3} and v is proportional to R and to \sqrt{ρ}.

8.3, #11. Slice the rectangular wall of the dam into horizontal strips of width Δy. A strip y feet from the top of the dam has area 1000Δy ft^{2}. The pressure at that level is 62.4y lb/ft^{2}. So the strip exerts a force of 62.4y × 1000Δy pounds. Summing over all strips gives ∑ 62400yΔy, a Riemann sum for ∫_{0}^{50} 62400y dy. The integral has value 31200(50)^{2} = 78, 000, 000 pounds.

8.3, #14. The car moves with velocity v(t) = 50t/(t + 1) mph at time t. On the time interval 2 ≤ t ≤ 3, its velocity increases from 100/3 = 33.33 mph to 150/4 = 37.5 mph. So the formula 25 + 0.1v giving miles per gallon holds for this interval of time. Slice the interval [2, 3] of time into pieces of width Δt. During the piece of time from t to t + Δt, the car goes approximately v(t)Δt miles using

\frac{v(t)Δt}{25 + 0.1v(t)} = \frac{50t Δt}{(t + 1) (25 + 5t/(t + 1))} = \frac{10t Δt}{6t + 5}

gallons of gasoline. The total amount of gasoline used is

10 ∫_{2}^{3} \frac{t dt}{6t + 5} = 10 ∫_{2}^{3}(\frac{1}{6} − \frac{5}{6(6t + 5)}) dt

= 10(\frac{t}{6} − \frac{5}{36} ln(6t + 5))|_{2}^{3}

= (5/3) − (25/18)(ln 23 − ln 17) = 1.2468 gallons.

8.3, #15. Slice the rod into pieces of width Δx. Since the density of the rod is 5/3 kg/m, the mass of each piece is 5/3Δx. A piece x meters from the axis of rotation moves with velocity 2x meters/sec. So its kinetic energy is (1/2)(5/3)(2x)^{2} Δx = (10/3)x^{2} Δx. The total energy of the rod is

∫_{−3}^{3} \frac{10x^{2}}{3} dx = 10/9x^{3}|_{−3}^{3} = 60kg m^{2}/sec^{2}.

8.3, #16. Slice the record into concentric rings of width Δr. A ring of radius r has approximate area 2πrΔr. Since the record has area (10)^{2}π = 100π cm^{2}, it follows the ring of radius r has approximate mass (2πrΔr/100π)50 = rΔr. Now consider the velocity of a point on the ring as the record spins. In 60 seconds, the record has completed 33\frac{1}{3} revolutions. With each revolution the point has gone 2πr cm. Thus the point has a velocity of (33\frac{1}{3})·(2πr)/60 = (10/9)πr cm/sec. So the ring has velocity (10/9)πr cm/sec and hence kinetic energy (1/2)rΔr[(10/9)πr^{2}]^{2}. Summing over all rings gives a Riemann sum for

∫_{0}^{10} \frac{50π^{2}r^{3}}{81} dr = \frac{25π^{2}r^{4}}{162} |_{0}^{10} = 15231 ergs,

where an erg=gram-cm/sec^{2}.

8.3, #17. Slice the rod into pieces of width Δx. Since the rod has length ℓ and mass M, the mass of each piece is M Δx/ℓ. Consider a piece of the rod at a distance x + a from the particle. So if x = 0, the piece is at the end of the rod nearest the particle, and if x = ℓ, the piece is at the end of the rod farthest from the particle. The gravitational force between the piece and the particle is (GMm Δx)/(ℓ(a + x)^{2}). Thus the total force is

\frac{GMm}{ℓ} ∫_{0}^{ℓ} \frac{dx}{(a + x)^{2}} = −\frac{GMm}{ℓ} \frac{1}{a + x}|_{0}^{ℓ}

= −\frac{GMm}{ℓ}(\frac{1}{a + ℓ} − \frac{1}{a})

= \frac{GMm}{ℓ} \frac{ℓ}{a(a + ℓ)} = \frac{GMm}{a(a + ℓ)}.

8.3, #18. Slice the first rod into pieces of width Δx. The mass of each piece is M_{1} Δx/ℓ_{1} (as computed in #17. Consider a piece of the first rod at a distance of x + a from the near end of the second rod. By the solution of #17 the gravitational force between this piece and the second rod is (GM_{2}M_{1} Δx)/(x + a)(x + a + ℓ_{2}). Thus the total force between the two rods is

GM_{1}M_{2} ∫_{0}^{ℓ_{1}} \frac{dx}{(x + a)(x + a + ℓ_{2})} = G\frac{M_{1}}{l_{1}} \frac{M_{2}}{l_{2}} ln \frac{(a + l_{1})(a + l_{2})}{a(a + l_{1} + l_{2})}