8.1, #1. Break up the rod by slicing it with planes perpendicular to the direction of the rod. A slice of thickness Δx at a point x meters from the left end of the rod has approximately mass ρ(x) Δdx = (2 + 6x) Δx g/m. So the rod has total mass

∫_{0}^{2}(2 + 6x) dx = 2x + 3x^{2}|_{0}^{2} = 16 grams.

Any Riemann sum for this integral answers the first part of the question.

8.1, #2. The moment of the rod with respect to its left end is

∫_{0}^{2} x(2 + 6x) dx = x^{2} + 2x^{3}|_{0}^{2} = 20.

Thus moment/mass= 20/16 = 1.25 and the center of mass of the rod is 1.25 meters from the left end.

8.1, #3. We are given a function ρ(x)m for the density of cars (in cars per mile). Between x and x + Δx there are appoximately ρ(x)Δx cars. The total number of cars in the twenty miles from x = 0 to x = 20 is approximately

Σ_{i=0}^{n−1}ρ(x_{i})Δx

where x_{i} = 20i/n and Δx = 20/n. This is a Riemann sum for the integral

300 ∫_{0}^{20} 2 + sin(4\sqrt{x + 0.15}) dx

Two Simpson’s approximations are S_{20} = 11505.3 and S_{40} = 11512.0. The integral can be evaluated in elementary terms by substituting u^{2} = 16(x + 0.15) leading to the answer 11512.85 cars.

8.1, #4. The city has radius 3 miles since the population density 10000(3 − r) is zero when r = 3. Partition the city into concentric rings around its center, where each ring has thickness Δr. Since the ring of inner radius r and thickness Δr has approximately area 2πrΔr, the population of the ring is 2πr(10000(3 − r))Δr. The total population is then

∫_{0}^{3} 2π10000(3 − r)r dr = 20000π(\frac{3r^{2}}{2} − \frac{r^{3}}{3} )|_{0}^{3} = 20000π(9/2) = 282743.

8.1, #7. Let x be the distance from the side of length 3. Then 0 ≤ x ≤ 3. The strip between x and x + Δx has area 5Δx and mass approximately 5/(1 + x^{4}) Δx. The total mass is approximately

Σ_{i=0}^{n−1}5/(1 + x^{4}) Δx.

This is a Riemann sum for the integral

∫_{0}^{3} 5/(1 + x^{4}) dx.

There are various ways to get the requested precision. The Simpson approximations S_{2} = 5.664, S_{4} = 5.486, S_{8} = 5.492, and S_{1}6 = 5.492 indicate the mass is 5.50. [The integral can be found in elementary terms by a use of partial fractions.]

8.1, #11. The cross-section of the pipe is a disk of radius 1 inch. At a point r inches from the center of the disk, water flows at the rate of 10(1 − r^{2}) inches/sec. Partition the cross-section into concentric rings of thickness Δr. The flow of water across a ring of inner radius r and thickness Δr is approximately (2πrΔr) × 10(1 − r^{2}) since the area of the ring is approximately 2πrΔr. Thus the total flow of water across the disk is

20π ∫_{0}^{1}(r − r^{3}) dr = 20π(\frac{r^{2}}{2} − \frac{r^{4}}{4} )|_{0}^{1} = 5π inches^{3}/sec.

8.2, #2. Let z be vertical displacement from the center of the sphere, so −r ≤ z ≤ r. The horizontal cut at height z is a disk of radius \sqrt{r^{2} − z^{2}} which has area π(r^{2} − z^{2}). The slice between the cuts at z and z + Δz has volume approximately π(r^{2} − z^{2})Δz. The resulting sum is a Riemann sum for the integral

∫_{−r}^{r} π(r^{2} − z^{2}) dz = \frac{4}{3}πr^{3}.

8.2, #3. Again a horizontal cut at height z above the base will be a disk of radius (3−z)/3 with area π(3−z)^{2}/9. The slice between z and z +Δz has volume approximately π(3 − z)^{2}/9 Δz so the total volume is given by the integral ∫_{0}^{3} π(3 − z)^{2}/9 dz = π.

8.2, #4. If the cone hs height h and base radius r then the radius at height z is b(1 − z/h). The integral in problem #3 becomes

∫_{0}^{b} πb^{2}(1 − z/h)^{2} dz = \frac{1}{3}πb^{2}h.

8.2, #5. Slice the solid by planes perpendicular to the axis of rotation. A slice meeting the axis at [x, x + Δx] has the approximate shape of a disk of radius sin x and thickness Δx. The volume of the disk is π sin^{2} x Δx, so the total volume of all the slices is ∑ π sin^{2} x Δx. Thus the solid has volume

∫_{0}^{π} π sin^{2} x dx = π(−\frac{1}{2} sin x cos x + \frac{1}{2}x)|_{0}^{π} = (π^{2}/2) = 4.9348,

using formula 17 on page 367.