Chapter 18 — Exercises

40 problems across area between curves, volumes (disk/washer/shell), arc length, surface area, work, hydrostatic force, and center of mass. The recurring instruction is the same as the chapter's: the setup is the hard part. For every problem, draw the picture, choose the slicing variable, write the slice's contribution, then integrate. Several problems ask you to set up only — that is deliberate. Answers to odd-numbered problems are in appendices/answers-to-selected.md.

Part A — Area Between Curves (§18.2)

18.1 ⭐ Find the area between $y = x^2$ and $y = 2x$. (Worked Example 18.2.1 derived $4/3$; reproduce the setup, naming top and bottom.)

18.2 ⭐ Find the area between $y = \cos x$ and $y = \sin x$ on $\left[0, \tfrac{\pi}{4}\right]$. State which curve is on top before integrating.

18.3 ⭐⭐ Find the total area enclosed between $y = x^3 - x$ and the $x$-axis. (The cubic crosses the axis three times; per §18.2's pitfall, split at every crossing and add absolute pieces.)

18.4 ⭐⭐ Find the area enclosed between $y = x^2$ and $y = \sqrt{x}$.

18.5 ⭐⭐ Find the area enclosed between $y = x$ and $y = x^3$ over $[-1, 1]$. Exploit symmetry, but explain why a single signed integral $\int_{-1}^{1}(x - x^3)\,dx$ gives the wrong area.

18.6 ⭐⭐⭐ Set up and evaluate the area of the region bounded on the left by $x = y^2$ and on the right by $x = y + 2$. Use horizontal strips and explain (per §18.2) why vertical strips would force a split.

18.7 ⭐⭐⭐ A region is bounded by $y = x^2$ and $y = 8 - x^2$. Decide whether vertical or horizontal strips give the cleaner single integral, justify the choice, then evaluate.

Part B — Volumes by Disks and Washers (§18.3)

18.8 ⭐ Rotate the region under $y = x^2$ on $[0, 1]$ about the $x$-axis. Use the disk method.

18.9 ⭐ Rotate the region under $y = \sqrt{x}$ on $[0, 4]$ about the $x$-axis. (You should get $8\pi$; keep this value for Exercise 18.16.)

18.10 ⭐⭐ Verify the sphere volume formula by rotating $y = \sqrt{R^2 - x^2}$ on $[-R, R]$ about the $x$-axis. Confirm $\tfrac{4}{3}\pi R^3$. (Compare Worked Example 18.3.2.)

18.11 ⭐⭐ Derive the cone formula a second way: rotate $y = \tfrac{r}{h}(h - x)$ on $[0, h]$ about the $x$-axis and confirm $\tfrac{1}{3}\pi r^2 h$. (Worked Example 18.3.1 used $y = \tfrac{r}{h}x$; explain why both generating lines give the same solid.)

18.12 ⭐⭐ Rotate the region between $y = x$ (lower) and $y = \sqrt{x}$ (upper) on $[0,1]$ about the $x$-axis. Identify the outer and inner radii, then apply the washer method. (Worked Example 18.3.3 — confirm $\pi/6$.)

18.13 ⭐⭐⭐ The region bounded by $y^2 = 4x$ and $x = 2$ is rotated about the $y$-axis. Set up the volume by washers in $y$ (the outer radius is the constant $x = 2$, the inner radius is $x = y^2/4$). Evaluate.

18.14 ⭐⭐⭐ Rotate the region between $y = x$ and $y = x^2$ on $[0,1]$ about the line $y = 2$. Identify the new outer and inner radii (both measured from $y = 2$) and set up the washer integral. Evaluate. (Watch the pitfall in §18.3: square each radius before subtracting.)

Part C — Volumes by Cylindrical Shells (§18.4)

18.15 ⭐ Rotate the region under $y = x^2$ on $[0, 1]$ about the $y$-axis using shells. (Worked Example 18.4.1 derived $\pi/2$.)

18.16 ⭐⭐ Rotate the region under $y = \sqrt{x}$ on $[0,4]$ about the $x$-axis using shells (slice in $y$), and confirm it matches the disk answer $8\pi$ from Exercise 18.9. (This is the disk-vs-shell consistency check promised in §18.4's Key Insight.)

18.17 ⭐⭐ Rotate the region under $y = \sin x$ on $[0, \pi]$ about the $y$-axis using shells. (The integrand $2\pi x \sin x$ needs integration by parts, Ch. 15.)

18.18 ⭐⭐ Rotate the region under $y = e^{-x^2}$ on $[0, 2]$ about the $y$-axis. Set up by shells and evaluate via the substitution $u = x^2$. (The §18.4 Check Your Understanding showed the extra factor of $x$ makes this elementary; confirm $\pi(1 - e^{-4})$.)

18.19 ⭐⭐⭐ Rotate the region under $y = x^2$ on $[0, 1]$ about the line $x = -1$. The shell radius is now $x + 1$. Set up and evaluate. (Compare Worked Example 18.4.2, which used $x = 2$.)

18.20 ⭐⭐⭐ The region bounded by $y = x$ and $y = x^2$ on $[0,1]$ is rotated about the $y$-axis. Choose shells, write the shell height as (top curve − bottom curve), and evaluate.

Part D — Arc Length and Surface Area (§§18.5–18.6)

18.21 ⭐⭐ Find the arc length of $y = x^{3/2}$ on $[0, 4]$. (Same family as Worked Example 18.5.1, but a longer interval; substitute $u = 1 + \tfrac{9}{4}x$.)

18.22 ⭐⭐ Find the arc length of the catenary $y = \cosh x$ on $[-1, 1]$. (Use the identity $1 + \sinh^2 x = \cosh^2 x$ — this is the rare arc length that simplifies exactly. Answer: $2\sinh 1$.)

18.23 ⭐⭐⭐ Set up only the arc length of $y = x^2$ on $[0, 1]$, and explain, citing the §18.5 warning, why it has no elementary antiderivative. State which technique (Ch. 16) would be required and what kind of function appears in the closed form.

18.24 ⭐⭐⭐ Set up the arc length of one arch of $y = \sin x$ on $[0, \pi]$. Explain why this is the elliptic-integral situation of §18.5 and note that it must be evaluated numerically (§18.10).

18.25 ⭐⭐ Verify the sphere surface-area formula $S = 4\pi R^2$ by rotating $y = \sqrt{R^2 - x^2}$ on $[-R, R]$ about the $x$-axis. (Worked Example 18.6.1 — show the radius and radical cancel.)

18.26 ⭐⭐⭐ Rotate $y = x^3$ on $[0, 1]$ about the $x$-axis. Set up the surface-area integral $\int_0^1 2\pi x^3 \sqrt{1 + 9x^4}\,dx$, then evaluate it exactly using the substitution $u = 1 + 9x^4$. (This one is elementary; confirm $\tfrac{\pi}{27}\big(10^{3/2} - 1\big)$.)

Part E — Work (§18.7)

18.27 ⭐ Find the work to stretch a spring from its rest length to $0.2\ \text{m}$ if $k = 100\ \text{N/m}$. (Worked Example 18.7.1 — $W = \tfrac12 k L^2$.)

18.28 ⭐⭐⭐ A cylindrical tank of radius $2\ \text{m}$ and height $5\ \text{m}$ is full of water. Find the work to pump all the water over the top rim. (Worked Example 18.7.2; take $\rho = 1000$, $g = 9.81$. Confirm $\approx 1.5\ \text{MJ}$.) Re-derive the lift distance carefully per the §18.7 pitfall.

18.29 ⭐⭐⭐ A conical tank with its apex at the bottom has top radius $3\ \text{m}$ and height $4\ \text{m}$, full of water pumped over the top. Setup is the whole problem: with $y$ measured from the bottom, the slab radius is $r(y) = \tfrac{3}{4}y$ (similar triangles), the slab volume is $\pi r(y)^2\,dy$, and the lift distance is $4 - y$. Write and evaluate $W = \int_0^4 \rho g\,\pi\big(\tfrac34 y\big)^2 (4 - y)\,dy$.

18.30 ⭐⭐⭐ A $50$-ft cable weighing $5\ \text{lb/ft}$ hangs fully extended off a roof. Find the work to wind the entire cable up. (Worked Example 18.7.3 — confirm $6250\ \text{ft·lb}$, then verify with the center-of-mass shortcut.)

18.31 ⭐⭐⭐⭐ A $100\ \text{kg}$ satellite is lifted from Earth's surface ($R_0 = 6.37\times10^6\ \text{m}$) to an altitude of $1000\ \text{km}$ against gravity $F(r) = \tfrac{GMm}{r^2}$, with $GM = g R_0^2$ and $g = 9.81$. (Physics — orbital mechanics.) Set up $W = \int_{R_0}^{R_0 + 10^6} \tfrac{GMm}{r^2}\,dr$, evaluate, and compare to the naive (constant-$g$) estimate $mg \cdot 10^6$. Explain the discrepancy.

Part F — Hydrostatic Force (§18.8)

18.32 ⭐⭐ Find the hydrostatic force on a vertical rectangular dam $50\ \text{m}$ wide and $30\ \text{m}$ deep, with water level at the top. (Worked Example 18.8.1 — constant width $w(h) = 50$.)

18.33 ⭐⭐⭐ A circular porthole of radius $0.5\ \text{m}$ is set in a vertical wall, its center $100\ \text{m}$ below the surface. Set up the force integral. The strip width at depth $h$ is $w = 2\sqrt{0.25 - (h - 100)^2}$. Argue by symmetry that the force equals (pressure at center) × (area) $= \rho g(100)\cdot\pi(0.5)^2$, and explain why the symmetry shortcut works here but not for the triangular plate of 18.34.

18.34 ⭐⭐⭐ A triangular plate (apex at top, base $4\ \text{m}$, height $3\ \text{m}$) is submerged vertically with its apex at the water surface. Setup is the point: by similar triangles the width at depth $h$ is $w(h) = \tfrac{4}{3}h$. Write $F = \int_0^3 \rho g\,h\cdot\tfrac{4}{3}h\,dh$ and evaluate.

Part G — Center of Mass and Centroids (§18.9)

18.35 ⭐⭐ Find the center of mass of a rod on $[0, 2]$ with linear density $\rho(x) = 1 + x^2$. (Compare Worked Example 18.9.1, which used $\rho = 1 + x$.)

18.36 ⭐⭐⭐ Find the centroid $(\bar x, \bar y)$ of the region bounded above by $y = 4$ and below by $y = x^2$. Use the §18.9 formulas; exploit the symmetry that fixes $\bar x = 0$ and confirm it falls out of the integral.

18.37 ⭐⭐⭐⭐ Pappus's theorem (proof-flavored). A circle of radius $r$ centered at $(R, 0)$ with $R > r$ is revolved about the $y$-axis, sweeping out a torus. Using the centroid of the disk (at distance $R$ from the axis) and Pappus's theorem $V = 2\pi\bar d\,A$ from §18.9, show $V = 2\pi^2 R r^2$. Then set up the shell integral for the same torus and confirm Pappus agrees. (You need only argue the shell setup; the integral uses a circular cross-section.)

Part H — Applied Synthesis: Modeling Across Fields (⭐⭐⭐⭐)

18.38 — Biology: tumor volume from imaging. An MRI gives a tumor's radial profile as $r(x) = 2 - \tfrac{1}{2}x^2$ cm for $-2 \le x \le 2$ (with $r$ in cm). Modeling the tumor as a solid of revolution about its central axis, set up and evaluate $V = \int_{-2}^{2}\pi[r(x)]^2\,dx$. State which §18.3 method you used and why a true 3D tumor would need the Chapter 32 machinery.

18.39 — Economics: consumer surplus as area between curves. For linear demand $D(q) = 100 - 5q$ and market price $p^* = 50$, the equilibrium quantity solves $D(q^*) = p^*$. Find $q^*$, then compute consumer surplus $\int_0^{q^*}\big[D(q) - p^*\big]\,dq$. Explain how this is literally the §18.2 area-between-curves integral wearing an economic costume.

18.40 — Data science / physics synthesis. A probability density $p(x) = 2e^{-2x}$ on $[0, \infty)$ describes wait times. (a) Confirm it integrates to $1$ (an improper integral, Ch. 17). (b) Compute the mean $\mu = \int_0^\infty x\,p(x)\,dx$ as a center-of-mass integral (§18.9) — treat $p$ as a "mass distribution." (c) Compute $P(0 \le X \le 1) = \int_0^1 p(x)\,dx$ as an area. Note how one §18.9 moment formula doubles as the definition of expected value.

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