Chapter 18 — Self-Assessment Quiz

Q: The area of the region between and on , where throughout, is:

B. Each vertical strip has height "top minus bottom" and width . The strips' heights add up to the area regardless of whether the curves sit above or below the -axis. (§18.2.)

Q: The curves and cross inside . To find the total enclosed area you must: Evaluate directly. Split at the crossing points and integrate "larger minus smaller" on each piece. Take the absolute value of the final number. Use the washer method.

B. A single signed integral lets negative pieces cancel positive ones, giving the wrong (often zero) answer. Find the crossings ( and ), then integrate top-minus-bottom on each subinterval. (§18.2, Common Pitfall.)

Q: Rotating the region under on about the -axis gives the volume:

C. The disk at position has radius , face area , and thickness . (§18.3.)

Q: When rotating the region between an outer curve and inner curve (with ) about the -axis, the washer integrand is:

B. A washer's area is (outer circle) − (inner circle) , which cannot be factored into . Squaring the gap is the single most common volume error. (§18.3, Common Pitfall.)

Q: You want the volume when the region under on is rotated about the -axis. The smart method is: Disks, because the axis is vertical. Washers in , after solving for . Shells, because they keep the original variable and avoid inverting the function. It cannot be done by single-variable calculus.

C. Shells give , which is elementary via (answer ). Disks/washers would force the messy inversion . The shell's extra factor of even rescues the antiderivative. (§18.4, Check Your Understanding.)

Q: The arc-length element for is . The surface area swept when that curve is rotated about the -axis is:

B. Each band has circumference and slant width , not flat width . Dropping the radical computes a stack of flat rings and undercounts a tilted surface. (§18.6, Common Pitfall.)

Q: The arc length of on : Equals . Equals . Has no elementary closed form. Equals exactly .

C. Setting it up gives , which needs trigonometric substitution (Ch. 16) and yields a formula containing a logarithm and . Most arc lengths — and every ellipse perimeter — escape elementary functions. (§18.5.)

Q: A spring with stiffness is stretched from rest. The work stored is:

A. . Energy grows with the square of the stretch. (§18.7, Worked Example 18.7.1.)

Q: In a pump-the-tank problem with measured from the bottom and water exiting over the rim at height , the lift distance for the slab at height is:

C. Two distances are in play — the slab's position and its lift distance. A slab starting at height must rise to the exit at , so it travels . Confusing the two silently corrupts the whole integral. (§18.7, Common Pitfall.)

Q: The hydrostatic force on a submerged vertical plate is — not (pressure) × (total area) — because: Pressure is the same everywhere on the plate. Pressure varies with depth, so it cannot be pulled out of the integral as a constant. The width is always constant. Force and pressure have the same units.

B. Pressure grows linearly with depth, so each strip feels a different pressure; only an integral captures that. (The width also varies for non-rectangular plates — read it off the geometry.) The center of mass of a rod, by contrast, is — moment over mass, §18.9. (§18.8, Warning.)

DataField.Dev

Chapter 18 — Self-Assessment Quiz

10 questions, ~20 minutes. A mix of recall and setup judgment. The point of this chapter is choosing the right slice, so several questions ask you to decide a method, not just plug into a formula. Work each one before opening the answer.

1. The area of the region between $y = f(x)$ and $y = g(x)$ on $[a,b]$, where $f \ge g$ throughout, is:

A) $\displaystyle\int_a^b (f + g)\,dx$
B) $\displaystyle\int_a^b (f - g)\,dx$
C) $\displaystyle\int_a^b (g - f)\,dx$
D) $\displaystyle\left|\int_a^b f\,dx\right| - \left|\int_a^b g\,dx\right|$

Answer

**B.** Each vertical strip has height "top minus bottom" $= f - g$ and width $dx$. The strips' heights add up to the area regardless of whether the curves sit above or below the $x$-axis. *(§18.2.)*

2. The curves $y = \sin x$ and $y = \cos x$ cross inside $[0, 2\pi]$. To find the total enclosed area you must:

A) Evaluate $\int_0^{2\pi}(\sin x - \cos x)\,dx$ directly.
B) Split at the crossing points and integrate "larger minus smaller" on each piece.
C) Take the absolute value of the final number.
D) Use the washer method.

Answer

**B.** A single signed integral lets negative pieces cancel positive ones, giving the wrong (often zero) answer. Find the crossings ($x = \pi/4$ and $5\pi/4$), then integrate top-minus-bottom on each subinterval. *(§18.2, Common Pitfall.)*

3. Rotating the region under $y = f(x) \ge 0$ on $[a,b]$ about the $x$-axis gives the volume:

A) $\displaystyle\int_a^b 2\pi x f(x)\,dx$
B) $\displaystyle\int_a^b \pi f(x)\,dx$
C) $\displaystyle\int_a^b \pi [f(x)]^2\,dx$
D) $\displaystyle\int_a^b \pi [f(x)]^2 \sqrt{1 + [f'(x)]^2}\,dx$

Answer

**C.** The disk at position $x$ has radius $f(x)$, face area $\pi[f(x)]^2$, and thickness $dx$. *(§18.3.)*

4. When rotating the region between an outer curve $f$ and inner curve $g$ (with $f \ge g \ge 0$) about the $x$-axis, the washer integrand is:

A) $\pi (f - g)^2$
B) $\pi (f^2 - g^2)$
C) $\pi f^2 - g^2$
D) $2\pi (f - g)$

Answer

**B.** A washer's area is (outer circle) − (inner circle) $= \pi f^2 - \pi g^2$, which **cannot** be factored into $\pi(f - g)^2$. Squaring the gap is the single most common volume error. *(§18.3, Common Pitfall.)*

5. You want the volume when the region under $y = e^{-x^2}$ on $[0, 2]$ is rotated about the $y$-axis. The smart method is:

A) Disks, because the axis is vertical.
B) Washers in $y$, after solving $y = e^{-x^2}$ for $x$.
C) Shells, because they keep the original variable $x$ and avoid inverting the function.
D) It cannot be done by single-variable calculus.

Answer

**C.** Shells give $\int_0^2 2\pi x\,e^{-x^2}\,dx$, which is elementary via $u = x^2$ (answer $\pi(1 - e^{-4})$). Disks/washers would force the messy inversion $x = \sqrt{-\ln y}$. The shell's extra factor of $x$ even rescues the antiderivative. *(§18.4, Check Your Understanding.)*

6. The arc-length element for $y = f(x)$ is $ds = \sqrt{1 + [f'(x)]^2}\,dx$. The surface area swept when that curve is rotated about the $x$-axis is:

A) $\displaystyle\int_a^b 2\pi f(x)\,dx$
B) $\displaystyle\int_a^b 2\pi f(x)\sqrt{1 + [f'(x)]^2}\,dx$
C) $\displaystyle\int_a^b \pi [f(x)]^2\,dx$
D) $\displaystyle\int_a^b \sqrt{1 + [f'(x)]^2}\,dx$

Answer

**B.** Each band has circumference $2\pi f(x)$ and *slant* width $ds$, not flat width $dx$. Dropping the radical computes a stack of flat rings and undercounts a tilted surface. *(§18.6, Common Pitfall.)*

7. The arc length of $y = x^2$ on $[0, 1]$:

A) Equals $\sqrt{2}$.
B) Equals $\pi/4$.
C) Has no elementary closed form.
D) Equals exactly $1$.

Answer

**C.** Setting it up gives $\int_0^1\sqrt{1 + 4x^2}\,dx$, which needs trigonometric substitution (Ch. 16) and yields a formula containing a logarithm and $\sqrt{5}$. Most arc lengths — and every ellipse perimeter — escape elementary functions. *(§18.5.)*

8. A spring with stiffness $k = 200\ \text{N/m}$ is stretched $10\ \text{cm}$ from rest. The work stored is:

A) $1\ \text{J}$
B) $2\ \text{J}$
C) $10\ \text{J}$
D) $20\ \text{J}$

Answer

**A.** $W = \tfrac12 k L^2 = \tfrac12 (200)(0.1)^2 = \tfrac12(200)(0.01) = 1\ \text{J}$. Energy grows with the *square* of the stretch. *(§18.7, Worked Example 18.7.1.)*

9. In a pump-the-tank problem with $y$ measured from the bottom and water exiting over the rim at height $H$, the lift distance for the slab at height $y$ is:

A) $y$
B) $H$
C) $H - y$
D) $H + y$

Answer

**C.** Two distances are in play — the slab's *position* $y$ and its *lift distance*. A slab starting at height $y$ must rise to the exit at $H$, so it travels $H - y$. Confusing the two silently corrupts the whole integral. *(§18.7, Common Pitfall.)*

10. The hydrostatic force on a submerged vertical plate is $F = \int \rho g\,h\,w(h)\,dh$ — not (pressure) × (total area) — because:

A) Pressure is the same everywhere on the plate.
B) Pressure $\rho g h$ varies with depth, so it cannot be pulled out of the integral as a constant.
C) The width $w(h)$ is always constant.
D) Force and pressure have the same units.

Answer

**B.** Pressure grows linearly with depth, so each strip feels a different pressure; only an integral captures that. (The width $w(h)$ also varies for non-rectangular plates — read it off the geometry.) The center of mass of a rod, by contrast, is $\bar x = \int x\rho\,dx \,/\, \int \rho\,dx$ — moment over mass, §18.9. *(§18.8, Warning.)*

Scoring Guide

9–10 correct — Excellent. You command both the formulas and the method-choice judgment that makes this chapter hard. Move on to Chapter 19.
7–8 correct — Good. Re-read the section behind each miss; the formula is rarely the issue — the setup decision (which slice, which radius, which distance) usually is.
5–6 correct — Fair. Re-work the worked examples in §§18.3–18.4 (disk/washer/shell choice) and §18.7 (lift distance), then retake.
Below 5 — Return to §18.11, the universal "slice and sum" table, and re-derive each integrand from a picture before attempting problems again. The framework, not the formulas, is what you are missing.