Chapter 32 — Quiz
Ten questions covering the limit definition, Fubini's theorem, general regions and reversing order, polar/cylindrical/spherical coordinates, and applications to mass, center of mass, and probability. Try each before opening the answer.
1. Write the limit definition of the double integral $\iint_R f(x,y)\,dA$ and state what each factor in a single term of the Riemann sum represents.
Answer
$$\iint_R f(x,y)\,dA = \lim_{\|P\|\to 0}\sum_{i,j} f(x_i^*, y_j^*)\,\Delta A_{ij}.$$ A single term $f(x_i^*, y_j^*)\,\Delta A_{ij}$ is the volume of a thin column: $\Delta A_{ij}$ is the floor-area of a small patch of $R$, and $f(x_i^*, y_j^*)$ is the height of the surface $z = f(x,y)$ above a sample point in that patch. (§32.1)2. State Fubini's theorem for a rectangle $R = [a,b]\times[c,d]$. What two separate guarantees does it make?
Answer
$$\iint_R f\,dA = \int_a^b\!\!\int_c^d f\,dy\,dx = \int_c^d\!\!\int_a^b f\,dx\,dy.$$ It guarantees (1) that an iterated integral *equals* the double integral, and (2) that the *order* of integration does not matter — you get the same number either way (for continuous $f$). (§32.2)3. For the factoring shortcut $\iint g(x)h(y)\,dA = \big(\int g\,dx\big)\big(\int h\,dy\big)$ to apply, what two conditions must hold?
Answer
The integrand must **factor** as a product $g(x)\,h(y)$, **and** the region must be a **rectangle with constant limits**. The shortcut fails the moment a limit depends on the other variable, or the integrand is a sum rather than a product. (§32.2)4. A region is described by $0 \le x \le 1$, $x^2 \le y \le 1$. Reverse the order of integration: write the new limits.
Answer
The region lies under $y = 1$ and above $y = x^2$, for $0 \le x \le 1$. Solving $y = x^2$ for $x$ gives $x = \sqrt{y}$. With $y$ outside: $0 \le y \le 1$, and for each $y$, $x$ runs from the left edge $x = 0$ to $x = \sqrt{y}$. So $$\int_0^1\!\!\int_0^{\sqrt{y}} f(x,y)\,dx\,dy.$$ Re-derived from the region, not by shuffling old limits. (§32.3)5. Why is $\int_0^1\!\int_x^1 \sin(y^2)\,dy\,dx$ impossible to evaluate as written, and what makes the reversed order tractable?
Answer
$\int \sin(y^2)\,dy$ is a Fresnel integral with **no elementary antiderivative** (Chapter 17), so the inner integral cannot be done in this order. Reversing gives $\int_0^1\!\int_0^y \sin(y^2)\,dx\,dy = \int_0^1 y\sin(y^2)\,dy$; the extra factor $y$ (the strip length) is exactly the chain-rule factor for $u = y^2$, making it elementary. (§32.3)6. What is the polar area element, and what is the single most common error students make with it? Give a one-line sanity check.
Answer
$dA = r\,dr\,d\theta$. The most common error is **dropping the $r$** and writing $dA = dr\,d\theta$. Sanity check: the unit disk's area must be $\pi$, and $\int_0^{2\pi}\!\int_0^1 r\,dr\,d\theta = 2\pi\cdot\tfrac12 = \pi$ ✓; without the $r$ you get $2\pi$, which is wrong. (§32.5)7. Outline the polar trick that evaluates $\int_{-\infty}^{\infty} e^{-x^2}\,dx = \sqrt{\pi}$. Why must we square first?
Answer
Call the value $I$. Since $e^{-x^2}$ has no elementary antiderivative, we square: $I^2 = \iint_{\mathbb{R}^2} e^{-(x^2+y^2)}\,dA$. In polar this is $\int_0^{2\pi}\!\int_0^\infty e^{-r^2} r\,dr\,d\theta = 2\pi\cdot\tfrac12 = \pi$, so $I = \sqrt{\pi}$. Squaring promotes the 1D integral to 2D, where rotational symmetry produces the magic factor $r$ that makes $u = r^2$ work. (§32.5)8. Give the volume elements for cylindrical and spherical coordinates, and name the symmetry each is suited to.
Answer
Cylindrical: $dV = r\,dr\,d\theta\,dz$ — for an **axis** of symmetry (cylinder, cone, paraboloid). Spherical: $dV = \rho^2\sin\phi\,d\rho\,d\phi\,d\theta$ — for a **point** of symmetry (ball, shell, anything with $x^2+y^2+z^2$). (§32.7–§32.8)9. Write the formulas for the mass and the $x$-coordinate of the center of mass of a flat plate $R$ with density $\rho(x,y)$. When does the center of mass reduce to the centroid?
Answer
$$M = \iint_R \rho\,dA, \qquad \bar x = \frac{1}{M}\iint_R x\,\rho\,dA.$$ When the density is **constant**, $\rho$ cancels between numerator and denominator and the center of mass becomes the purely geometric **centroid** $\bar x = \frac{1}{\text{Area}}\iint_R x\,dA$. (§32.10)10. A point $(X,Y)$ is uniformly distributed on $[0,1]^2$, so $p(x,y) = 1$ there. Express $P(X^2 + Y^2 \le 1)$ as a double integral and evaluate it.
Answer
$$P(X^2+Y^2\le 1) = \iint_R 1\,dA,$$ where $R$ is the part of the unit square inside the unit circle — a quarter-disk of radius $1$. Since the density is the constant $1$, the probability equals the area of that quarter-disk: $\frac14\pi(1)^2 = \frac{\pi}{4}\approx 0.785$. (§32.11)Scoring Guide
- 9–10 correct: Excellent. You command both the conceptual core (limit definition, Fubini, the $r$ factor) and the computational toolkit. Move on to Chapter 33's general change of variables with confidence.
- 7–8 correct: Solid. Re-read the sections flagged on any missed question, especially polar coordinates (§32.5) and reversing order (§32.3) — the two highest-leverage skills.
- 5–6 correct: Partial. Redo Examples 2, 3, and 5 from
index.mdby hand, then retake. Focus on sketching the region first every time. - 0–4 correct: Revisit the chapter from §32.1. The single biggest payoff is internalizing that a multiple integral is always "value times patch-size, summed in the limit" — every formula is bookkeeping on that one idea.