Chapter 38 — Exercises
32 problems on differential forms and the generalized Stokes' theorem. ⭐ to ⭐⭐⭐⭐.
This is a synthesis and conceptual chapter (a preview of differential forms, Section 38.1 and Section 38.9), so most problems ask you to classify, explain, show, and connect — not to grind out arithmetic. The handful of computations (wedge products, exterior derivatives) exist to make the abstractions concrete. Lean into the "explain in words" prompts: being able to say why gradient, curl, and divergence are one operator is the real learning objective.
A note on honesty: when a problem asks you to "verify" a master-theorem identity, you are confirming that the classical theorem you already proved — FTC in Chapter 14, the line-integral theorem and Green's in Chapter 35, Stokes' and Divergence in Chapter 37 — is the special case the form-language predicts. You are not proving the master theorem itself; that belongs to a course in differential geometry.
Difficulty tiers: ⭐ recall / direct classification · ⭐⭐ single-step computation or explanation · ⭐⭐⭐ multi-step synthesis · ⭐⭐⭐⭐ deep conceptual / connect-across-chapters.
| Tier | Count | Problems |
|---|---|---|
| ⭐ | 7 | A1–A7 |
| ⭐⭐ | 9 | B1–B9 |
| ⭐⭐⭐ | 9 | C1–C9 |
| ⭐⭐⭐⭐ | 7 | D1–D7 |
| Total | 32 |
Answers to selected problems appear in the back-of-book answers-to-selected.md.
Part A — Classification and Recall (⭐)
A1. Classify each object as a $0$-form, $1$-form, $2$-form, or $3$-form in $\mathbb{R}^3$ (use the dictionary in Section 38.2): (a) $f = x^2 y z$ (b) $x\,dy - y\,dx$ (c) $x\,dy\wedge dz + z\,dx\wedge dy$ (d) $\sin(xyz)\,dx\wedge dy\wedge dz$
Solution
(a) $0$-form (a function). (b) $1$-form. (c) $2$-form. (d) $3$-form. The degree equals the number of wedged differentials.A2. Match each form-degree to the dimension of the region it is meant to be integrated over: $0$-form, $1$-form, $2$-form, $3$-form ↔ point, curve, surface, solid.
Solution
$0$-form ↔ point (evaluate); $1$-form ↔ curve; $2$-form ↔ surface; $3$-form ↔ solid. A $k$-form integrates over a $k$-dimensional region (Section 38.2).A3. State the single defining rule of the wedge product, and use it to write $dy \wedge dx$ in terms of $dx \wedge dy$, and the value of $dz \wedge dz$. (Section 38.2)
Solution
Antisymmetry: $dx\wedge dy = -dy\wedge dx$. Hence $dy\wedge dx = -dx\wedge dy$, and $dz\wedge dz = 0$.A4. Fill in the blank: applying the exterior derivative $d$ to a $k$-form produces a $\underline{\quad}$-form. (Section 38.3)
Solution
$(k+1)$-form. The exterior derivative raises degree by one.A5. Name the classical vector operator that the exterior derivative $d$ reproduces when applied to (a) a $0$-form, (b) a $1$-form, (c) a $2$-form. (Section 38.3)
Solution
(a) gradient $\nabla f$; (b) curl $\nabla\times\mathbf{F}$; (c) divergence $\nabla\cdot\mathbf{F}$.A6. Write the generalized (master) Stokes' theorem, naming what each symbol $\partial M$, $M$, $\omega$, and $d\omega$ represents. (Section 38.4)
Solution
$\int_{\partial M}\omega = \int_M d\omega$. $M$ is an oriented manifold with boundary; $\partial M$ is its boundary; $\omega$ is an $(n-1)$-form (right thing to integrate over the boundary); $d\omega$ is its exterior derivative, an $n$-form (right thing to integrate over $M$).A7. State the identity $d^2 = 0$ in words, and name the two classical vector identities it encodes. (Section 38.6)
Solution
Applying $d$ twice always gives zero: $d(d\omega) = 0$. It encodes $\nabla\times(\nabla f) = \mathbf{0}$ (curl of a gradient) and $\nabla\cdot(\nabla\times\mathbf{F}) = 0$ (divergence of a curl).Part B — Single-Step Computation and Explanation (⭐⭐)
B1. Using only antisymmetry, simplify $(dx + 2\,dy)\wedge(3\,dx - dy)$. Show the coefficient of $dx\wedge dy$ equals the determinant $\begin{vmatrix} 1 & 2 \\ 3 & -1\end{vmatrix}$.
Solution
Expand: $3\,dx\wedge dx - dx\wedge dy + 6\,dy\wedge dx - 2\,dy\wedge dy$. The $dx\wedge dx$ and $dy\wedge dy$ terms vanish; $dy\wedge dx = -dx\wedge dy$, so we get $-dx\wedge dy - 6\,dx\wedge dy = -7\,dx\wedge dy$. The determinant is $(1)(-1) - (2)(3) = -1 - 6 = -7$. They match.B2. Simplify $(dx + dy + dz)\wedge(dx + dy + dz)$. Explain in one sentence why the answer is what it is.
Solution
The answer is $0$. Any $1$-form wedged with itself vanishes: the diagonal terms $dx\wedge dx$ etc. are zero, and the off-diagonal terms cancel in antisymmetric pairs ($dx\wedge dy + dy\wedge dx = 0$). Geometrically, the parallelogram spanned by a vector with itself is degenerate.B3. Compute $df$ for the $0$-form $f = x^2 y - z^3$. Confirm its coefficients are the components of $\nabla f$. (Section 38.3)
Solution
$df = 2xy\,dx + x^2\,dy - 3z^2\,dz$. The coefficients $(2xy,\ x^2,\ -3z^2)$ are exactly $\nabla f$.B4. For $\omega = y\,dx + z\,dy + x\,dz$, compute $d\omega$ and read off the curl of $\mathbf{F} = (y, z, x)$. (Section 38.3)
Solution
With $P=y,\,Q=z,\,R=x$: $d\omega = (R_y - Q_z)\,dy\wedge dz + (P_z - R_x)\,dz\wedge dx + (Q_x - P_y)\,dx\wedge dy = (0-1)\,dy\wedge dz + (0-1)\,dz\wedge dx + (0-1)\,dx\wedge dy$. So $\nabla\times\mathbf{F} = (-1, -1, -1)$.B5. For $\eta = x\,dy\wedge dz + y\,dz\wedge dx + z\,dx\wedge dy$, compute $d\eta$ and identify the divergence it produces. (Section 38.3)
Solution
With $A=x,\,B=y,\,C=z$: $d\eta = (A_x + B_y + C_z)\,dx\wedge dy\wedge dz = (1+1+1)\,dx\wedge dy\wedge dz = 3\,dx\wedge dy\wedge dz$. The divergence of $\mathbf{F} = (x,y,z)$ is $3$.B6. Explain why "integrating a $0$-form over a point" amounts to evaluation, and how the orientation of the two endpoints of an interval supplies the minus sign in $f(b) - f(a)$. (Section 38.5)
Solution
A point is $0$-dimensional, so "integrating" a function there just reads off its value. The boundary $\partial[a,b] = \{a,b\}$ is oriented so $b$ counts with $+$ and $a$ with $-$ (the interval exits at $b$, enters at $a$). Summing the signed evaluations gives $f(b) - f(a)$.B7. The chapter says Green's and Stokes' are "the same case" of the master theorem. In terms of the dimension of $M$ and the degree of $\omega$, explain what they share and what distinguishes them. (Section 38.5)
Solution
Both have $\dim M = 2$ with $\omega$ a $1$-form and $d\omega$ a $2$-form. They differ only in the ambient space: Green's region lives flat in the plane (so $d\omega$ has one component, $Q_x - P_y$), while Stokes' surface lives in $\mathbb{R}^3$ (so $d\omega$ is the full three-component curl).B8. Which row of the master-theorem table (Section 38.5) corresponds to a $1$-dimensional $M$ bent into a curve in space? Name the theorem and write its statement.
Solution
The Line-integral FTC (Chapter 35): $\int_C \nabla f\cdot d\mathbf{r} = f(B) - f(A)$. Here $\omega = f$ is a $0$-form and $d\omega = \nabla f\cdot d\mathbf{r}$.B9. Explain why the cross product satisfies $\mathbf{u}\times\mathbf{v} = -\mathbf{v}\times\mathbf{u}$, tracing the antisymmetry back to oriented area and the wedge product. (Section 38.2)
Solution
The wedge product encodes *oriented* area: swapping the two spanning directions reverses orientation, so $dx\wedge dy = -dy\wedge dx$. The cross product inherits this — it is built from the same signed-area/determinant structure — so swapping its arguments flips the sign.Part C — Multi-Step Synthesis (⭐⭐⭐)
C1. Build the master-theorem table from memory if you can. For each of the five classical theorems give: region $M$, its dimension, boundary $\partial M$, the degree of $\omega$, and the degree of $d\omega$. Then state in one sentence the pattern all five rows share. (Section 38.5)
Solution
FTC: $M=[a,b]$, dim 1, $\partial M = \{a,b\}$, $\omega$ a $0$-form, $d\omega$ a $1$-form. Line-integral FTC: curve $C$, dim 1, two endpoints, $0$-form, $1$-form. Green's: plane region, dim 2, boundary curve, $1$-form, $2$-form. Stokes': surface in $\mathbb{R}^3$, dim 2, boundary curve, $1$-form, $2$-form. Divergence: solid in $\mathbb{R}^3$, dim 3, boundary surface, $2$-form, $3$-form. Pattern: every row says $\int_{\partial M}\omega = \int_M d\omega$ — boundary integral of $\omega$ equals region integral of its derivative.C2. Take $M = [a,b]$ and $\omega = f$. Walk through how $\int_{\partial M}\omega = \int_M d\omega$ reduces to the ordinary FTC from Chapter 14, and identify exactly where the orientation of the boundary points enters. (Section 38.5)
Solution
$d\omega = f'(x)\,dx$, so the right side is $\int_a^b f'(x)\,dx$. The left side is the signed evaluation over $\partial M = \{a,b\}$: with $b$ positive and $a$ negative, $\int_{\partial M} f = f(b) - f(a)$. Equating gives FTC. The orientation (exit at $b$, enter at $a$) is what makes the boundary term a *difference* rather than a sum.C3. Let $f(x,y) = x^2 - y^2$. Compute $df$, then $d(df)$, showing the wedge cancellation. Explain how the zero answer is an instance of $d^2 = 0$ and which Chapter 37 identity it equals. (Section 38.6)
Solution
$df = 2x\,dx - 2y\,dy$. Then $d(df) = d(2x)\wedge dx + d(-2y)\wedge dy = 2\,dx\wedge dx - 2\,dy\wedge dy = 0$ (both diagonal wedges vanish). This is $d^2 f = 0$, i.e., $\nabla\times(\nabla f) = \mathbf{0}$: the curl of a gradient is zero.C4. Let $\omega = P\,dx + Q\,dy$ in the plane. Compute $d\omega$ and show the master theorem reproduces Green's Theorem (Chapter 35). Where does $dA$ come from? (Section 38.5)
Solution
$d\omega = dP\wedge dx + dQ\wedge dy = (P_y\,dy)\wedge dx + (Q_x\,dx)\wedge dy = (Q_x - P_y)\,dx\wedge dy$. The master theorem gives $\oint_{\partial D} P\,dx + Q\,dy = \iint_D (Q_x - P_y)\,dx\wedge dy$. The $2$-form $dx\wedge dy$ *is* the oriented area element $dA$, so the right side is $\iint_D(Q_x - P_y)\,dA$ — Green's Theorem.C5. Verify by direct computation that the vortex $1$-form $$\omega = \frac{-y\,dx + x\,dy}{x^2 + y^2}$$ on $\mathbb{R}^2 \setminus \{0\}$ is closed ($d\omega = 0$). [Let $P = \frac{-y}{x^2+y^2}$, $Q = \frac{x}{x^2+y^2}$, check $Q_x - P_y$.] (Section 38.6)
Solution
$Q_x = \frac{(x^2+y^2) - x(2x)}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}$. $P_y = \frac{-(x^2+y^2) - (-y)(2y)}{(x^2+y^2)^2} = \frac{-x^2 - y^2 + 2y^2}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}$. Hence $Q_x - P_y = 0$, so $d\omega = 0$.C6. Continuing C5: it is a fact that $\oint_C \omega = 2\pi$ for any circle $C$ around the origin. Explain why this shows $\omega$ is not exact, even though it is closed. Connect to the Poincaré lemma and the role of the hole. (Section 38.6)
Solution
If $\omega = d\eta$ for a $0$-form $\eta$ on the punctured plane, then by the line-integral FTC $\oint_C \omega = 0$ around any closed loop. But $\oint_C\omega = 2\pi \neq 0$, contradiction — so $\omega$ is not exact. The Poincaré lemma guarantees closed implies exact only on simply connected domains; the puncture (hole) at the origin breaks simple connectivity, and the nonzero integral *counts* that hole.C7. Using $\partial\partial = 0$ ("the boundary of a boundary is empty"), explain why a sphere has no boundary and a closed loop has no boundary. Then explain how $\partial\partial = 0$ mirrors the algebraic $d^2 = 0$. (Section 38.6)
Solution
A solid ball's boundary is a sphere; the sphere is a closed surface with no edge, so its boundary is empty. A filled disk's boundary is a loop; the loop's endpoints cancel ($+$ and $-$ at each shared vertex), so its boundary is empty. The master theorem $\int_{\partial M}\omega = \int_M d\omega$ pairs the geometric identity $\partial\partial = 0$ with the algebraic $d^2 = 0$: they are two faces of the same truth, bridged by Stokes.C8. A student says "$d$ is just another name for the gradient, so I'll treat all three operators the same." Construct a nonsensical expression this permits, and explain how the degree bookkeeping of forms forbids it. (Section 38.3)
Solution
It permits "the gradient of a vector field" or "the curl of a function" — neither is defined. In the form language $d$ acts on a *specific degree*: $d$ of a $0$-form (function) is a $1$-form (gradient); $d$ of a $1$-form is a $2$-form (curl). You cannot apply the "gradient version" of $d$ to a $1$-form. The typed degree prevents mismatched, meaningless compositions.C9. The chapter notes div and curl have "ugly formulas in spherical coordinates," while $d$ does not. Explain why the form-language is coordinate-free and what advantage that gives on a sphere or in curved spacetime. (Section 38.8)
Solution
The exterior derivative $d$ and wedge $\wedge$ are defined without reference to any coordinate system — $x,y,z$ appear only as temporary labels. So $d$ has the same simple rules in spherical coordinates as in Cartesian, and the master theorem holds verbatim on any manifold. Vector calculus formulas (div, curl) bake in Cartesian assumptions, which is why they balloon in spherical coordinates; forms make that ugliness disappear because it was an artifact of coordinates.Part D — Deep Conceptual / Connect-Across-Chapters (⭐⭐⭐⭐)
D1. Write a short essay titled "One Theorem, Five Times." Trace $\int_{\partial M}\omega = \int_M d\omega$ through FTC (Chapter 14), the line-integral FTC and Green's (Chapter 35), and Stokes' and Divergence (Chapter 37), naming the dimension and form-degree at each step. Conclude with what the reader who "learned five theorems" actually learned. (Section 38.5)
Solution
A complete answer fixes $n = 1$ ($\omega$ a $0$-form) for FTC and the line-integral FTC; $n = 2$ ($\omega$ a $1$-form) for Green's (flat) and Stokes' (curved surface); $n = 3$ ($\omega$ a $2$-form) for Divergence. In every case $d$ raises the degree to match $\dim M$. The reader learned *one* theorem stated five times in rising dimension.D2. Explain, with reference to Clairaut's theorem on mixed partials from Chapter 30, why $d^2 = 0$ holds. Connect $f_{xy} = f_{yx}$ to the antisymmetry $dx\wedge dy = -dy\wedge dx$ and show the mixed-partial terms cancel in pairs. (Section 38.6)
Solution
For a $0$-form, $df = f_x\,dx + f_y\,dy$, so $d(df)$ produces terms like $f_{xy}\,dy\wedge dx + f_{yx}\,dx\wedge dy$. Antisymmetry rewrites the first as $-f_{xy}\,dx\wedge dy$, giving $(f_{yx} - f_{xy})\,dx\wedge dy$. Clairaut's theorem says $f_{xy} = f_{yx}$, so each such pair cancels and $d^2 f = 0$. The pattern repeats for higher-degree forms.D3. In form language Maxwell's four equations collapse to $dF = 0$ and $d{\star}F = J$ (Section 38.8). Explain which equation is "automatic" from $d^2 = 0$ once one writes $F = dA$, and connect this to the absence of magnetic monopoles. (You need not master the Hodge star $\star$.)
Solution
If $F = dA$ for a potential $1$-form $A$, then $dF = d(dA) = 0$ automatically by $d^2 = 0$. So $dF = 0$ — the two source-free Maxwell laws (Gauss for $\mathbf{B}$, Faraday) — is built in, and $\nabla\cdot\mathbf{B} = 0$ means no magnetic monopoles. The equation with sources, $d{\star}F = J$, is the independent physical input.D4. The change-of-variables formula (Chapter 33) introduced the Jacobian as the local area/volume scaling factor. Explain the conceptual link between the Jacobian determinant and the wedge product as "oriented area made algebraic." Why is the shared determinant not a coincidence? (Section 38.2)
Solution
Under a coordinate change, $du\wedge dv = \det(J)\,dx\wedge dy$, where $J$ is the Jacobian matrix — the wedge of the transformed differentials produces exactly the Jacobian determinant. Both the Jacobian and the wedge measure how an oriented area/volume element scales, with sign tracking orientation. The determinant appears in both because both *are* the oriented-volume function; the Jacobian factor in change-of-variables is the wedge product computing the new area element.D5. Define closed and exact forms, and prove the easy half: every exact form is closed. Then explain, using the vortex form of C5–C6, why the converse can fail on a region with a hole, and what $\{\text{closed}\}/\{\text{exact}\}$ (de Rham cohomology) measures. (Section 38.6)
Solution
Closed: $d\omega = 0$. Exact: $\omega = d\eta$. If $\omega = d\eta$ then $d\omega = d(d\eta) = 0$ by $d^2 = 0$, so exact $\Rightarrow$ closed. The vortex form is closed but not exact on the punctured plane (its loop integral is $2\pi \neq 0$), so the converse fails when there is a hole. The quotient $H^k = \{\text{closed}\}/\{\text{exact}\}$ counts these failures — it measures the $k$-dimensional holes of the space.D6. The unification "took roughly three hundred years to come into focus" (Section 38.5). Sketch the timeline from Newton/Leibniz (FTC) through Green/Gauss/Stokes (special cases) to Cartan (master theorem). Argue why the general result was harder to see than its special cases.
Solution
Newton and Leibniz had FTC by the 1680s; Green, Gauss, and Stokes found the planar and three-dimensional special cases in the early-to-mid 1800s; Cartan assembled the single statement $\int_{\partial M}\omega = \int_M d\omega$ in the early 1900s. The general result was harder because it required *inventing the language of differential forms* — recognizing that line/surface/volume integrands are one family, and that grad/curl/div are one operator $d$. Without that abstraction the rhyme among the theorems looks like analogy, not identity.D7. Topological data analysis (Section 38.8) uses a discrete boundary operator $\partial$ obeying $\partial\partial = 0$. Explain the analogy: what plays the role of "forms," what plays the role of "holes," and why a loop that persists across many scales signals genuine structure. Connect to the continuous de Rham picture of Section 38.6.
Solution
In TDA, chains of simplices play the role of forms (dually), and persistent homology classes play the role of de Rham cohomology classes detecting holes. The discrete $\partial$ obeys $\partial\partial = 0$ for the same reason a sphere has no edge. A loop that survives across many scales is a robust closed-but-not-bounding cycle — a genuine hole in the data's shape (periodicity, a ring of states) rather than sampling noise. It is the discrete cousin of a nonzero de Rham class.Reflection
The synthesis of FTC, the line-integral theorem, Green's, Stokes', and Divergence into the single statement $\int_{\partial M}\omega = \int_M d\omega$ is one of the great unifications in mathematics. What felt like five separate theorems is one theorem read in rising dimension. Mastering this conceptual capstone — forms as integrands, $d$ as the universal derivative, $d^2 = 0$ as the universal identity — is the real payoff of vector calculus.