02.10.E1 · analysis / harmonic

Vector calculus exercise pack (Apostol Vol. 2 Ch. 10-12 supplement)

shippedIntermediate-onlyLean: nonepending prereqs

Anchor (Master):

Formal definition of the pack Intermediate

Apostol Vol. 2 Chapters 10-12 develop the integral calculus of vector fields in $R^{2}$ and $R^{3}$ . Chapter 10 covers the line integral $\int_{γ} P d x + Q d y + R d z$ , the work integral $\int_{γ} F \cdot d r$ , and the equivalence of path independence, the conservative condition $F = \nabla ϕ$ , and vanishing of every closed-loop integral. Chapter 11 builds the multiple integral, Fubini's theorem, and the change-of-variables formula with the Jacobian factor, including polar, cylindrical, and spherical coordinates. Chapter 12 treats parametric surfaces, surface area, flux integrals, and the three integral theorems: Green's in the plane, Stokes's on a surface in $R^{3}$ , and the divergence theorem.

This pack collects ten problems from those chapters — three easy, four medium, three hard — each with a hint and a complete worked solution. It tests operational competence: evaluating a line integral, recognising and integrating a conservative field, converting a multiple integral to polar or spherical coordinates, computing a flux, and applying Green's, Stokes's, and the divergence theorems.

Conventions follow the prerequisite units and Apostol: $F$ a vector field; $F \cdot d r$ the work form; $\nabla ϕ$ a gradient field; $\nabla \times F$ the curl, $\nabla \cdot F$ the divergence; $d S = n d S$ the oriented area element on a surface. A field is conservative on a simply connected region iff $\nabla \times F = 0$ there.

Key theorem with full solution Intermediate

Before the pack proper, we work one problem in full as an exemplar of the format. The remaining nine follow the same structure (problem, hint, full answer in <details> blocks).

Lead problem. Divergence theorem on a sphere. Compute the outward flux of $F (x, y, z) = (x, y, z)$ through the sphere $S$ of radius $a$ centred at the origin, both directly and via the divergence theorem.

Solution. Direct. On the sphere of radius $a$ , the outward unit normal is $n = \frac{1}{a} (x, y, z)$ , so $F \cdot n = \frac{1}{a} (x^{2} + y^{2} + z^{2}) = \frac{a ^{2}}{a} = a$ (constant on $S$ ). The flux is $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iint_S a, dS = a\cdot\operatorname{area}(S) = a\cdot 4\pi a^2 = 4\pi a^3. $$

Via the divergence theorem. The divergence is $\nabla \cdot F = \partial_{x} x + \partial_{y} y + \partial_{z} z = 3$ . The divergence theorem converts the flux to a volume integral over the enclosed ball $B$ : $$ \iint_S \mathbf{F}\cdot d\mathbf{S} = \iiint_B \nabla\cdot\mathbf{F},dV = \iiint_B 3,dV = 3\cdot\operatorname{vol}(B) = 3\cdot\tfrac{4}{3}\pi a^3 = 4\pi a^3. $$

Both routes give $4 π a^{3}$ . $□$

The field $F = (x, y, z)$ is the position field, with constant divergence $3$ ; its flux through any closed surface equals $3$ times the enclosed volume. The divergence theorem turns a two-dimensional flux computation into a one-line volume calculation — its operational payoff (Apostol Ch. 12).

Exercises Intermediate

Exercise 7 (medium, proof). Curl-free is not conservative on a punctured plane.

Let $F = (\frac{- y}{x ^{2} + y ^{2}}, \frac{x}{x ^{2} + y ^{2}})$ on $R^{2} ∖ {0}$ . Show $\nabla \times F = 0$ everywhere it is defined, yet $\oint_{C} F \cdot d r = 2 π \neq = 0$ around the unit circle.

Hint

Compute the scalar curl $Q_{x} - P_{y}$ . For the loop integral, parametrise the unit circle and recognise $F \cdot d r = d θ$ .

Answer

Write $P = \frac{- y}{x ^{2} + y ^{2}}$ , $Q = \frac{x}{x ^{2} + y ^{2}}$ . Compute (away from the origin) $$ Q_x = \frac{(x^2+y^2) - x\cdot 2x}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}, \qquad P_y = \frac{-(x^2+y^2) + y\cdot 2y}{(x^2+y^2)^2} = \frac{y^2 - x^2}{(x^2+y^2)^2}. $$ So $Q_{x} - P_{y} = 0$ : the field is curl-free on $R^{2} ∖ {0}$ .

But around the unit circle $C$ , parametrised $x = cos t$ , $y = sin t$ , one finds $F \cdot d r = d θ$ (this is the angular form $d θ$ ), so $$ \oint_C \mathbf{F}\cdot d\mathbf{r} = \int_0^{2\pi} dt = 2\pi \neq 0. $$ A nonzero loop integral means $F$ is not conservative. There is no contradiction with Green's theorem: the domain $R^{2} ∖ {0}$ is not simply connected, and the origin is excluded, so the closed form $F$ need not be exact. This is the generator of $H_{dR}^{1}$ of the punctured plane.

Exercise 8 (hard, numeric). Stokes's theorem.

Verify Stokes's theorem for $F = (y, z, x)$ over the surface $S$ that is the part of the plane $x + y + z = 1$ in the first octant, with boundary triangle $C$ , oriented so the normal has positive $z$ -component.

Hint

Compute $\nabla \times F$ and integrate its flux through $S$ (use the unit normal $\frac{1}{3} (1, 1, 1)$ and the triangle's area). Then independently compute $\oint_{C} F \cdot d r$ around the three edges.

Answer

Curl side. $\nabla \times F = (\partial_{y} x - \partial_{z} z, \partial_{z} y - \partial_{x} x, \partial_{x} z - \partial_{y} y) = (0 - 1, 0 - 1, 0 - 1) = (- 1, - 1, - 1)$ . The plane $x + y + z = 1$ has unit normal $n = \frac{1}{3} (1, 1, 1)$ , so $(\nabla \times F) \cdot n = \frac{- 3}{3} = - 3$ . The triangle has vertices $(1, 0, 0), (0, 1, 0), (0, 0, 1)$ ; its area is $\frac{3}{2}$ . Flux of curl: $- 3 \cdot \frac{3}{2} = - \frac{3}{2}$ .

Boundary side. Traverse $C$ as $(1, 0, 0) \to (0, 1, 0) \to (0, 0, 1) \to (1, 0, 0)$ (this matches the upward normal by the right-hand rule). On edge $1$ , $r (t) = (1 - t, t, 0)$ , $d r = (- 1, 1, 0) d t$ , $F = (y, z, x) = (t, 0, 1 - t)$ , so $F \cdot d r = - t d t$ , integrating to $- \frac{1}{2}$ . By the cyclic symmetry of $F = (y, z, x)$ under $(x, y, z) \to (y, z, x)$ , the other two edges each also give $- \frac{1}{2}$ . Total: $- \frac{3}{2}$ .

Both sides equal $- \frac{3}{2}$ , verifying Stokes's theorem.

Exercise 9 (hard, numeric). Change of variables with a nontrivial Jacobian.

Evaluate $\iint_{D} (x + y) d A$ where $D$ is the parallelogram with vertices $(0, 0), (2, 1), (3, 3), (1, 2)$ , using the substitution $x = 2 u + v$ , $y = u + 2 v$ .

Hint

Check that $(u, v) \in [0, 1]^{2}$ maps onto $D$ . Compute the Jacobian determinant, transform the integrand, and integrate over the unit square.

Answer

The map $(u, v) \mapsto (2 u + v, u + 2 v)$ sends $(0, 0) \mapsto (0, 0)$ , $(1, 0) \mapsto (2, 1)$ , $(1, 1) \mapsto (3, 3)$ , $(0, 1) \mapsto (1, 2)$ — exactly the four vertices of $D$ . So the unit square $[0, 1]^{2}$ maps onto $D$ .

Jacobian: $\frac{\partial ( x , y )}{\partial ( u , v )} = det (2112) = 3$ , so $d A = 3 d u d v$ . The integrand $x + y = (2 u + v) + (u + 2 v) = 3 u + 3 v = 3 (u + v)$ . Therefore $$ \iint_D (x+y),dA = \int_0^1!!\int_0^1 3(u+v)\cdot 3,du,dv = 9\int_0^1!!\int_0^1 (u+v),du,dv = 9\cdot 1 = 9, $$ using $\int_{0}^{1} \int_{0}^{1} (u + v) d u d v = \frac{1}{2} + \frac{1}{2} = 1$ .

Exercise 10 (hard, proof). Divergence of a curl is zero, and the consequence for flux.

Prove that $\nabla \cdot (\nabla \times F) = 0$ for any $C^{2}$ field $F$ , and deduce that the flux of a curl through any closed surface vanishes.

Hint

Write out $\nabla \times F$ in components and take the divergence; mixed partials cancel by equality of second partials. For the flux, apply the divergence theorem.

Answer

Let $F = (F_{1}, F_{2}, F_{3})$ . The curl is $$ \nabla\times\mathbf{F} = (\partial_y F_3 - \partial_z F_2,; \partial_z F_1 - \partial_x F_3,; \partial_x F_2 - \partial_y F_1). $$ Take the divergence: $$ \nabla\cdot(\nabla\times\mathbf{F}) = \partial_x(\partial_y F_3 - \partial_z F_2) + \partial_y(\partial_z F_1 - \partial_x F_3) + \partial_z(\partial_x F_2 - \partial_y F_1). $$ Group the six terms into pairs: $\partial_{x} \partial_{y} F_{3} - \partial_{y} \partial_{x} F_{3}$ , $- \partial_{x} \partial_{z} F_{2} + \partial_{z} \partial_{x} F_{2}$ , $\partial_{y} \partial_{z} F_{1} - \partial_{z} \partial_{y} F_{1}$ . Each pair vanishes by equality of mixed second partials (Clairaut/Schwarz, valid since $F$ is $C^{2}$ ). So $\nabla \cdot (\nabla \times F) = 0$ .

Consequence: for any closed surface $S$ bounding a region $V$ , the divergence theorem gives $$ \iint_S (\nabla\times\mathbf{F})\cdot d\mathbf{S} = \iiint_V \nabla\cdot(\nabla\times\mathbf{F}),dV = \iiint_V 0,dV = 0. $$ In differential-forms language this is $d^{2} = 0$ : the curl corresponds to $d$ on $1$ -forms, the divergence to $d$ on $2$ -forms, and their composite is zero.

Exercise pack. Apostol Vol. 2 Chapters 10-12 supplement: line integrals and conservative fields, multiple integrals and change of variables in polar/cylindrical/spherical coordinates, surface integrals and flux, and the Green/Stokes/divergence theorems.

Prerequisites

02.10.05 pending
03.04.05 pending
03.04.20 pending
03.04.21 pending

Tier anchors

beginner
intermediate: Apostol Vol. 2 exercises (Ch. 10 §10.5-10.20, Ch. 11 §11.18-11.34, Ch. 12 §12.4-12.21): line integrals, conservative fields, multiple integrals and change of variables, surface integrals, Green/Stokes/divergence theorems
master

References

Apostol — Calculus, Volume II: Multi-Variable Calculus and Linear Algebra · Ch. 10 exercises (line integrals, path independence, conservative fields, scalar potentials), Ch. 11 exercises (multiple integrals, change of variables, polar/cylindrical/spherical), Ch. 12 exercises (surface integrals, Green's, Stokes's, and the divergence theorem)
Marsden-Tromba — Vector Calculus · Ch. 4-8 — line integrals, multiple integrals, integral theorems
Schey — Div, Grad, Curl, and All That · Ch. II-IV — flux, divergence, circulation, curl, and the integral theorems

Estimated time

intermediate: 5h