Chapter 1.10

Integration of Differential Forms

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Problem 1.10.1

Let HH be a compact convex set in Rk\mathbb {R}^{k}, with nonempty interior. Let f∈C(H)f \in \mathscr {C}(H), put f(x)=0f(\mathbf{x}) = 0 in the complement of HH, and define ∫Hf\int_{H} f as in Definition 10.3.

Prove that ∫Hf\int_{H} f is independent of the order in which the kk integrations are carried out.

Hint: Approximate ff by functions that are continuous on Rk\mathbb {R}^{k} and whose supports are in HH, as was done in Example 10.4.

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Problem 1.10.2

For i=1,2,3,…i = 1,2,3, \ldots, let φi∈C(R1)\varphi_{i} \in \mathscr {C}\left(\mathbb {R}^{1}\right) have support in (2−i,21−i)\left(2^{-i}, 2^{1-i}\right), such that ∫φi=1\int \varphi_{i} = 1. Put

f(x,y)=∑i=1∞[φi(x)−φi+1(x)]φi(y) f(x, y) = \sum _{i = 1}^{\infty }\left[\varphi _{i}(x)-\varphi _{i+1}(x)\right] \varphi _{i}(y)

Then ff has compact support in R2,f\mathbb {R}^{2}, f is continuous except at (0,0)(0,0), and

∫dy∫f(x,y)dx=0 but ∫dx∫f(x,y)dy=1 \int d y \int f(x, y) d x = 0 \quad \text{ but } \quad \int d x \int f(x, y) d y = 1

Observe that ff is unbounded in every neighborhood of (0,0)(0,0).

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Problem 1.10.3
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(a)

If FF is as in Theorem 10.7, put A=F′(0),F1(x)=A−1F(x)A = F^{\prime }(0), F_{1}(\mathbf{x}) = A^{-1} F(\mathbf{x}). Then F1′(0)=IF_{1}^{\prime }(0) = I. Show that

F1(x)=Gn∘Gn−1∘⋯∘G1(x) \mathbf{F}_{1}(\mathbf{x}) = \mathbf{G}_{n} \circ \mathbf{G}_{n-1} \circ \cdots \circ \mathbf{G}_{1}(\mathbf{x})

in some neighborhood of 0\mathbf{0}, for certain primitive mappings G1,…,Gn\mathbf{G}_{1}, \ldots , \mathbf{G}_{n}. This gives another version of Theorem 10.7:

F(x)=F′(0)Gn∘Gn−1∘⋯∘G1(x) \mathbf{F}(\mathbf{x}) = \mathbf{F}^{\prime }(\mathbf{0}) \mathbf{G}_{n} \circ \mathbf{G}_{n-1} \circ \cdots \circ \mathbf{G}_{1}(\mathbf{x})
(b)

Prove that the mapping (x,y)→(y,x)(x, y) \rightarrow (y, x) of R2\mathbb {R}^{2} onto R2\mathbb {R}^{2} is not the composition of any two primitive mappings, in any neighborhood of the origin. (This shows that the flips BiB_{i} cannot be omitted from the statement of Theorem 10.7).

Problem 1.10.4

For (x,y)∈R2(x, y) \in \mathbb {R}^{2}, define

F(x,y)=(excos⁡y−1,exsin⁡y) \mathbf{F}(x, y) = \left(e^{x} \cos y-1, e^{x} \sin y\right)

Prove that F=G2∘G1\mathbf{F} = \mathbf{G}_{2} \circ \mathbf{G}_{1}, where

G1(x,y)=(excos⁡y−1,y)G2(u,v)=(u,(1+u)tan⁡v) \begin{aligned} & \mathbf{G}_{1}(x, y) = \left(e^{x} \cos y-1, y\right) \\ & \mathbf{G}_{2}(u, v) = (u,(1+u) \tan v) \end{aligned}

are primitive in some neighborhood of (0,0)(0,0). Compute the Jacobians of G1,G2,F\mathbf{G}_{1}, \mathbf{G}_{\mathbf{2}}, \mathbf{F} at (0,0)(0,0). Define

H2(x,y)=(x,exsin⁡y) \mathbf{H}_{2}(x, y) = \left(x, e^{x} \sin y\right)

and find

H1(u,v)=(h(u,v),v) \mathbf{H}_{1}(u, v) = (h(u, v), v)

so that F=H1∘H2F = H_{1} \circ H_{2} is some neighborhood of (0,0)(0,0).

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Problem 1.10.5

Formulate and prove an analogue of Theorem 10.8, in which KK is a compact subset of an arbitrary metric space. (Replace the functions φi\varphi_{i} that occur in the proof of Theorem 10.8 by functions of the type constructed in Exercise 22 of Chap. 4).

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Problem 1.10.6

Strengthen the conclusion of Theorem 10.8 by showing that the functions ψi\psi_{i} can be made differentiable, and even infinitely differentiable. (Use Exercise 1 of Chap. 8 in the construction of the auxiliary functions φi\varphi_{i}).

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Problem 1.10.7
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(a)

Show that the simplex QkQ^{k} is the smallest convex subset of Rk\mathbb {R}^{k} that contains 0,e1,…,ek\mathbf{0}, \mathbf{e}_{1}, \ldots , \mathbf{e}_{k}.

(b)

Show that affine mappings take convex sets to convex sets.

Problem 1.10.8

Let HH be the parallelogram in R2\mathbb {R}^{2} whose vertices are (1,1),(3,2),(4,5),(2,4)(1,1),(3,2),(4,5),(2,4). Find the affine map TT which sends (0,0)(0,0) to (1,1),(1,0)(1,1),(1,0) to (3,2),(0,1)(3,2),(0,1) to (2,4)(2,4). Show that JT=5J_{T} = 5. Use TT to convert the integral

α=∫Hex−ydxdy \alpha = \int _{H} e^{x-y} d x d y

to an integral over I2I^{2} and thus compute α\alpha.

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Problem 1.10.9

Define (x,y)=T(r,θ)(x, y) = T(r, \theta ) on the rectangle

0≤r≤a,0≤θ≤2π 0 \leq r \leq a, \quad 0 \leq \theta \leq 2 \pi

by the equations

x=rcos⁡θ,y=rsin⁡θ x = r \cos \theta , \quad y = r \sin \theta

Show that TT maps this rectangle onto the closed disc DD with center at (0,0)(0,0) and radius aa, that TT is one-to-one in the interior of the rectangle, and that JT(r,θ)=rJ_{T}(r, \theta ) = r. If f∈C(D)f \in \mathscr {C}(D), prove the formula for integration in polar coordinates:

∫Df(x,y)dxdy=∫0a∫02πf(T(r,θ))rdrdθ \int _{D} f(x, y) d x d y = \int _{0}^{a} \int _{0}^{2 \pi } f(T(r, \theta )) r d r d \theta

Hint: Let D0D_{0} be the interior of DD, minus the interval from (0,0)(0,0) to (0,a)(0, a). As it stands, Theorem 10.9 applies to continuous functions ff whose support lies in D0D_{0}. To remove this restriction, proceed as in Example 10.4.

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Problem 1.10.10

Let a→∞a \rightarrow \infty in Exercise 9 and prove that

∫R2f(x,y)dxdy=∫0∞∫02πf(T(r,θ))rdrdθ \int _{\mathbb {R}^{2}} f(x, y) d x d y = \int _{0}^{\infty } \int _{0}^{2 \pi } f(T(r, \theta )) r d r d \theta

for continuous functions ff that decrease sufficiently rapidly as ∣x∣+∣y∣→∞\left|x\right|+\left|y\right| \rightarrow \infty. (Find a more precise formulation). Apply this to

f(x,y)=exp⁡(−x2−y2) f(x, y) = \exp \left(-x^{2}-y^{2}\right)

to derive formula (101) of Chap. 8.

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Problem 1.10.11

Define (u,v)=T(s,t)(u, v) = T(s, t) on the strip

0<s<∞,0<t<1 0 < s < \infty , \quad 0 < t < 1

by setting u=s−st,v=stu = s-s t, v = s t. Show that TT is a 1−11-1 mapping of the strip onto the positive quadrant QQ in R2\mathbb {R}^{2}. Show that JT(s,t)=sJ_{T}(s, t) = s.

For x>0,y>0x > 0, y > 0, integrate

ux−1e−uvy−1e−v u^{x-1} e^{-u} v^{y-1} e^{-v}

over QQ, use Theorem 10.9 to convert the integral to one over the strip, and derive formula (96) of Chap. 8 in this way. (For this application, Theorem 10.9 has to be extended so as to cover certain improper integrals. Provide this extension).

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Problem 1.10.12

Let IkI^{k} be the set of all u=(u1,…,uk)∈Rk\mathbf{u} = \left(u_{1}, \ldots , u_{k}\right) \in \mathbb {R}^{k} with 0≤ui≤10 \leq u_{i} \leq 1 for all ii; let QkQ^{k} be the set of all x=(x1,…,xk)∈Rk\mathrm{x} = \left(x_{1}, \ldots , x_{k}\right) \in \mathbb {R}^{k} with xl≥0,Σxl≤1x_{l} \geq 0, \Sigma x_{l} \leq 1. (IkI^{k} is the unit cube; QkQ^{k} is the standard simplex in Rk\mathbb {R}^{k}). Define x=T(u)\mathbf{x} = T(\mathbf{u}) by

x1=u1x2=(1−u1)u2⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯xk=(1−u1)⋯(1−uk−1)uk \begin{aligned} & x_{1} = u_{1} \\ & x_{2} = \left(1-u_{1}\right) u_{2} \\ & \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \cdots \\ & x_{k} = \left(1-u_{1}\right) \cdots \left(1-u_{k-1}\right) u_{k} \end{aligned}

Show that

∑i=1kxl=1−∏i=1k(1−ul) \sum _{i = 1}^{k} x_{l} = 1-\prod _{i = 1}^{k}\left(1-u_{l}\right)

Show that TT maps IkI^{k} onto QkQ^{k}, that TT is 1−11-1 in the interior of IkI^{k}, and that its inverse SS is defined in the interior of QkQ^{k} by u1=x1u_{1} = x_{1} and

ul=x11−x1−⋯−xl−1 u_{l} = \frac{x_{1}}{1-x_{1}-\cdots -x_{l-1}}

for i=2,…,ki = 2, \ldots , k. Show that

JT(u)=(1−u1)k−1(1−u2)k−2⋯(1−uk−1) J_{T}(u) = \left(1-u_{1}\right)^{k-1}\left(1-u_{2}\right)^{k-2} \cdots \left(1-u_{k-1}\right)

and

Js(x)=[(1−x1)(1−x1−x2)⋯(1−x1−⋯−xk−1)]−1 J_{s}(\mathbf{x}) = \left[\left(1-x_{1}\right)\left(1-x_{1}-x_{2}\right) \cdots \left(1-x_{1}-\cdots -x_{k-1}\right)\right]^{-1}
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Problem 1.10.13

Let r1,…,rkr_{1}, \ldots , r_{k} be nonnegative integers, and prove that

∫Qkx1r1⋯xkrkdx=r1!⋯rk!(k+r1+⋯+rk)! \int _{\mathbf{Q}^{k}} x_{1}^{r_{1}} \cdots x_{k}^{r_{k}} d x = \frac{r_{1}!\cdots r_{k}!}{\left(k+r_{1}+\cdots +r_{k}\right)!}

Hint: Use Exercise 12, Theorems 10.9 and 8.20. Note that the special case r1=⋯=rk=0r_{1} = \cdots = r_{k} = 0 shows that the volume of QkQ^{k} is 1/k!1 / k!.

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Problem 1.10.14

Prove formula (46).

Formula (46) of Section 10.14: every kk-tuple {j1,…,jk}\left\{ j_{1}, \ldots , j_{k}\right\} of distinct integers 1≤ji≤n1 \leq j_{i} \leq n can be converted to an increasing kk-index JJ by a finite number of interchanges of pairs, each of which amounts to a multiplication by −1-1; hence

dxj1∧⋯∧dxjk=ε(j1,…,jk)dxJ,(45) d x_{j_{1}} \wedge \cdots \wedge d x_{j_{k}} = \varepsilon \left(j_{1}, \ldots , j_{k}\right) d x_{J}, \tag {45}

where ε(j1,…,jk)\varepsilon \left(j_{1}, \ldots , j_{k}\right) is 11 or −1-1, depending on the number of interchanges that are needed. Formula (46) asserts that

ε(j1,…,jk)=s(j1,…,jk),(46) \varepsilon \left(j_{1}, \ldots , j_{k}\right) = s\left(j_{1}, \ldots , j_{k}\right), \tag {46}

where ss is as in Definition 9.33: for an ordered kk-tuple of integers, s(j1,…,jk)=∏p<qsgn⁡(jq−jp)s\left(j_{1}, \ldots , j_{k}\right) = \prod_{p<q} \operatorname {sgn}\left(j_{q}-j_{p}\right), with sgn⁡x=1\operatorname {sgn} x = 1 if x>0x > 0, sgn⁡x=−1\operatorname {sgn} x = -1 if x<0x < 0, sgn⁡x=0\operatorname {sgn} x = 0 if x=0x = 0. (Here, for an increasing kk-index I={i1,…,ik}I = \left\{ i_{1}, \ldots , i_{k}\right\}, 1≤i1<⋯<ik≤n1 \leq i_{1} < \cdots < i_{k} \leq n, one writes dxI=dxi1∧⋯∧dxikd x_{I} = d x_{i_{1}} \wedge \cdots \wedge d x_{i_{k}}.)

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Problem 1.10.15

If ω\omega and λ\lambda are kk - and mm-forms, respectively, prove that

ω∧λ=(−1)kmλ∧ω \omega \wedge \lambda = (-1)^{k m} \lambda \wedge \omega
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Problem 1.10.16

If k≥2k \geq 2 and σ=[p0,p1,…,pk]\sigma = \left[p_{0}, p_{1}, \ldots , p_{k}\right] is an oriented affine kk-simplex, prove that ∂2σ=0\partial^{2} \sigma = 0, directly from the definition of the boundary operator ∂\partial. Deduce from this that ∂2Ψ=0\partial^{2} \Psi = 0 for every chain Ψ\Psi.

Hint: For orientation, do it first for k=2,k=3k = 2, k = 3. In general, if i<ji < j, let σij\sigma_{i j} be the (k−2)(k-2)-simplex obtained by deleting pi\mathbf{p}_{i} and pj\mathbf{p}_{j} from σ\sigma. Show that each σij\sigma_{i j} occurs twice in ∂2σ\partial^{2} \sigma, with opposite sign.

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Problem 1.10.17

Put J2=τ1+τ2J^{2} = \tau_{1}+\tau_{2}, where

τ1=[0,e1,e1+e2],τ2=−[0,e2,e2+e1] \tau _{1} = \left[\mathbf{0}, \mathbf{e}_{1}, \mathbf{e}_{1}+\mathbf{e}_{2}\right], \quad \tau _{2} = -\left[\mathbf{0}, \mathbf{e}_{2}, \mathbf{e}_{2}+\mathbf{e}_{1}\right]

Explain why it is reasonable to call J2J^{2} the positively oriented unit square in R2\mathbb {R}^{2}. Show that ∂J2\partial J^{2} is the sum of 4 oriented affine 1 -simplexes. Find these. What is ∂(τ1−τ2)\partial \left(\tau_{1}-\tau_{2}\right)?

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Problem 1.10.18

Consider the oriented affine 3 -simplex

σ1=[0,e1,e1+e2,e1+e2+e3] \sigma _{1} = \left[0, e_{1}, e_{1}+e_{2}, e_{1}+e_{2}+e_{3}\right]

in R3\mathbb {R}^{3}. Show that σ1\sigma_{1} (regarded as a linear transformation) has determinant 1. Thus σ1\sigma_{1} is positively oriented.

Let σ2,…,σ6\sigma_{2}, \ldots , \sigma_{6} be five other oriented 3 -simplexes, obtained as follows: There are five permutations (i1,i2,i3)\left(i_{1}, i_{2}, i_{3}\right) of (1,2,3)(1,2,3), distinct from (1,2,3)(1,2,3). Associate with each (i1,i2,i3)\left(i_{1}, i_{2}, i_{3}\right) the simplex

s(i1,i2,i3)[0,ei1,ei1+ei2,ei1+ei2+ei3] s\left(i_{1}, i_{2}, i_{3}\right)\left[0, \mathbf{e}_{i_{1}}, \mathbf{e}_{i_{1}}+\mathbf{e}_{i_{2}}, \mathbf{e}_{i_{1}}+\mathbf{e}_{i_{2}}+\mathbf{e}_{i_{3}}\right]

where ss is the sign that occurs in the definition of the determinant. (This is how τ2\tau_{2} was obtained from τ1\tau_{1} in Exercise 17).

Show that σ2,…,σ6\sigma_{2}, \ldots , \sigma_{6} are positively oriented. Put J3=σ1+⋯+σ6J^{3} = \sigma_{1}+\cdots +\sigma_{6}. Then J3J^{3} may be called the positively oriented unit cube in R3\mathbb {R}^{3}.

Show that ∂J3\partial J^{3} is the sum of 12 oriented affine 2 -simplexes. (These 12 triangles cover the surface of the unit cube I3I^{3}).

Show that x=(x1,x2,x3)\mathrm{x} = \left(x_{1}, x_{2}, x_{3}\right) is in the range of σ1\sigma_{1} if and only if 0≤x3≤x2≤x1≤10 \leq x_{3} \leq x_{2} \leq x_{1} \leq 1.

Show that the ranges of σ1,…,σ6\sigma_{1}, \ldots , \sigma_{6} have disjoint interiors, and that their union covers I3I^{3}. (Compare with Exercise 13; note that 3!=63! = 6).

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Problem 1.10.19

Let J2J^{2} and J3J^{3} be as in Exercise 17 and 18. Define

B01(u,v)=(0,u,v),B11(u,v)=(1,u,v)B02(u,v)=(u,0,v),B12(u,v)=(u,1,v)B03(u,v)=(u,v,0),B13(u,v)=(u,v,1) \begin{array}{ll} B_{01}(u, v) = (0, u, v), & B_{11}(u, v) = (1, u, v) \\ B_{02}(u, v) = (u, 0, v), & B_{12}(u, v) = (u, 1, v) \\ B_{03}(u, v) = (u, v, 0), & B_{13}(u, v) = (u, v, 1) \end{array}

These are affine, and map R2\mathbb {R}^{2} into R3\mathbb {R}^{3}. Put βri=Bri(J2)\beta_{r i} = B_{r i}\left(J^{2}\right), for r=0,1,i=1,2,3r = 0,1, i = 1,2,3. Each βri\beta_{r i} is an affine-oriented 2-chain. (See Sec. 10.30). Verify that

∂J3=∑t=13(−1)t(β0t−β1i), \partial J^{3} = \sum _{t = 1}^{3}(-1)^{t}\left(\beta _{0 t}-\beta _{1 i}\right),

in agreement with Exercise 18.

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Problem 1.10.20

State conditions under which the formula

∫⊕fdω=∫ΔΦfω−∫Φ(df)∧ω \int _{\oplus } f d \omega = \int _{\Delta \Phi } f \omega -\int _{\Phi }(d f) \wedge \omega

is valid, and show that it generalizes the formula for integration by parts. Hint: d(fω)=(df)∧ω+fdωd(f \omega ) = (d f) \wedge \omega +f d \omega.

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Problem 1.10.21

As in Example 10.36, consider the 1 -form

η=xdy−ydxx2+y2 \eta = \frac{x d y-y d x}{x^{2}+y^{2}}

in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\}.

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(a)

Carry out the computation that leads to formula (113), and prove that dη=0d \eta = 0.

(b)

Let γ(t)=(rcos⁡t,rsin⁡t)\gamma (t) = (r \cos t, r \sin t), for some r>0r > 0, and let Γ\Gamma be a C′′\mathscr {C}^{\prime \prime }-curve in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\}, with parameter interval [0,2π][0,2 \pi ], with Γ(0)=Γ(2π)\Gamma (0) = \Gamma (2 \pi ), such that the intervals [γ(t)[\gamma (t), Γ(t)]\Gamma (t)] do not contain 0 for any t∈[0,2π]t \in [0,2 \pi ]. Prove that

∫Γη=2π \int _{\Gamma } \eta = 2 \pi

Hint: For 0≤t≤2π,0≤u≤10 \leq t \leq 2 \pi , 0 \leq u \leq 1, define

Φ(t,u)=(1−u)Γ(t)+uγ(t). \Phi (t, u) = (1-u) \Gamma (t)+u \gamma (t).

Then Φ\Phi is a 2 -surface in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\} whose parameter domain is the indicated rectangle. Because of cancellations (as in Example 10.32),

∂Φ=Γ−γ \partial \Phi = \Gamma -\gamma

Use Stokes' theorem to deduce that

∫Γη=∫γη \int _{\Gamma } \eta = \int _{\gamma } \eta

because dη=0d \eta = 0.

(c)

Take Γ(t)=(acos⁡t,bsin⁡t)\Gamma (t) = (a \cos t, b \sin t) where a>0,b>0a > 0, b > 0 are fixed. Use part (b) to show that

∫02πaba2cos⁡2t+b2sin⁡2tdt=2π \int _{0}^{2 \pi } \frac{a b}{a^{2} \cos ^{2} t+b^{2} \sin ^{2} t} d t = 2 \pi
(d)

Show that

η=d(arctan⁡yx) \eta = d\left(\arctan \frac{y}{x}\right)

in any convex open set in which x≠0x \neq 0, and that

η=d(−arctan⁡xy) \eta = d\left(-\arctan \frac{x}{y}\right)

in any convex open set in which y≠0y \neq 0. Explain why this justifies the notation η=dθ\eta = d \theta, in spite of the fact that η\eta is not exact in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\}.

(e)

Show that (b) can be derived from (d).

(f)

If Γ\Gamma is any closed C′\mathscr {C}^{\prime }-curve in R2−{0}\mathbb {R}^{2}-\left\{ 0\right\}, prove that

12π∫Γη=Ind⁡(Γ) \frac{1}{2 \pi } \int _{\Gamma } \eta = \operatorname {Ind}(\Gamma )

(See Exercise 23 of Chap. 8 for the definition of the index of a curve).

Problem 1.10.22

As in Example 10.37, define ζ\zeta in R3−{0}\mathbb {R}^{3}-\left\{ 0\right\} by

ζ=xdy∧dz+ydz∧dx+zdx∧dyr3 \zeta = \frac{x d y \wedge d z+y d z \wedge d x+z d x \wedge d y}{r^{3}}

where r=(x2+y2+z2)1/2r = \left(x^{2}+y^{2}+z^{2}\right)^{1 / 2}, let DD be the rectangle given by 0≤u≤π,0≤v≤2π0 \leq u \leq \pi , 0 \leq v \leq 2 \pi, and let Σ\Sigma be the 2 -surface in R3\mathbb {R}^{3}, with parameter domain DD, given by

x=sin⁡ucos⁡v,y=sin⁡usin⁡v,z=cos⁡u x = \sin u \cos v, \quad y = \sin u \sin v, \quad z = \cos u
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(a)

Prove that dζ=0d \zeta = 0 in R3−{0}\mathbb {R}^{3}-\left\{ 0\right\}

(b)

Let SS denote the restriction of Σ\Sigma to a parameter domain E⊂DE \subset D. Prove that

∫Sζ=∫Esin⁡ududv=A(S) \int _{S} \zeta = \int _{E} \sin u d u d v = A(S)

where AA denotes area, as in Sec. 10.43. Note that this contains (115) as a special case.

(c)

Suppose g,h1,h2,h3g, h_{1}, h_{2}, h_{3}, are C′′\mathscr {C}^{\prime \prime }-functions on [ 0,1],g>0], g > 0. Let (x,y,z)=Φ(s,t)(x, y, z) = \Phi (s, t) define a 2 -surface Φ\Phi, with parameter domain I2I^{2}, by

x=g(t)h1(s),y=g(t)h2(s),z=g(t)h3(s) x = g(t) h_{1}(s), \quad y = g(t) h_{2}(s), \quad z = g(t) h_{3}(s)

Prove that

∫Φζ=0 \int _{\Phi } \zeta = 0

directly from (35). Note the shape of the range of Φ\Phi : For fixed s,Φ(s,t)s, \Phi (s, t) runs over an interval on a line through 0. The range of Φ\Phi thus lies in a "cone" with vertex at the origin.

(d)

Let EE be a closed rectangle in DD, with edges parallel to those of DD. Suppose f∈C′′(D),f>0f \in \mathscr {C}^{\prime \prime }(D), f > 0. Let Ω\Omega be the 2 -surface with parameter domain EE, defined by

Ω(u,v)=f(u,v)Σ(u,v) \Omega (u, v) = f(u, v) \Sigma (u, v)

Define SS as in (b)(b) and prove that

∫Ωζ=∫Sζ=A(S) \int _{\Omega } \zeta = \int _{S} \zeta = A(S)

(Since SS is the "radial projection" of Ω\Omega into the unit sphere, this result makes it reasonable to call ∫Ωζ\int_{\Omega } \zeta the "solid angle" subtended by the range of Ω\Omega at the origin).

Hint: Consider the 3-surface Ψ\Psi given by

Ψ(t,u,v)=[1−t+tf(u,v)]Σ(u,v) \Psi (t, u, v) = [1-t+t f(u, v)] \Sigma (u, v)

where (u,v)∈E,0≤t≤1(u, v) \in E, 0 \leq t \leq 1. For fixed vv, the mapping (t,u)→Ψ(t,u,v)(t, u) \rightarrow \Psi (t, u, v) is a 2 -sur- face Φ\Phi to which (c) can be applied to show that ∫Φζ=0\int_{\Phi } \zeta = 0. The same thing holds when uu is fixed. By (a) and Stokes' theorem,

∫∂Ψζ=∫Ψdζ=0 \int _{\partial \Psi } \zeta = \int _{\Psi } d \zeta = 0
(e)

Put λ=−(z/r)η\lambda = -(z / r) \eta, where

η=xdy−ydxx2+y2 \eta = \frac{x d y-y d x}{x^{2}+y^{2}}

as in Exercise 21. Then λ\lambda is a 1 -form in the open set V⊂R3V \subset \mathbb {R}^{3} in which x2+y2>0x^{2}+y^{2} > 0. Show that ζ\zeta is exact in VV by showing that

ζ=dλ \zeta = d \lambda
(f)

Derive (d) from (e), without using (c). Hint: To begin with, assume 0<u<π0 < u < \pi on EE. By (e),

∫Ωζ=∫∂Ωλ and ∫Sζ=∫∂Sλ \int _{\Omega } \zeta = \int _{\partial \Omega } \lambda \quad \text{ and } \quad \int _{S} \zeta = \int _{\partial S} \lambda

Show that the two integrals of λ\lambda are equal, by using part (d) of Exercise 21, and by noting that z/rz / r is the same at Σ(u,v)\Sigma (u, v) as at Ω(u,v)\Omega (u, v).

(g)

Is ζ\zeta exact in the complement of every line through the origin?

Problem 1.10.23

Fix nn. Define rk=(x12+⋯+xk2)1/2r_{k} = \left(x_{1}^{2}+\cdots +x_{k}^{2}\right)^{1 / 2} for 1≤k≤n1 \leq k \leq n, let EkE_{k} be the set of all x∈Rn\mathbf{x} \in \mathbb {R}^{n} at which rk>0r_{k} > 0, and let ωk\omega_{k} be the (k−1)(k-1)-form defined in EkE_{k} by

ωk=(rk)−k∑i=1k(−1)i−1xidx1∧⋯∧dxi−1∧dxi+1∧⋯∧dxk \omega _{k} = \left(r_{k}\right)^{-k} \sum _{i = 1}^{k}(-1)^{i-1} x_{i} d x_{1} \wedge \cdots \wedge d x_{i-1} \wedge d x_{i+1} \wedge \cdots \wedge d x_{k}

Note that ω2=η,ω3=ζ\omega_{2} = \eta , \omega_{3} = \zeta, in the terminology of Exercises 21 and 22. Note also that

E1⊂E2⊂⋯⊂En=Rn−{0} E_{1} \subset E_{2} \subset \cdots \subset E_{n} = \mathbb {R}^{n}-\left\{ 0\right\}
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(a)

Prove that dωk=0d \omega_{k} = 0 in EkE_{k}.

(b)

For k=2,…,nk = 2, \ldots , n, prove that ωk\omega_{k} is exact in Ek−1E_{k-1}, by showing that

ωk=d(fkωk−1)=(dfk)∧ωk−1 \omega _{k} = d\left(f_{k} \omega _{k-1}\right) = \left(d f_{k}\right) \wedge \omega _{k-1}

where fk(x)=(−1)kgk(xk/rk)f_{k}(\mathbf{x}) = (-1)^{k} g_{k}\left(x_{k} / r_{k}\right) and

gk(t)=∫−1t(1−s2)(k−3)/2ds(−1<t<1) g_{k}(t) = \int _{-1}^{t}\left(1-s^{2}\right)^{(k-3) / 2} d s \quad (-1 < t < 1)

Hint: fkf_{k} satisfies the differential equations

x⋅(∇fk)(x)=0 \mathbf{x} \cdot \left(\nabla f_{k}\right)(\mathbf{x}) = 0

and

(Dkfk)(x)=(−1)k(rk−1)k−1(rk)k \left(D_{k} f_{k}\right)(\mathbf{x}) = \frac{(-1)^{k}\left(r_{k-1}\right)^{k-1}}{\left(r_{k}\right)^{k}}
(c)

Is ωn\omega_{n} exact in EnE_{n}?

(d)

Note that (b) is a generalization of part (e) of Exercise 22. Try to extend some of the other assertions of Exercises 21 and 22 to ωn\omega_{n}, for arbitrary nn.

Problem 1.10.24

Let ω=Σai(x)dxi\omega = \Sigma a_{i}(\mathbf{x}) d x_{i} be a 1 -form of class C′′\mathscr {C}^{\prime \prime } in a convex open set E⊂RnE \subset \mathbb {R}^{n}. Assume dω=0d \omega = 0 and prove that ω\omega is exact in EE, by completing the following outline:

Fix p∈E\mathbf{p} \in E. Define

f(x)=∫[p,x]ω(x∈E) f(\mathbf{x}) = \int _{[\mathbf{p}, \mathbf{x}]} \omega \quad (\mathbf{x} \in E)

Apply Stokes' theorem to affine-oriented 2-simplexes [p, x, y] in E. Deduce that

f(y)−f(x)=∑i=1n(yi−xi)∫01ai((1−t)x+ty)dt f(\mathbf{y})-f(\mathbf{x}) = \sum _{i = 1}^{n}\left(y_{i}-x_{i}\right) \int _{0}^{1} a_{i}((1-t) \mathbf{x}+t \mathbf{y}) d t

for x∈E,y∈E\mathbf{x} \in E, \mathbf{y} \in E. Hence (Dif)(x)=ai(x)\left(D_{i} f\right)(\mathbf{x}) = a_{i}(\mathbf{x}).

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Problem 1.10.25

Assume that ω\omega is a 1-form in an open set E⊂RnE \subset \mathbb {R}^{n} such that

∫γω=0 \int _{\gamma } \omega = 0

for every closed curve γ\gamma in EE, of class C′\mathscr {C}^{\prime }. Prove that ω\omega is exact in EE, by imitating part of the argument sketched in Exercise 24.

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Problem 1.10.26

Assume ω\omega is a 1 -form in R3−{0}\mathbb {R}^{3}-\left\{ 0\right\}, of class C′\mathscr {C}^{\prime } and dω=0d \omega = 0. Prove that ω\omega is exact in R3−{0}\mathbb {R}^{3}-\left\{ 0\right\}.

Hint: Every closed continuously differentiable curve in R3−{0}\mathbb {R}^{3}-\left\{ 0\right\} is the boundary of a 2-surface in R3−{0}\mathbb {R}^{3}-\left\{ 0\right\}. Apply Stokes' theorem and Exercise 25.

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Problem 1.10.27

Let EE be an open 3-cell in R3\mathbb {R}^{3}, with edges parallel to the coordinate axes. Suppose (a,b,c)∈E,fi∈C′(E)(a, b, c) \in E, f_{i} \in \mathscr {C}^{\prime }(E) for i=1,2,3i = 1,2,3,

ω=f1dy∧dz+f2dz∧dx+f3dx∧dy \omega = f_{1} d y \wedge d z+f_{2} d z \wedge d x+f_{3} d x \wedge d y

and assume that dω=0d \omega = 0 in EE. Define

λ=g1dx+g2dy \lambda = g_{1} d x+g_{2} d y

where

g1(x,y,z)=∫czf2(x,y,s)ds−∫byf3(x,t,c)dtg2(x,y,z)=−∫czf1(x,y,s)ds \begin{aligned} & g_{1}(x, y, z) = \int _{c}^{z} f_{2}(x, y, s) d s-\int _{b}^{y} f_{3}(x, t, c) d t \\ & g_{2}(x, y, z) = -\int _{c}^{z} f_{1}(x, y, s) d s \end{aligned}

for (x,y,z)∈E(x, y, z) \in E. Prove that dλ=ωd \lambda = \omega in EE. Evaluate these integrals when ω=ζ\omega = \zeta and thus find the form λ\lambda that occurs in part (e)(e) of Exercise 22.

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Problem 1.10.28

Fix b>a>0b > a > 0, define

Φ(r,θ)=(rcos⁡θ,rsin⁡θ) \Phi (r, \theta ) = (r \cos \theta , r \sin \theta )

for a≤r≤b,0≤θ≤2πa \leq r \leq b, 0 \leq \theta \leq 2 \pi. (The range of Φ\Phi is an annulus in R2\mathbb {R}^{2}). Put ω=x3dy\omega = x^{3} d y, and compute both

∫Φdω and ∫∂Φω \int _{\Phi } d \omega \quad \text{ and } \quad \int _{\partial \Phi } \omega

to verify that they are equal.

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Problem 1.10.29

Prove the existence of a function α\alpha with the properties needed in the proof of Theorem 10.38, and prove that the resulting function FF is of class C′\mathscr {C}^{\prime }. (Both assertions become trivial if EE is an open cell or an open ball, since α\alpha can then be taken to be a constant. Refer to Theorem 9.42).

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Problem 1.10.30

If N\mathbf{N} is the vector given by (135), prove that

det⁡[α1β1α2β3−α3β2α2β2α3β1−α1β3α3β3α1β2−α2β1]=∣N∣2 \operatorname {det}\left[\begin{smallmatrix} \alpha _{1} & \beta _{1} & \alpha _{2} \beta _{3}-\alpha _{3} \beta _{2} \\ \alpha _{2} & \beta _{2} & \alpha _{3} \beta _{1}-\alpha _{1} \beta _{3} \\ \alpha _{3} & \beta _{3} & \alpha _{1} \beta _{2}-\alpha _{2} \beta _{1} \end{smallmatrix}\right] = \left|\mathbf{N}\right|^{2}

Also, verify Eq. (137).

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Problem 1.10.31

Let E⊂R3E \subset \mathbb {R}^{3} be open, suppose g∈C′′(E),h∈C′′(E)g \in \mathscr {C}^{\prime \prime }(E), h \in \mathscr {C}^{\prime \prime }(E), and consider the vector field

F=g∇h \mathbf{F} = g \nabla h
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(a)

Prove that

∇⋅F=g∇2h+(∇g)⋅(∇h) \nabla \cdot F = g \nabla ^{2} h+(\nabla g) \cdot (\nabla h)

where ∇2h=∇⋅(∇h)=Σ∂2h/∂xi2\nabla^{2} h = \nabla \cdot (\nabla h) = \Sigma \partial^{2} h / \partial x_{i}^{2} is the so-called "Laplacian" of hh.

(b)

If Ω\Omega is a closed subset of EE with positively oriented boundary ∂Ω\partial \Omega (as in Theorem 10.51), prove that

∫Ω[g∇2h+(∇g)⋅(∇h)]dV=∫∂Ωg∂h∂ndA \int _{\Omega }\left[g \nabla ^{2} h+(\nabla g) \cdot (\nabla h)\right] d V = \int _{\partial \Omega } g \frac{\partial h}{\partial n} d A

where (as is customary) we have written ∂h/∂n\partial h / \partial n in place of (∇h)⋅n(\nabla h) \cdot \mathbf{n}. (Thus ∂h/∂n\partial h / \partial n is the directional derivative of hh in the direction of the outward normal to ∂Ω\partial \Omega, the so-called normal derivative of hh). Interchange gg and hh, subtract the resulting formula from the first one, to obtain

∫Ω(g∇2h−h∇2g)dV=∫∂Ω(g∂h∂n−h∂g∂n)dA \int _{\Omega }\left(g \nabla ^{2} h-h \nabla ^{2} g\right) d V = \int _{\partial \Omega }\left(g \frac{\partial h}{\partial n}-h \frac{\partial g}{\partial n}\right) d A

These two formulas are usually called Green's identities.

(c)

Assume that hh is harmonic in EE; this means that ∇2h=0\nabla^{2} h = 0. Take g=1g = 1 and conclude that

∫∂Ω∂h∂ndA=0 \int _{\partial \Omega } \frac{\partial h}{\partial n} d A = 0

Take g=hg = h, and conclude that h=0h = 0 in Ω\Omega if h=0h = 0 on ∂Ω\partial \Omega.

(d)

Show that Green's identities are also valid in R2\mathbb {R}^{2}.

Problem 1.10.32

Fix δ,0<δ<1\delta , 0 < \delta < 1. Let DD be the set of all (θ,t)∈R2(\theta , t) \in \mathbb {R}^{2} such that 0≤θ≤π,−δ≤t≤δ0 \leq \theta \leq \pi ,-\delta \leq t \leq \delta. Let Φ\Phi be the 2 -surface in R3\mathbb {R}^{3}, with parameter domain DD, given by

x=(1−tsin⁡θ)cos⁡2θy=(1−tsin⁡θ)sin⁡2θz=tcos⁡θ \begin{aligned} & x = (1-t \sin \theta ) \cos 2 \theta \\ & y = (1-t \sin \theta ) \sin 2 \theta \\ & z = t \cos \theta \end{aligned}

where (x,y,z)=Φ(θ,t)(x, y, z) = \Phi (\theta , t). Note that Φ(π,t)=Φ(0,−t)\Phi (\pi , t) = \Phi (0,-t), and that Φ\Phi is one-to-one on the rest of DD.

The range M=Φ(D)M = \Phi (D) of Φ\Phi is known as a Möbius band. It is the simplest example of a nonorientable surface.

Prove the various assertions made in the following description: Put p1=(0,−δ),p2=(π,−δ),p3=(π,δ),p4=(0,δ),p5=p1\mathbf{p}_{1} = (0,-\delta ), \mathbf{p}_{2} = (\pi ,-\delta ), \mathbf{p}_{3} = (\pi , \delta ), \mathbf{p}_{4} = (0, \delta ), \mathbf{p}_{5} = \mathbf{p}_{1}. Put γi=[pi,pi+1]\gamma_{i} = \left[\mathbf{p}_{i}, \mathbf{p}_{i+1}\right], i=1,…,4i = 1, \ldots , 4, and put Γi=Φ∘γi\Gamma_{i} = \Phi \circ \gamma_{i}. Then

∂Φ=Γ1+Γ2+Γ3+Γ4 \partial \Phi = \Gamma _{1}+\Gamma _{2}+\Gamma _{3}+\Gamma _{4}

Put a=(1,0,−δ),b=(1,0,δ)\mathbf{a} = (1,0,-\delta ), \mathbf{b} = (1,0, \delta ). Then

Φ(p1)=Φ(p3)=a,Φ(p2)=Φ(p4)=b \Phi \left(\mathbf{p}_{1}\right) = \Phi \left(\mathbf{p}_{3}\right) = \mathbf{a}, \quad \Phi \left(\mathbf{p}_{2}\right) = \Phi \left(\mathbf{p}_{4}\right) = \mathbf{b}

and ∂Φ\partial \Phi can be described as follows. Γ1\Gamma_{1} spirals up from a to b\mathbf{b}; its projection into the (x,y)(x, y)-plane has winding number +1 around the origin. (See Exercise 23, Chap. 8). Γ2=[b,a]\Gamma_{2} = [b, a]. Γ3\Gamma_{3} spirals up from a to b\mathbf{b}; its projection into the (x,y)(x, y) plane has winding number -1 around the origin. Γ4=[b,a]\Gamma_{4} = [b, a]. Thus ∂Φ=Γ1+Γ3+2Γ2\partial \Phi = \Gamma_{1}+\Gamma_{3}+2 \Gamma_{2}. If we go from a\mathbf{a} to bb along Γ1\Gamma_{1} and continue along the "edge" of MM until we return to a, the curve traced out is

Γ=Γ1−Γ3, \Gamma = \Gamma _{1}-\Gamma _{3},

which may also be represented on the parameter interval [0,2π][0,2 \pi ] by the equations

x=(1+δsin⁡θ)cos⁡2θy=(1+δsin⁡θ)sin⁡2θz=−δcos⁡θ. \begin{aligned} & x = (1+\delta \sin \theta ) \cos 2 \theta \\ & y = (1+\delta \sin \theta ) \sin 2 \theta \\ & z = -\delta \cos \theta . \end{aligned}

It should be emphasized that Γ≠∂Φ\Gamma \neq \partial \Phi : Let η\eta be the 1 -form discussed in Exercises 21 and 22. Since dη=0d \eta = 0, Stokes' theorem shows that

∫∂Φη=0. \int _{\partial \Phi } \eta = 0.

But although Γ\Gamma is the "geometric" boundary of MM, we have

∫Γη=4π \int _{\Gamma } \eta = 4 \pi

In order to avoid this possible source of confusion, Stokes' formula (Theorem 10.50) is frequently stated only for orientable surfaces Φ\Phi.

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