University Mathematics — Year 2 · Bachelor Year 2

20Line Integrals and Multiple Integrals

This chapter extends integration from intervals to curves and to domains of the plane and of space. Line integrals integrate a differential form $P\,\dd x + Q\,\dd y$ along an oriented arc; double and triple integrals integrate functions over two- and three-dimensional domains. The two theories meet in the Green–Riemann theorem, the two-dimensional fundamental theorem of calculus, and the main computational tool throughout is the change of variables formula, whose distortion factor is the absolute Jacobian determinant.

20.1 Line integrals

Definition 20.1 (Differential form; line integral)

Let $U \subseteq \R^2$ be open. A differential form of degree $1$ and class $\mathcal{C}^0$ on $U$ is an expression $\omega = P\,\dd x + Q\,\dd y$ with $P, Q \colon U \to \R$ continuous — formally, a continuous map from $U$ into the dual of $\R^2$ , $\omega(M) = P(M)\,e_1^* + Q(M)\,e_2^*$ . For a $\mathcal{C}^1$ arc $\gamma \colon [a, b] \to U$ , $\gamma(t) = (x(t), y(t))$ , the line integral of $\omega$ along $\gamma$ is

\int_\gamma \omega = \int_a^b \Bigl(P(\gamma(t))\,x'(t) + Q(\gamma(t))\,y'(t)\Bigr)\,\dd t .

The definitions extend verbatim to $\R^3$ (forms $P\,\dd x + Q\,\dd y + R\,\dd z$ ) and to piecewise $\mathcal{C}^1$ arcs (sum over the pieces).

Proposition 20.2 (Invariance and orientation)

The line integral is unchanged under an increasing $\mathcal{C}^1$ change of parameter, and changes sign under a decreasing one. It therefore only depends on the oriented geometric arc.

Proof. If $\theta \colon [c, d] \to [a, b]$ is a change of parameter and $\tilde\gamma = \gamma \circ \theta$ , then by the chain rule and the one-variable change of variables $t = \theta(u)$ ,

\int_{\tilde\gamma}\omega = \int_c^d \bigl(P(\gamma(\theta(u)))\,x'(\theta(u)) + Q(\gamma(\theta(u)))\,y'(\theta(u))\bigr)\,\theta'(u)\,\dd u = \pm\int_a^b \bigl(Px' + Qy'\bigr)(t)\,\dd t ,

with sign $+$ if $\theta$ is increasing ( $\theta(c) = a$ ) and $-$ if decreasing (the bounds swap). ∎

Example 20.3 (Work of a force; circulation)

If $F = (P, Q)$ is a force field, $\int_\gamma P\dd x + Q\dd y = \int_a^b \langle F(\gamma(t)), \gamma'(t)\rangle\,\dd t$ is the work of $F$ along $\gamma$ . For $\omega = -y\,\dd x + x\,\dd y$ along the counterclockwise unit circle $\gamma(t) = (\cos t, \sin t)$ :

\int_\gamma \omega = \int_0^{2\pi}\bigl((-\sin t)(-\sin t) + \cos t\cos t\bigr)\,\dd t = 2\pi ,

twice the enclosed area — a first hint of Green–Riemann.

Example 20.4 (One integral, two parametrizations, one sign trap)

Compute $\int_\gamma x\,\dd y$ along the upper unit semicircle from $(1, 0)$ to $(-1, 0)$ . With $\gamma(t) = (\cos t, \sin t)$ , $t \in \intcc0\pi$ :

\int_0^\pi\cos t\cdot\cos t\,\dd t = \frac\pi2 .

With the graph parametrization $x \mapsto (x, \sqrt{1 - x^2})$ , $x$ from $1$ to $-1$ (note the direction!):

\int_1^{-1}x\cdot\frac{-x}{\sqrt{1 - x^2}}\,\dd x = \int_{-1}^{1}\frac{x^2}{\sqrt{1 - x^2}}\,\dd x = \frac\pi2

( $x = \sin u$ reduces it to a Wallis integral). Same value, as Proposition 20.2 guarantees — but only because both runs go from $(1,0)$ to $(-1,0)$ ; reversing the travel flips the sign. Closing the path along the $x$ -axis (where $\dd y = 0$ ) adds nothing, and the total $\frac\pi2$ is the area of the half-disk: the first instance of the boundary-area formulas of Green–Riemann below.

Definition 20.5 (Exact and closed forms)

The form $\omega = P\,\dd x + Q\,\dd y$ of class $\mathcal{C}^0$ is exact on $U$ if there is $f \in \mathcal{C}^1(U)$ (a potential) with $\omega = \dd f$ , i.e. $P = f_x$ and $Q = f_y$ . A $\mathcal{C}^1$ form is closed if $P_y = Q_x$ on $U$ .

Theorem 20.6 (Fundamental theorem for line integrals)

If $\omega = \dd f$ is exact and $\gamma$ is a piecewise $\mathcal{C}^1$ arc in $U$ from $A$ to $B$ , then

\int_\gamma \omega = f(B) - f(A) .

In particular the integral of an exact form along any closed arc is zero, and every exact $\mathcal{C}^1$ form is closed.

Proof. $\frac{\dd}{\dd t}f(\gamma(t)) = f_x(\gamma(t))x'(t) + f_y(\gamma(t))y'(t)$ by the chain rule (Chapter 15), so the integrand in Definition 20.1 is the derivative of $t \mapsto f(\gamma(t))$ , and the fundamental theorem of calculus gives the result on each piece; the intermediate values telescope. Closedness of exact $\mathcal{C}^1$ forms is Schwarz’s theorem: $P_y = f_{xy} = f_{yx} = Q_x$ . ∎

Example 20.7 (Reconstructing a potential)

Let $\omega = y\,\eu^{xy}\,\dd x + (x\,\eu^{xy} + 2y)\,\dd y$ on $\R^2$ . It is closed: both cross derivatives equal $\eu^{xy}(1 + xy)$ . To find a potential, integrate $P$ in $x$ at fixed $y$ :

f(x, y) = \int y\,\eu^{xy}\,\dd x = \eu^{xy} + c(y),

then adjust $c$ by matching $f_y$ : $x\,\eu^{xy} + c'(y) = x\,\eu^{xy} + 2y$ gives $c(y) = y^2$ . So $f(x,y) = \eu^{xy} + y^2$ , and for any piecewise $\mathcal C^1$ arc from $(0,0)$ to $(1,1)$ ,

\int_\gamma\omega = f(1,1) - f(0,0) = (\eu + 1) - 1 = \eu ,

independently of the path — the two-step recipe (integrate in $x$ , correct in $y$ ) is the practical converse of Theorem 20.6 on domains where closed forms are exact.

Example 20.8 (Closed does not imply exact)

On $U = \R^2 \setminus \{0\}$ , the angle form

\omega = \frac{-y\,\dd x + x\,\dd y}{x^2 + y^2}

is closed (direct computation: both $P_y$ and $Q_x$ equal $\frac{y^2 - x^2}{(x^2+y^2)^2}$ ), but its integral along the unit circle is $2\pi \neq 0$ (same computation as Example 20.3, divided by $1$ ): $\omega$ is not exact on $U$ . Locally, $\omega = \dd\theta$ for a determination $\theta$ of the polar angle; the failure is global — the angle cannot be defined continuously around the puncture. On domains without holes the pathology disappears: on a star-shaped open set, every closed $\mathcal{C}^1$ form is exact (Poincaré’s lemma, Exercise 20.8).

20.2 Double integrals

We take for granted the theory of the Riemann integral in one variable (Year 1 volume, and Chapter 9) and sketch its two-variable version. A function $f$ continuous on a rectangle $R = [a, b] \times [c, d]$ has a double integral $\iint_R f$ , defined by Riemann sums over grids exactly as in one variable, and computed by iteration:

Theorem 20.9 (Fubini on a rectangle)

For $f$ continuous on $R = [a,b] \times [c,d]$ ,

\iint_R f = \int_a^b \Bigl(\int_c^d f(x, y)\,\dd y\Bigr)\dd x = \int_c^d \Bigl(\int_a^b f(x, y)\,\dd x\Bigr)\dd y .

Proof. Set $F(x) = \int_c^d f(x, y)\,\dd y$ . Uniform continuity of $f$ on the compact $R$ makes $F$ continuous (dominated estimate: $\abs{F(x) - F(x')} \leq (d - c)\sup_y\abs{f(x,y) - f(x',y)}$ ). Now subdivide $[a,b]$ and $[c,d]$ into $n$ equal parts, giving a grid of cells $R_{ij}$ of area $\Delta x\,\Delta y$ . On each cell, $\inf_{R_{ij}} f \cdot \Delta x \Delta y \leq \int_{x_{i-1}}^{x_i}\int_{y_{j-1}}^{y_j} f(x,y)\,\dd y\,\dd x \leq \sup_{R_{ij}} f \cdot \Delta x \Delta y$ by monotonicity of the one-variable integral (applied twice). Summing over cells, the iterated integral $\int_a^b F$ is squeezed between the lower and upper Riemann sums of the grid; by uniform continuity both sums converge to the common value defining $\iint_R f$ as $n \to \infty$ . The same argument applies with the roles of $x$ and $y$ exchanged, so both iterated integrals equal $\iint_R f$ . ∎

Remark 20.10

Continuity on a compact rectangle is doing real work in Fubini’s proof: it provides the uniform continuity that squeezes the Riemann sums. For wilder integrands the statement genuinely fails — there are functions whose two iterated integrals exist and differ. The honest general theorem, with integrability as the only hypothesis, is Fubini’s theorem for the Lebesgue integral, proved in the Year 3 volume; everything in this chapter stays within the continuous setting where the elementary proof above is complete.

Definition 20.11 (Elementary domains)

A domain $D \subseteq \R^2$ is $y$ -elementary if

D = \{(x, y) : a \leq x \leq b,\ \varphi_1(x) \leq y \leq \varphi_2(x)\}

with $\varphi_1 \leq \varphi_2$ continuous on $[a,b]$ ( $x$ -elementary: symmetrically). For $f$ continuous on a $y$ -elementary $D$ ,

\iint_D f = \int_a^b\Bigl( \int_{\varphi_1(x)}^{\varphi_2(x)} f(x,y)\,\dd y\Bigr)\dd x ,

and one checks (by extending $f$ by an approximation argument, or by subdividing) that when $D$ is elementary in both directions the two iterated integrals agree. Domains cut into finitely many elementary pieces are handled by additivity, and the area of $D$ is $\operatorname {Area}(D) = \iint_D 1$ .

Example 20.12

On the triangle $D = \{0 \leq x \leq 1,\ 0 \leq y \leq x\}$ :

\iint_D xy \,\dd x\,\dd y = \int_0^1 x\Bigl(\int_0^x y\,\dd y\Bigr)\dd x = \int_0^1 x\cdot\frac{x^2}{2}\,\dd x = \frac18 .

Swapping the order ( $x$ from $y$ to $1$ ): $\int_0^1 y\bigl(\int_y^1 x\,\dd x\bigr)\dd y = \int_0^1 y\,\frac{1 - y^2}{2}\,\dd y = \frac18$ — same value, different computation: choosing the order of integration well is half the craft.

Example 20.13 (When only one order works)

Compute $I = \displaystyle\int_0^1\!\!\int_x^1 \eu^{y^2}\,\dd y\,\dd x$ . As written, the inner integral $\int\eu^{y^2}\dd y$ has no elementary antiderivative: the computation is stuck. But the domain is the triangle $0 \leq x \leq y \leq 1$ , which is elementary in both directions; swapping the order,

I = \int_0^1\!\!\int_0^y \eu^{y^2}\,\dd x\,\dd y = \int_0^1 y\,\eu^{y^2}\,\dd y = \Bigl[\tfrac12\eu^{y^2}\Bigr]_0^1 = \frac{\eu - 1}{2} .

The inner variable $x$ appeared nowhere in the integrand, so integrating it first produced exactly the factor $y$ that makes the outer integral immediate. Moral: Fubini is not just a license to iterate — it is a license to choose, and the right order can turn an impossible integral into a one-liner. Always sketch the domain and read off both descriptions before starting.

Theorem 20.14 (Change of variables)

Let $\Phi \colon U' \to U$ be a $\mathcal{C}^1$ diffeomorphism between open sets of $\R^2$ , let $K \subseteq U$ be a compact domain cut into elementary pieces with $K' = \Phi^{-1}(K)$ , and let $f$ be continuous on $K$ . Then

\iint_K f(x, y)\,\dd x\,\dd y = \iint_{K'} f\bigl(\Phi(u, v)\bigr)\, \abs{\det J_\Phi(u, v)}\,\dd u\,\dd v .

Proof. Admitted at this level. ∎

Remark 20.15

The complete proof — approximating $\Phi$ by its differential on a fine grid and controlling the boundary cells — is long though not deep; it is done in full in the measure theory of Year 3, as a consequence of the Lebesgue theory. The heuristic is the picture already used for surface area: a small square of side $\dd u$ at $(u, v)$ is mapped, to first order, onto the parallelogram spanned by $\Phi_u\,\dd u$ and $\Phi_v\,\dd v$ , whose area is $\abs{\det J_\Phi}\,\dd u\,\dd v$ (Lemma 19.18).

Remark 20.16 (Method: choosing the change of variables)

Three reflexes cover most cases. Symmetry of the integrand: $x^2 + y^2$ calls for polar, a product structure calls for keeping Cartesian axes. Shape of the boundary: boundaries $u(x,y) = c_1$ , $v(x,y) = c_2$ beg for the coordinates $(u, v)$ themselves, as in the hyperbolic region example below — the domain becomes a rectangle, which is the whole victory. Linear structure: expressions in $x + y$ and $x - y$ invite the $45$ -degree rotation or a shear (Example 20.18). In all cases, three boxes to tick before integrating: the map is a bijection of the new domain onto the old one; its Jacobian is computed in the direction actually used (invert at the end if easier); and the Jacobian enters with its absolute value.

Example 20.17 (Polar coordinates)

$\Phi(\rho, \alpha) = (\rho\cos\alpha,\ \rho\sin\alpha)$ has

J_\Phi = \begin{pmatrix} \cos\alpha & -\rho\sin\alpha\\ \sin\alpha & \rho\cos\alpha \end{pmatrix}, \qquad \det J_\Phi = \rho ,

so $\dd x\,\dd y = \rho\,\dd\rho\,\dd\alpha$ . For the disk $D_R$ of radius $R$ :

\iint_{D_R} e^{-(x^2 + y^2)}\,\dd x\,\dd y = \int_0^{2\pi}\!\!\int_0^R e^{-\rho^2}\rho\,\dd\rho\,\dd\alpha = \pi\bigl(1 - e^{-R^2}\bigr) \xrightarrow[R\to\infty]{} \pi .

Comparing with the square $[-R, R]^2$ (which is squeezed between the disks $D_R$ and $D_{R\sqrt2}$ , all integrands positive) gives $\bigl(\int_{-\infty}^\infty e^{-x^2}\dd x\bigr)^2 = \pi$ :

\boxed{\ \int_{-\infty}^{+\infty} e^{-x^2}\,\dd x = \sqrt{\pi}\ }

— the Gauss integral again, now by its most famous proof (compare the one-variable derivation in Chapter 9).

Example 20.18 (Affine changes of variables)

For an affine map $\Phi(u, v) = M(u, v)^{\mathsf T} + C$ with $M$ invertible, the Jacobian is the constant matrix $M$ : areas are multiplied by the constant factor $\abs{\det M}$ — the promise made in Chapter 17 is now a theorem. Two immediate uses. The ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} \leq 1$ is the image of the unit disk under $(u, v) \mapsto (au, bv)$ , so its area is $ab \cdot \pi$ — no computation. And for the integral of $f(x + y)$ over the square $K = \intcc01^2$ , the shear $\Phi(u, v) = (u - v, v)$ (determinant $1$ ) turns it into an integral of $f(u)$ over a parallelogram, which Fubini slices at constant $u$ : with $f = \exp$ ,

\iint_K \eu^{x+y}\,\dd x\,\dd y = \Bigl(\int_0^1 \eu^x\,\dd x\Bigr)^2 = (\eu - 1)^2,

as the product structure confirms. Choosing coordinates adapted to the integrand — not to the domain — is the other half of the craft.

Example 20.19 (Coordinates adapted to a curvilinear domain)

Let $D$ be the region of the first quadrant bounded by the hyperbolas $xy = 1$ and $xy = 3$ and the lines $y = x$ and $y = 3x$ . In the coordinates $u = xy$ , $v = y/x$ the domain becomes the square $\intcc13 \times \intcc13$ ; inverting,

x = \sqrt{u/v}, \qquad y = \sqrt{uv}, \qquad \det J = x_uy_v - x_vy_u = \frac{1}{2v}

(a two-line computation with $x = u^{1/2}v^{-1/2}$ , $y = u^{1/2}v^{1/2}$ ). Hence

\operatorname{Area}(D) = \int_1^3\!\!\int_1^3\frac{\dd u\,\dd v}{2v} = 2\cdot\frac{\ln 3}{2} = \ln 3 \approx 1.10 .

Trying to slice $D$ in Cartesian coordinates means cutting it into three pieces with hyperbolic and linear boundaries — feasible, joyless, and error-prone. The moral repeats Example 20.18 at full strength: read the boundary equations, and let them choose the coordinates; the Jacobian then converts the curvilinear mesh’s cell area, exactly as $\rho$ did for polar coordinates.

Example 20.20 (Average values)

The mean value of $f$ over a domain $D$ is $\frac1{\operatorname{Area}(D)}\iint_Df$ . Sample: the average distance from the center for a uniformly chosen point of the disk of radius $R$ is

\frac{1}{\pi R^2}\int_0^{2\pi}\!\!\int_0^R \rho\cdot\rho\,\dd\rho\,\dd\alpha = \frac{2\pi R^3/3}{\pi R^2} = \frac{2R}3 ,

not $R/2$ : uniform area puts more mass at large radii (the annulus at radius $\rho$ has weight proportional to $\rho$ ), so the average sits beyond the halfway mark. Getting this factor right is exactly the polar Jacobian at work, and the same weighting explains the centroid $\bar z = 3R/8$ of the half-ball computed later in the chapter rather than $R/2$ .

20.3 The Green–Riemann theorem

Theorem 20.21 (Green–Riemann)

Let $K \subseteq \R^2$ be a compact domain which is elementary in both directions (or a finite union of such glued along segments), with boundary $\partial K$ a piecewise $\mathcal{C}^1$ closed curve oriented counterclockwise (the domain stays on the left). For $P, Q$ of class $\mathcal{C}^1$ on a neighbourhood of $K$ :

\oint_{\partial K} P\,\dd x + Q\,\dd y = \iint_K \Bigl(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\Bigr)\,\dd x\,\dd y .

Proof. First, both sides are additive under cutting $K$ along a segment into two pieces $K_1, K_2$ : the double integrals add by additivity of $\iint$ ; as for the boundary integrals, the counterclockwise boundaries of $K_1$ and $K_2$ each traverse the interior cut once, in opposite directions, so in the sum

\oint_{\partial K_1} + \oint_{\partial K_2} = \oint_{\partial K} + (\text{cut, both ways}) = \oint_{\partial K},

the two passes along the cut cancel (Proposition 20.2) and only the outer boundary survives. Iterating finitely many cuts, it suffices to treat an elementary domain. We prove $\oint P\,\dd x = -\iint_K P_y$ on a $y$ -elementary domain $D = \{a \leq x \leq b,\ \varphi_1(x) \leq y \leq \varphi_2(x)\}$ ; the identity $\oint Q\,\dd y = \iint_K Q_x$ is symmetric ( $x$ -elementary), and the theorem is their sum.

Compute the double integral by Fubini and the one-variable fundamental theorem:

\iint_D \frac{\partial P}{\partial y}\,\dd x\,\dd y = \int_a^b \bigl(P(x, \varphi_2(x)) - P(x, \varphi_1(x))\bigr) \,\dd x .

Now the boundary of $D$ , counterclockwise, consists of: the lower graph $y = \varphi_1(x)$ traversed left to right, the right vertical segment $x = b$ (upward), the upper graph $y = \varphi_2(x)$ traversed right to left, the left vertical segment $x = a$ (downward). Along the vertical segments $x$ is constant, so they contribute $0$ to $\oint P\,\dd x$ ; the graphs, parametrized by $x$ , give

\oint_{\partial D} P\,\dd x = \int_a^b P(x, \varphi_1(x))\,\dd x - \int_a^b P(x, \varphi_2(x))\,\dd x = -\iint_D \frac{\partial P}{\partial y}\,\dd x\,\dd y . \qedhere

∎

Corollary 20.22 (Area by the boundary)

Under the hypotheses of Theorem 20.21,

\operatorname{Area}(K) = \oint_{\partial K} x\,\dd y = -\oint_{\partial K} y\,\dd x = \frac12\oint_{\partial K} x\,\dd y - y\,\dd x .

Proof. Apply Green–Riemann to $(P, Q) = (0, x)$ , $(-y, 0)$ and $\frac12(-y, x)$ : each time $Q_x - P_y = 1$ . ∎

Remark 20.23 (Choosing among the three area formulas)

The three boundary formulas are equal, not interchangeable in practice. Use $\oint x\,\dd y$ when the parametrization makes $\dd y$ simple (graphs over the $y$ -axis), $-\oint y\,\dd x$ symmetrically, and the symmetric half-sum when the parametrization treats $x$ and $y$ evenly — for the ellipse it produced a constant integrand, with no trigonometric linearization at all. On polygonal boundaries the half-sum becomes the shoelace formula of Exercise 20.12, the surveyors’ algorithm. And when the boundary is traversed clockwise by the given parametrization, all three formulas return minus the area: a negative result is not a computational error but an orientation report — flip the sign, or the parametrization.

Example 20.24 (Area of the ellipse)

For $x = a\cos t$ , $y = b\sin t$ , $t \in [0, 2\pi]$ :

\operatorname{Area} = \frac12\int_0^{2\pi}\bigl(a\cos t \cdot b\cos t - b\sin t\cdot(-a\sin t)\bigr)\,\dd t = \frac{ab}{2}\int_0^{2\pi}\dd t = \pi ab .

Example 20.25 (Green–Riemann as a cross-check)

Take $P = -y^3$ , $Q = x^3$ on the closed unit disk $D$ . Boundary side, with $\gamma(t) = (\cos t, \sin t)$ :

\oint_{\partial D}P\,\dd x + Q\,\dd y = \int_0^{2\pi}\bigl(\sin^4 t + \cos^4 t\bigr)\dd t = 2\pi\cdot\Bigl(\frac38 + \frac38\Bigr) = \frac{3\pi}2 ,

by linearization ( $\sin^4 + \cos^4 = \tfrac34 + \tfrac14\cos4t$ ). Interior side:

\iint_D(Q_x - P_y)\,\dd x\,\dd y = \iint_D 3(x^2 + y^2)\,\dd x\,\dd y = 3\int_0^{2\pi}\!\!\int_0^1\rho^3\,\dd\rho\,\dd\alpha = \frac{3\pi}2 .

Same number, two very different computations — and that is the practical use: whichever side of Green’s identity is easier becomes the computation, the other a verification. For circulations of polynomial fields around closed curves, the double integral is almost always the easy side.

Remark 20.26

Green–Riemann explains Example 20.8: for a closed form ( $Q_x = P_y$ ), the integral around the boundary of any domain contained in $U$ vanishes. The angle form fails to be exact only because the puncture at the origin prevents the disk bounded by the unit circle from lying inside $U$ — line integrals of closed forms detect the holes of the domain. (Pushed further, this observation becomes de Rham cohomology.)

20.4 Triple integrals

The theory extends to three variables with no new idea: Fubini reduces $\iiint$ to three one-variable integrals (either by slicing: $\iiint_K f = \int\bigl(\iint_{K_z} f\bigr)\dd z$ over the horizontal slices $K_z$ , or by stacking: integrating in $z$ first along vertical sticks), and the change of variables formula holds with the $3 \times 3$ Jacobian.

Example 20.27 (Cylindrical and spherical coordinates)

Cylindrical $(x, y, z) = (\rho\cos\alpha, \rho\sin\alpha, z)$ : $\dd x\,\dd y\,\dd z = \rho\,\dd\rho\,\dd\alpha\,\dd z$ . Spherical $(x, y, z) = (r\cos\theta\cos\varphi,\ r\sin\theta\cos\varphi,\ r\sin\varphi)$ ( $\theta$ longitude, $\varphi \in [-\frac\pi2, \frac\pi2]$ latitude): expanding the $3 \times 3$ determinant along the last row,

\det J = r^2\cos\varphi , \qquad \dd x\,\dd y\,\dd z = r^2\cos\varphi\;\dd r\,\dd\theta\,\dd\varphi .

Volume of the ball of radius $R$ :

V = \int_0^R\!\!\int_0^{2\pi}\!\!\int_{-\pi/2}^{\pi/2} r^2\cos\varphi\;\dd\varphi\,\dd\theta\,\dd r = \frac{R^3}{3}\cdot 2\pi \cdot 2 = \boxed{\frac43\pi R^3} ,

discharging at last the formula admitted in the volume chapters of the earlier books.

Example 20.28 (The tetrahedron, twice)

The volume of $T = \{x, y, z \geq 0,\ x + y + z \leq 1\}$ , by stacking: for fixed $(x, y)$ in the triangle $x + y \leq 1$ , $z$ runs over $\intcc0{1 - x - y}$ , so

V = \int_0^1\!\!\int_0^{1-x}(1 - x - y)\,\dd y\,\dd x = \int_0^1\frac{(1 - x)^2}{2}\,\dd x = \frac16 .

By slicing: the section at height $z$ is the triangle $\{x, y \geq 0,\ x + y \leq 1 - z\}$ , of area $\frac{(1-z)^2}2$ , and $V = \int_0^1\frac{(1-z)^2}2\,\dd z = \frac16$ again — the two computations are the same integrals in a different order, which is all Fubini claims. The value $\frac16 = \frac13\cdot\frac12\cdot1$ is the cone formula (Example 20.30) with triangular base, and the $n$ -dimensional version $1/n!$ is proved by exactly this slicing in the weekend problem.

Example 20.29 (Centroid of a half-ball)

For the upper half-ball $H$ of radius $R$ ( $z \geq 0$ ), the centroid height is $\bar z = \frac1{V}\iiint_H z$ , with $V = \frac23\pi R^3$ . In spherical coordinates ( $z = r\sin\varphi$ , $\varphi \in \intcc0{\pi/2}$ ):

\iiint_H z = \int_0^R r^3\,\dd r\int_0^{2\pi}\dd\theta \int_0^{\pi/2}\sin\varphi\cos\varphi\,\dd\varphi = \frac{R^4}4\cdot2\pi\cdot\frac12 = \frac{\pi R^4}4 ,

\bar z = \frac{\pi R^4/4}{2\pi R^3/3} = \frac{3R}8 :

the balance point of a solid hemisphere sits three-eighths of the radius above the flat face — below the half-height $R/2$ , as it must, since the solid is fatter near the base. Every centroid computation has this shape: one moment integral, one volume, one ratio, and a plausibility check against the geometry.

Example 20.30 (Volume by slicing: the cone)

A cone of base area $A$ and height $h$ (apex up, base at $z = 0$ ): the slice at height $z$ is the base scaled by the factor $(1 - z/h)$ , of area $A(1 - z/h)^2$ . Hence

V = \int_0^h A\Bigl(1 - \frac zh\Bigr)^2\dd z = \frac{Ah}{3} :

the one-third of the school formulas, valid for any base shape — slicing turns it into the integral of a square.

Example 20.31 (Integrability thresholds in the plane)

For which $\alpha > 0$ does $\iint_{D}\rho^{-\alpha}\,\dd x\,\dd y$ converge on the punctured unit disk $D$ (limit over annuli $\varepsilon \leq \rho \leq 1$ )? In polar coordinates,

\int_0^{2\pi}\!\!\int_\varepsilon^1\rho^{-\alpha}\, \rho\,\dd\rho\,\dd\alpha = 2\pi\int_\varepsilon^1\rho^{1-\alpha}\,\dd\rho ,

which converges as $\varepsilon \to 0$ iff $1 - \alpha > -1$ , i.e. $\alpha < 2$ : in dimension $2$ the critical singularity exponent is the dimension itself, the extra $\rho$ from the Jacobian softening the singularity by one power. (Likewise $\alpha < 3$ for a point singularity in space, via $r^2$ .) Radial bookkeeping of this kind is how integrability is decided at a glance in the Lebesgue framework of Year 3 — and it is the reason $\iiint 1/r$ converged effortlessly in Exercise 20.7.

Remark 20.32 (Common pitfalls)

(i) Orientation: a line integral changes sign with the direction of travel, and Green–Riemann requires the boundary counterclockwise (domain on the left); for a domain with a hole, the inner boundary is traversed clockwise. (ii) The Jacobian enters with an absolute value: change of variables never produces negative area, and forgetting $\abs{\det}$ typically flips signs exactly when the map reverses orientation. (iii) The polar factor $\rho$ : $\dd x\,\dd y = \rho\,\dd\rho\,\dd\alpha$ , not $\dd\rho\,\dd\alpha$ — the most common error in the whole chapter; dimensional analysis catches it, since $\dd\rho\,\dd\alpha$ has the dimension of a length, not an area. (iv) Improper double integrals: limits over growing disks and growing squares agree here because the integrands are positive (squeezing); for sign-changing integrands the limit may depend on the exhaustion, and no claim is made without absolute convergence. (v) Domains versus integrands: a product integrand on a non-product domain does not factor the integral — factorization needs both, as in Example 20.17’s square.

Remark 20.33 (Perspectives within this volume)

The Gauss integral computed here is quietly everywhere in the probability chapters: the constant $\sqrt\pi$ inside Stirling’s formula (Theorem 6.13) is this chapter’s integral, and through Stirling it fixes the $1/\sqrt{\pi n}$ asymptotics of the random walk’s return probabilities in Chapter 21. The Wallis integrals of the weekend problem reappear there too, driving the same central-binomial estimates. In the other direction, the area and volume elements of this chapter complete the geometry of Chapter 19, and Green’s formula recomputes the envelope areas of Chapter 18 (the astroid, in Exercise 20.5). One chapter, three services: measure for geometry, constants for probability, and the change-of-variables discipline used by both.

20.5 Exercises

Exercise 20.1 ★

Compute $\int_\gamma y^2\,\dd x + x\,\dd y$ along: (a) the segment from $(0,0)$ to $(1,1)$ ; (b) the parabola arc $y = x^2$ from $(0,0)$ to $(1,1)$ . Is the form exact?

Solution

Solution of Exercise 20.1.

(a) Segment $\gamma(t) = (t, t)$ , $t \in [0,1]$ :

\int_\gamma y^2\dd x + x\dd y = \int_0^1 (t^2 + t)\,\dd t = \frac13 + \frac12 = \frac56 .

(b) Parabola $\gamma(t) = (t, t^2)$ :

\int_0^1 \bigl(t^4\cdot 1 + t\cdot 2t\bigr)\dd t = \frac15 + \frac23 = \frac{13}{15} .

The two values differ, so the integral is path-dependent: the form is not exact — consistently, $P_y = 2y \neq 1 = Q_x$ , so it is not even closed.

Exercise 20.2 ★

Show that $\omega = (2xy + y^3)\,\dd x + (x^2 + 3xy^2 + 1)\,\dd y$ is closed on $\R^2$ , find a potential, and compute $\int_\gamma\omega$ along any arc from $(0, 0)$ to $(1, 2)$ .

Solution

Solution of Exercise 20.2.

$P = 2xy + y^3$ , $Q = x^2 + 3xy^2 + 1$ : $P_y = 2x + 3y^2 = Q_x$ , closed on $\R^2$ . Search for $f$ with $f_x = P$ : $f = x^2y + xy^3 + g(y)$ ; then $f_y = x^2 + 3xy^2 + g'(y) = Q$ forces $g'(y) = 1$ , say $g(y) = y$ . So

f(x, y) = x^2y + xy^3 + y

is a potential ( $\R^2$ is star-shaped, so a potential had to exist by Poincaré’s lemma — but exhibiting it is quicker). By Theorem 20.6, for any arc from $(0,0)$ to $(1,2)$ :

\int_\gamma\omega = f(1, 2) - f(0, 0) = 2 + 8 + 2 = 12 .

Exercise 20.3 ★

Compute $\iint_D (x + y)\,\dd x\,\dd y$ where $D$ is the domain bounded by $y = x^2$ and $y = x$ ( $0 \leq x \leq 1$ ), in both orders of integration.

Solution

Solution of Exercise 20.3.

The domain is $\{0 \leq x \leq 1,\ x^2 \leq y \leq x\}$ . $y$ -first:

\int_0^1\!\int_{x^2}^{x}(x + y)\,\dd y\,\dd x = \int_0^1\Bigl(x(x - x^2) + \frac{x^2 - x^4}{2}\Bigr)\dd x = \int_0^1\Bigl(\frac{3x^2}{2} - x^3 - \frac{x^4}{2}\Bigr)\dd x = \frac12 - \frac14 - \frac1{10} = \frac{3}{20} .

$x$ -first: the slice at height $y \in [0, 1]$ is $y \leq x \leq \sqrt y$ , so

\int_0^1\!\int_{y}^{\sqrt y}(x + y)\,\dd x\,\dd y = \int_0^1\Bigl(\frac{y - y^2}{2} + y(\sqrt y - y)\Bigr)\dd y = \frac14 - \frac16 + \frac25 - \frac13 = \frac{3}{20} .

Exercise 20.4 ★★

Using polar coordinates, compute $\iint_D \frac{\dd x\,\dd y}{(1 + x^2 + y^2)^2}$ over the whole plane (as a limit over disks), and $\iint_{D'} xy\,\dd x\,\dd y$ over the quarter disk $D' = \{x, y \geq 0,\ x^2 + y^2 \leq 1\}$ .

Solution

Solution of Exercise 20.4.

First integral. On the disk $D_R$ , in polar coordinates:

\iint_{D_R}\frac{\dd x\,\dd y}{(1 + x^2 + y^2)^2} = \int_0^{2\pi}\!\!\int_0^R \frac{\rho\,\dd\rho\,\dd\alpha}{(1 + \rho^2)^2} = 2\pi\Bigl[-\frac{1}{2(1 + \rho^2)}\Bigr]_0^R = \pi\Bigl(1 - \frac{1}{1 + R^2}\Bigr) \xrightarrow[R \to \infty]{} \pi .

Second integral. The quarter disk is $0 \leq \alpha \leq \frac\pi2$ , $0 \leq \rho \leq 1$ , and $xy = \rho^2\cos\alpha\sin\alpha$ :

\iint_{D'}xy\,\dd x\,\dd y = \int_0^{\pi/2}\!\!\cos\alpha\sin\alpha\,\dd\alpha \int_0^1 \rho^3\,\dd\rho = \frac12\cdot\frac14 = \frac18 .

Exercise 20.5 ★★

Compute the area enclosed by the astroid $x = \cos^3 t$ , $y = \sin^3 t$ , $t \in [0, 2\pi]$ , using Corollary 20.22. (Linearize $\sin^2 t\cos^2 t$ .)

Solution

Solution of Exercise 20.5.

By Corollary 20.22 with $x = \cos^3 t$ , $y = \sin^3 t$ : $x' = -3\cos^2 t\sin t$ , $y' = 3\sin^2 t\cos t$ , so

xy' - yx' = 3\cos^4 t\sin^2 t + 3\sin^4 t\cos^2 t = 3\sin^2 t\cos^2 t = \frac{3}{4}\sin^2 2t = \frac{3}{8}(1 - \cos 4t) .

Hence

A = \frac12\int_0^{2\pi}\frac38(1 - \cos 4t)\,\dd t = \frac{3}{16}\cdot 2\pi = \frac{3\pi}{8} .

(The astroid fits in the unit disk of area $\pi$ ; three eighths of $\pi$ is plausible for its four-cusped star shape.)

Exercise 20.6 ★★

Compute the volume of the solid bounded below by the paraboloid $z = x^2 + y^2$ and above by the plane $z = 1$ , by both methods: stacking (integrate $1 - x^2 - y^2$ over the unit disk, polar coordinates) and slicing (horizontal slices are disks of radius $\sqrt z$ ).

Solution

Solution of Exercise 20.6.

Stacking: above each $(x, y)$ of the unit disk $D$ , $z$ runs from $x^2 + y^2$ to $1$ :

V = \iint_D \bigl(1 - x^2 - y^2\bigr)\dd x\,\dd y = \int_0^{2\pi}\!\!\int_0^1 (1 - \rho^2)\rho\,\dd\rho\,\dd\alpha = 2\pi\Bigl(\frac12 - \frac14\Bigr) = \frac\pi2 .

Slicing: the slice at height $z \in [0, 1]$ is the disk $x^2 + y^2 \leq z$ , of area $\pi z$ :

V = \int_0^1 \pi z\,\dd z = \frac\pi2 .

Exercise 20.7 ★★

(Gravitational attraction of a ball — Newton’s theorem, special case) Show that the volume of the spherical shell $a \leq r \leq b$ is $\frac43\pi(b^3 - a^3)$ and compute $\iiint_{B} \frac{\dd x\,\dd y\,\dd z}{r}$ over the ball $B$ of radius $R$ ( $r$ the distance to the origin). (Spherical coordinates.)

Solution

Solution of Exercise 20.7.

In spherical coordinates the volume element is $r^2\cos\varphi\,\dd r\,\dd\theta\,\dd\varphi$ (Example 20.27), and the angular part integrates to $4\pi$ ( $2\pi$ from $\theta$ , $\int_{-\pi/2}^{\pi/2}\cos = 2$ ). The shell volume is

\int_a^b 4\pi r^2\,\dd r = \frac43\pi\bigl(b^3 - a^3\bigr).

For the second integral, the integrand $1/r$ depends only on $r$ :

\iiint_B \frac{\dd x\,\dd y\,\dd z}{r} = \int_0^R 4\pi r^2\cdot\frac1r\,\dd r = 4\pi\,\frac{R^2}{2} = 2\pi R^2 .

(The integrand blows up at the origin, but harmlessly: $r^2/r = r$ is continuous — the integral over the shells $\varepsilon \leq r \leq R$ converges as $\varepsilon \to 0$ , which is the precise sense of the statement. This kind of computation is the first step toward Newton’s theorem that a homogeneous ball attracts like a point mass at its center.)

Exercise 20.8 ★★★

(Poincaré’s lemma, star-shaped case) Let $U$ be star-shaped with respect to $0$ (i.e. $M \in U \Rightarrow [0, M] \subseteq U$ ) and $\omega = P\dd x + Q\dd y$ a closed $\mathcal{C}^1$ form on $U$ . Define

f(x, y) = \int_0^1 \bigl(x\,P(tx, ty) + y\,Q(tx, ty)\bigr)\dd t .

Using differentiation under the integral sign (Chapter 9) and $P_y = Q_x$ , show that $f_x = P$ and $f_y = Q$ : every closed form on a star-shaped open set is exact.

Solution

Solution of Exercise 20.8.

The integrand $g(t; x, y) = xP(tx, ty) + yQ(tx, ty)$ is $\mathcal{C}^1$ in $(x, y)$ , continuous in $t$ , with partial derivatives continuous on $[0,1] \times U$ ; differentiation under the integral sign (Chapter 9, applied on the compact $t$ -interval $[0,1]$ , domination being automatic there) gives

f_x(x, y) = \int_0^1 \bigl(P(tx, ty) + tx\,P_x(tx, ty) + ty\,Q_x(tx, ty)\bigr)\dd t .

Using closedness $Q_x = P_y$ :

tx\,P_x(tx, ty) + ty\,P_y(tx, ty) = t\,\frac{\dd}{\dd t}\bigl[P(tx, ty)\bigr] ,

so the integrand is $P(tx, ty) + t\frac{\dd}{\dd t}P(tx, ty) = \frac{\dd}{\dd t}\bigl[t\,P(tx, ty)\bigr]$ and

f_x(x, y) = \Bigl[t\,P(tx, ty)\Bigr]_0^1 = P(x, y) .

Symmetrically $f_y = Q$ (same computation with $P_y = Q_x$ used the other way). Note where the hypothesis enters: $f$ is defined by integrating along the segment $[0, M]$ , which lies in $U$ precisely because $U$ is star-shaped.

Exercise 20.9 ★★★

(Dirichlet integral by double integration) Justify and exploit

\int_0^\infty\!\!\int_0^\infty e^{-xy}\sin x\;\dd y\,\dd x \quad\text{vs}\quad \int_0^\infty\!\!\int_0^\infty e^{-xy}\sin x\;\dd x\,\dd y

on $[0, A] \times [0, \infty)$ : show $\int_0^A \frac{\sin x}{x}\dd x = \frac\pi2 - \int_0^\infty e^{-Ay}\frac{y\sin A + \cos A}{1 + y^2}\dd y$ and recover $\int_0^\infty \frac{\sin x}{x}\,\dd x = \frac\pi2$ , comparing with the parameter-integral proof of Chapter 9.

Solution

Solution of Exercise 20.9.

On the strip $[0, A] \times [0, \infty)$ the function $(x, y) \mapsto e^{-xy}\sin x$ is not absolutely integrable up to $y = \infty$ uniformly in a naive sense, but each iterated integral converges and their equality follows from Fubini on $[0, A] \times [0, B]$ plus a limit $B \to \infty$ (the tail $\int_0^A\int_B^\infty e^{-xy}\abs{\sin x}\,\dd y\,\dd x \leq \int_0^A \frac{e^{-Bx}\abs{\sin x}}{x}\dd x \leq \int_0^A e^{-Bx} \dd x\to 0$ , using $\abs{\sin x} \leq x$ ).

$y$ first: $\int_0^\infty e^{-xy}\,\dd y = \frac1x$ for $x > 0$ , so the first integral is $\int_0^A \frac{\sin x}{x}\,\dd x$ .

$x$ first: two integrations by parts (or taking the imaginary part of $\int_0^A e^{(i - y)x}\dd x$ ) give

\int_0^A e^{-xy}\sin x\,\dd x = \frac{1 - e^{-Ay}(y\sin A + \cos A)}{1 + y^2} .

Integrating in $y$ over $[0, \infty)$ , the term $\int_0^\infty \frac{\dd y}{1 + y^2} = \frac\pi2$ splits off:

\int_0^A \frac{\sin x}{x}\,\dd x = \frac{\pi}{2} - \int_0^\infty e^{-Ay}\,\frac{y\sin A + \cos A}{1 + y^2}\,\dd y .

The remainder is bounded by $\int_0^\infty e^{-Ay}\frac{y + 1}{1 + y^2}\dd y \leq \int_0^\infty e^{-Ay}\cdot\frac{1+y}{1+y^2} \,\dd y \to 0$ as $A \to \infty$ (dominated convergence, or the crude bound $\frac{1 + y}{1 + y^2} \leq \frac32$ giving $\frac{3}{2A}$ ). Hence $\int_0^\infty\frac{\sin x}{x}\dd x = \frac\pi2$ — the same value obtained in Chapter 9 by differentiating a parameter integral; here Fubini does the work instead.

Exercise 20.10 ★★★

(Isoperimetric inequality via Wirtinger) Let $\gamma$ be a simple closed $\mathcal{C}^1$ curve of length $2\pi$ , parametrized by arc length on $[0, 2\pi]$ , enclosing area $A$ . Using Corollary 20.22, Parseval and the Wirtinger inequality (exercises of Chapter 14), prove $A \leq \pi$ , with equality for the circle. (Normalize $\int_0^{2\pi} x(s)\dd s = 0$ ; write $2A = \oint x\,\dd y - y\,\dd x$ and bound $2A \leq \int (x^2 + y'^2)$ carefully via $2A = \int_0^{2\pi}(xy' - yx')\dd s$ and $x^2 + y'^2 \geq 2xy'$ .)

Solution

Solution of Exercise 20.10.

Parametrize by arc length $s \in [0, 2\pi]$ , so $x'^2 + y'^2 = 1$ , and translate so that $\int_0^{2\pi} x(s)\,\dd s = 0$ . By Corollary 20.22,

2A = \oint x\,\dd y - y\,\dd x = \int_0^{2\pi}\bigl(xy' - yx'\bigr)\dd s .

Integrating $\oint y\,\dd x$ by parts over the period (boundary terms cancel by periodicity), $-\int yx' = \int y'x$ , so in fact $2A = 2\int_0^{2\pi}xy'\,\dd s$ . Then $2xy' \leq x^2 + y'^2$ gives

2A \leq \int_0^{2\pi}\bigl(x^2 + y'^2\bigr)\dd s = \int_0^{2\pi} x^2 + \int_0^{2\pi}\bigl(1 - x'^2\bigr) = 2\pi - \int_0^{2\pi}\bigl(x'^2 - x^2\bigr)\dd s .

Wirtinger’s inequality (Chapter 14 exercises: for a $2\pi$ -periodic $\mathcal{C}^1$ function with zero mean, $\int x^2 \leq \int x'^2$ ) makes the last integral nonnegative: $A \leq \pi$ . Equality requires equality in Wirtinger ( $x(s) = a\cos s + b\sin s$ ) and in $2xy' \leq x^2 + y'^2$ ( $y' = x$ pointwise), which forces $y = a\sin s - b\cos s + c$ : the curve is the unit circle (suitably centered). Since a curve of length $L$ rescales to length $2\pi$ , the general statement is $A \leq \frac{L^2}{4\pi}$ : among all closed curves of given perimeter, the circle encloses the largest area.

Exercise 20.11 ★★

(Moments of the ball) For the ball $B$ of radius $R$ in $\R^3$ , compute $\iiint_B z^2\,\dd x\,\dd y\,\dd z$ in spherical coordinates, and deduce $\iiint_B (x^2 + y^2 + z^2)\,\dd x\,\dd y\,\dd z$ by symmetry. Cross-check the latter against the shell computation $\int_0^R r^2\cdot4\pi r^2\,\dd r$ .

Solution

Solution of Exercise 20.11.

In spherical coordinates $z = r\sin\varphi$ and $\dd x\,\dd y\,\dd z = r^2\cos\varphi\,\dd r\,\dd\theta\,\dd\varphi$ :

\iiint_B z^2 = \int_0^R r^4\,\dd r\int_0^{2\pi}\dd\theta \int_{-\pi/2}^{\pi/2}\sin^2\varphi\cos\varphi\,\dd\varphi = \frac{R^5}5\cdot2\pi\cdot \Bigl[\frac{\sin^3\varphi}3\Bigr]_{-\pi/2}^{\pi/2} = \frac{4\pi R^5}{15}.

By the symmetry of the ball under permuting coordinates, $\iiint_B x^2 = \iiint_B y^2 = \iiint_B z^2$ , so $\iiint_B(x^2 + y^2 + z^2) = 3\cdot\frac{4\pi R^5}{15} = \frac{4\pi R^5}5$ . Shell check: $\int_0^R r^2\cdot 4\pi r^2\,\dd r = \frac{4\pi R^5}5$ — the integrand $r^2$ is constant on the sphere of radius $r$ , of area $4\pi r^2$ .

Exercise 20.12 ★★

(Shoelace formula) Let $K$ be a polygon with vertices $(x_1, y_1), \dots, (x_m, y_m)$ in counterclockwise order (indices mod $m$ ). Deduce from Corollary 20.22 that

\operatorname{Area}(K) = \frac12\sum_{i=1}^m \bigl(x_iy_{i+1} - x_{i+1}y_i\bigr) ,

and check the formula on the triangle $(0,0)$ , $(1,0)$ , $(0,1)$ .

Solution

Solution of Exercise 20.12.

Parametrize the edge from $(x_i, y_i)$ to $(x_{i+1}, y_{i+1})$ by $\gamma(t) = \bigl((1-t)x_i + tx_{i+1},\ (1-t)y_i + ty_{i+1}\bigr)$ . Its contribution to $\frac12\oint(x\,\dd y - y\,\dd x)$ is

\frac12\int_0^1\Bigl(\bigl((1-t)x_i + tx_{i+1}\bigr)(y_{i+1} - y_i) - \bigl((1-t)y_i + ty_{i+1}\bigr)(x_{i+1} - x_i)\Bigr)\dd t ,

and since $\int_0^1\bigl((1-t)u + tv\bigr)\dd t = \frac{u + v}2$ , this equals

\frac14\Bigl((x_i + x_{i+1})(y_{i+1} - y_i) - (y_i + y_{i+1})(x_{i+1} - x_i)\Bigr) = \frac12\bigl(x_iy_{i+1} - x_{i+1}y_i\bigr),

the cross terms cancelling. Summing over the $m$ edges gives the shoelace formula, by Corollary 20.22. Triangle $(0,0), (1,0), (0,1)$ : $\frac12\bigl((0\cdot0 - 1\cdot0) + (1\cdot1 - 0\cdot0) + (0\cdot0 - 0\cdot1)\bigr) = \frac12$ , the correct area.

20.6 Problem: the volume of the ball in dimension $n$

Problem 20.1

Weekend problem — $V_n = \pi^{n/2}/\Gamma(\frac n2 + 1)$ , and the strangeness of high dimensions

The disk has area $\pi$ , the ball volume $\frac43\pi$ — and then? This problem computes the volume of the unit ball of $\R^n$ for every $n$ , twice (by a slicing recursion driven by the Wallis integrals, then via the $\Gamma$ function and the Gauss integral of Example 20.17), and then reads off the geometry: the volumes peak in dimension five and rush to zero, and almost all of a high-dimensional ball hides in a thin shell near its boundary. For a continuous function on a ball of $\R^n$ , the integral is understood as the $n$ -fold iterated integral (slicing one coordinate at a time, as in the chapter for $n \leq 3$ ); we write $B_n(R)$ for the closed ball of radius $R$ centered at $0$ , $v_n(R)$ for its volume, and $V_n = v_n(1)$ , with $V_0 = 1$ by convention.

Part I — The slicing recursion.

Substituting $x_i = Ru_i$ in each of the $n$ iterated integrals, show $v_n(R) = V_nR^n$ .
Slicing $B_n(1)$ along its last coordinate, show

$V_n = V_{n-1}\int_{-1}^{1}(1 - t^2)^{\frac{n-1}2}\,\dd t .$
With $t = \sin\theta$ , identify the integral as a Wallis integral: $\int_{-1}^1(1 - t^2)^{\frac{n-1}2}\dd t = 2W_n$ , where $W_n = \int_0^{\pi/2}\cos^n\theta\,\dd\theta = \int_0^{\pi/2}\sin^n\theta\,\dd\theta$ .
Prove the two Wallis identities (integrate by parts; then telescope $nW_nW_{n-1}$ ):

$W_n = \frac{n-1}nW_{n-2} \quad (n \geq 2), \qquad W_nW_{n-1} = \frac{\pi}{2n} \quad (n \geq 1).$

Part II — The recursion solved.

Combine questions 2–4 into the two-step recursion

$V_n = \frac{2\pi}{n}\,V_{n-2} \qquad (n \geq 2).$
Deduce the closed forms, for $k \geq 0$ :

$V_{2k} = \frac{\pi^k}{k!}, \qquad V_{2k+1} = \frac{2^{k+1}\pi^k}{1\cdot3\cdot5\cdots (2k+1)} .$
Tabulate $V_1, \dots, V_7$ numerically. Using the ratio $V_n/V_{n-2} = 2\pi/n$ and the values of $2W_5$ and $2W_6$ , prove that the sequence $(V_n)$ increases up to its maximum $V_5 = \frac{8\pi^2}{15} \approx 5.26$ and decreases from then on.
Show that $V_n \to 0$ faster than any geometric sequence, and that $\sum_{n\geq1} V_n$ converges: all the unit balls together have finite total volume.
Prove the generating identity

$\sum_{k\geq0} V_{2k}\,x^{2k} = \eu^{\pi x^2} \qquad (x \in \R),$

and deduce $\sum_{k\geq0}V_{2k} = \eu^\pi \approx 23.14$ .

Part III — Second route: $\Gamma$ and the Gauss integral.

Show by Fubini (the integrand is a product) that

$I_n = \int_{\R^n}\eu^{-\norm x^2}\dd x = \Bigl(\int_{-\infty}^{+\infty} \eu^{-t^2}\dd t\Bigr)^{\!n} = \pi^{n/2},$

the $n$ -dimensional Gauss integral, understood as a limit over cubes $\intcc{-R}{R}^n$ .
Recall $\Gamma(s) = \int_0^\infty t^{s-1}\eu^{-t}\dd t$ (Definition 9.17). From $\Gamma(s+1) = s\,\Gamma(s)$ (Theorem 9.18) and $\Gamma(\tfrac12) = \sqrt\pi$ (substitute $t = u^2$ and invoke the Gauss integral), compute

$\Gamma(k + 1) = k!, \qquad \Gamma\Bigl(k + \frac32\Bigr) = \frac{1\cdot3\cdots(2k+1)}{2^{k+1}}\,\sqrt\pi .$
Prove, by induction through the recursion of question 5, the single formula

$V_n = \frac{\pi^{n/2}}{\Gamma\bigl(\frac n2 + 1\bigr)} \qquad (n \geq 1),$

and check it reproduces both closed forms of question 6.
Show $\int_0^\infty \eu^{-r^2}r^{n-1}\dd r = \tfrac12\Gamma\bigl(\tfrac n2\bigr)$ and deduce the identity

$I_n = n\,V_n\int_0^\infty \eu^{-r^2}\,r^{n-1}\,\dd r.$

Interpret it: the Gaussian mass of $\R^n$ is collected along spherical shells whose “ $(n-1)$ -dimensional area” at radius $r$ is $nV_nr^{n-1}$ — both sides are now proved independently, so the interpretation costs nothing.
Set $s_{n-1} = nV_n$ (the area of the unit sphere $S^{n-1}$ , consistently with $v_n(R) = \int_0^R s_{n-1}r^{n-1}\dd r$ ). Tabulate $s_0, \dots, s_3$ and check $s_1 = 2\pi$ , $s_2 = 4\pi$ , $s_3 = 2\pi^2$ .

Part IV — High dimensions are strange.

From Stirling’s formula (Theorem 6.13) applied to $k!$ , show for even $n = 2k$ :

$V_n \sim \frac{1}{\sqrt{\pi n}} \Bigl(\frac{2\pi\eu}{n}\Bigr)^{n/2} \qquad (n \to \infty, \ n \text{ even}),$

and explain why the same super-geometric decay bound extends to odd $n$ via the recursion.
The unit ball sits in the cube $\intcc{-1}1^n$ of volume $2^n$ . Compute the filling ratio $V_n/2^n$ for $n = 2, 3, 10$ , and show it tends to $0$ : in high dimension, essentially all of the cube lies in its corners.
Show that the fraction of $v_n(1)$ lying within distance $\varepsilon$ of the boundary sphere is $1 - (1 - \varepsilon)^n \to 1$ ; numerically, what fraction of a $100$ -dimensional ball lies in the outer shell of thickness $1\%$ ?
Prove the Wallis asymptotic $W_n \sim \sqrt{\dfrac{\pi}{2n}}$ (monotonicity of $(W_n)$ , the ratio $W_n/W_{n-2} \to 1$ , and $W_nW_{n-1} = \frac\pi{2n}$ ), and the lower bound $W_n \geq \sqrt{\dfrac{\pi}{2(n+1)}}$ for all $n$ .
(Concentration on a slab) The fraction of the unit ball with first coordinate beyond $\delta$ is $\int_\delta^1(1 - x^2)^{\frac{n-1}2}\dd x \,\big/\, (2W_n)$ . Using $1 - u \leq \eu^{-u}$ and the tail bound $\int_\delta^\infty \eu^{-a x^2}\dd x \leq \frac{\eu^{-a\delta^2}}{2a\delta}$ , show this fraction is at most

$\frac{\eu^{-(n-1)\delta^2/2}}{(n-1)\,\delta} \Big/ \sqrt{\frac{2\pi}{n+1}}$

and conclude: for $\delta = s/\sqrt{n-1}$ , all but a fraction $O(\eu^{-s^2/2}/s)$ of the ball lies in the slab $\abs{x_1} \leq s/\sqrt{n-1}$ . A ball of high dimension is, statistically, a thin pancake in every direction at once.
Assemble questions 16–19 into one paragraph: where the volume of $B_n(1)$ sits (near the boundary sphere, yet within $O(1/\sqrt n)$ slabs of every hyperplane through the center), and why these two statements do not contradict each other.

Part V — Other bodies, and synthesis.

(Simplex) Let $\Delta_n = \{x \in \R^n : x_i \geq 0,\ \sum x_i \leq 1\}$ . Show by slicing and induction that $\operatorname{vol}(\Delta_n) = \frac1{n!}$ .
(Cross-polytope) Deduce that $C_n = \{x : \sum\abs{x_i} \leq 1\}$ has volume $\frac{2^n}{n!}$ , and verify the sandwich $C_n \subseteq B_n(1) \subseteq \intcc{-1}1^n$ at the level of volumes: $\frac{2^n}{n!} \leq V_n \leq 2^n$ .
Compute $V_4$ a third way: slice $\R^4 = \R^2 \times \R^2$ , integrate the area of the $(z, w)$ -disk over the $(x, y)$ -disk in polar coordinates, and recover $V_4 = \frac{\pi^2}2$ .
(Monte Carlo in trouble) A point is drawn uniformly in the cube $\intcc{-1}1^{20}$ . Show that the probability that it lands in the inscribed ball is $V_{20}/2^{20} \approx 2.5\cdot10^{-8}$ , so about forty million draws are needed before the first hit is expected: estimating $V_n$ by rejection sampling collapses in high dimension (the curse of dimensionality).
Synthesis. Two independent derivations met at $V_n = \pi^{n/2}/\Gamma(\frac n2 + 1)$ : list which theorem of this chapter each one used (Fubini, change of variables, the polar Gauss integral), and which one-variable inputs (Wallis, $\Gamma$ , Stirling). Where does the Year 3 volume redo this computation with Lebesgue’s theory, and what does it add?

Solution

Solution of Problem 20.1.

1. The ball $B_n(R)$ is described by iterated bounds $-R \leq x_n \leq R$ , then $\abs{x_{n-1}} \leq \sqrt{R^2 - x_n^2}$ , and so on; substituting $x_i = Ru_i$ in each of the $n$ one-variable integrals multiplies each by $R$ and maps the bounds onto those of $B_n(1)$ : $v_n(R) = R^n\,v_n(1) = V_nR^n$ .

2. Slicing along $x_n = t$ : the slice of $B_n(1)$ is the ball $B_{n-1}\bigl(\sqrt{1 - t^2}\bigr)$ , so, by question 1,

V_n = \int_{-1}^1 v_{n-1}\bigl(\sqrt{1 - t^2}\bigr)\,\dd t = V_{n-1}\int_{-1}^{1}(1 - t^2)^{\frac{n-1}2}\,\dd t .

3. With $t = \sin\theta$ , $\dd t = \cos\theta\,\dd\theta$ and $(1 - t^2)^{\frac{n-1}2} = \cos^{n-1}\theta$ on $\intcc{-\pi/2}{\pi/2}$ :

\int_{-1}^1(1 - t^2)^{\frac{n-1}2}\dd t = \int_{-\pi/2}^{\pi/2}\cos^n\theta\,\dd\theta = 2\int_0^{\pi/2}\cos^n\theta\,\dd\theta = 2W_n,

and $\theta \mapsto \frac\pi2 - \theta$ interchanges the sine and cosine forms of $W_n$ .

4. Write $\sin^n = \sin^{n-2}(1 - \cos^2)$ and integrate $\int\sin^{n-2}\cos\cdot\cos$ by parts ( $v = \frac{\sin^{n-1}}{n-1}$ ):

W_n = W_{n-2} - \frac{W_n}{n-1} \quad\Longrightarrow\quad W_n = \frac{n-1}{n}W_{n-2}.

Hence $nW_nW_{n-1} = (n-1)W_{n-1}W_{n-2}$ : the sequence $(nW_nW_{n-1})$ is constant, equal to $1\cdot W_1W_0 = 1\cdot\frac\pi2$ , so $W_nW_{n-1} = \frac{\pi}{2n}$ .

5. Questions 2–3 give $V_n = 2W_nV_{n-1}$ , twice:

V_n = 2W_n\cdot 2W_{n-1}\,V_{n-2} = 4\,\frac{\pi}{2n}\,V_{n-2} = \frac{2\pi}n\,V_{n-2}.

6. From $V_0 = 1$ : $V_{2k} = \frac{2\pi}{2k}V_{2k-2} = \frac\pi kV_{2k-2}$ , so $V_{2k} = \frac{\pi^k}{k!}$ by induction. From $V_1 = 2$ : $V_{2k+1} = \frac{2\pi}{2k+1}V_{2k-1}$ , so

V_{2k+1} = 2\prod_{j=1}^k\frac{2\pi}{2j+1} = \frac{2^{k+1}\pi^k}{1\cdot3\cdot5\cdots(2k+1)} .

7. $V_1 = 2$ , $V_2 = \pi \approx 3.142$ , $V_3 = \frac{4\pi}3 \approx 4.189$ , $V_4 = \frac{\pi^2}2 \approx 4.935$ , $V_5 = \frac{8\pi^2}{15} \approx 5.264$ , $V_6 = \frac{\pi^3}6 \approx 5.168$ , $V_7 = \frac{16\pi^3}{105} \approx 4.725$ . The one-step ratio is $V_n/V_{n-1} = 2W_n$ , and $(W_n)$ is decreasing ( $\sin^n \leq \sin^{n-1}$ pointwise). Now $2W_5 = 2\cdot\frac45\cdot\frac23 = \frac{16}{15} > 1$ while $2W_6 = 2\cdot\frac56\cdot\frac34\cdot\frac12\cdot\frac\pi2 = \frac{5\pi}{16} < 1$ : the ratios exceed $1$ up to $n = 5$ and are below $1$ from $n = 6$ on — $(V_n)$ increases to its maximum $V_5$ and then decreases.

8. For $n \geq 13 > 4\pi$ : $V_n/V_{n-2} = 2\pi/n < \tfrac12$ , so $V_{n} \leq C\cdot 2^{-n/2}$ with a fixed constant; better, for any $q > 0$ , $2\pi/n < q^2$ for large $n$ , so $V_n/q^n \to 0$ : the decay beats every geometric sequence. Convergence of $\sum V_n$ follows from the ratio $V_n/V_{n-2} \to 0$ (compare with a geometric series from some rank on).

9. $\sum_{k\geq0}V_{2k}x^{2k} = \sum_{k\geq0}\frac{(\pi x^2)^k}{k!} = \eu^{\pi x^2}$ , the exponential series (Chapter 11), convergent for every $x$ . At $x = 1$ : $\sum_kV_{2k} = \eu^\pi \approx 23.14$ .

10. On the cube $\intcc{-R}R^n$ the integrand is the product $\prod_i\eu^{-x_i^2}$ , so the iterated integral factorizes: $\bigl(\int_{-R}^R\eu^{-t^2}\dd t\bigr)^n$ . Letting $R \to \infty$ and using $\int_\R\eu^{-t^2}\dd t = \sqrt\pi$ (Example 20.17): $I_n = \pi^{n/2}$ .

11. $t = u^2$ gives $\Gamma(\tfrac12) = \int_0^\infty t^{-1/2}\eu^{-t}\dd t = 2\int_0^\infty\eu^{-u^2}\dd u = \sqrt\pi$ . Iterating $\Gamma(s+1) = s\Gamma(s)$ : $\Gamma(k+1) = k!\,\Gamma(1) = k!$ , and

\Gamma\Bigl(k + \frac32\Bigr) = \Bigl(k + \frac12\Bigr)\Bigl(k - \frac12\Bigr)\cdots \frac12\cdot\Gamma\Bigl(\frac12\Bigr) = \frac{(2k+1)(2k-1)\cdots1}{2^{k+1}}\,\sqrt\pi .

12. Set $F_n = \pi^{n/2}/\Gamma(\frac n2 + 1)$ . Since $\Gamma(\frac n2 + 1) = \frac n2\,\Gamma(\frac n2) = \frac n2\,\Gamma(\frac{n-2}2 + 1)$ , we get $F_n = \frac{2\pi}nF_{n-2}$ : the same recursion as $V_n$ (question 5). Bases: $F_1 = \sqrt\pi/\Gamma(\frac32) = \sqrt\pi/(\frac{\sqrt\pi}2) = 2 = V_1$ and $F_2 = \pi/\Gamma(2) = \pi = V_2$ . By induction $V_n = F_n$ for all $n$ ; question 11 turns this back into the two closed forms of question 6.

13. With $r = \sqrt t$ : $\int_0^\infty \eu^{-r^2}r^{n-1}\dd r = \frac12\int_0^\infty t^{\frac n2 - 1}\eu^{-t}\dd t = \frac12\Gamma(\frac n2)$ . Hence

n\,V_n\int_0^\infty\eu^{-r^2}r^{n-1}\dd r = V_n\cdot\frac n2\,\Gamma\Bigl(\frac n2\Bigr) = V_n\,\Gamma\Bigl(\frac n2 + 1\Bigr) = \pi^{n/2} = I_n .

Both sides being proved, the identity can be read as the shell decomposition of the Gaussian integral: the sphere of radius $r$ carries area $nV_nr^{n-1}$ , and the Gaussian weight $\eu^{-r^2}$ is integrated over the shells.

14. $s_0 = V_1 = 2$ (the $0$ -sphere is two points), $s_1 = 2V_2 = 2\pi$ , $s_2 = 3V_3 = 4\pi$ , $s_3 = 4V_4 = 2\pi^2$ ; and $\int_0^R s_{n-1}r^{n-1}\dd r = V_nR^n = v_n(R)$ : area is the radial derivative of volume.

15. For $n = 2k$ , Stirling (Theorem 6.13) gives $k! \sim \sqrt{2\pi k}\,(k/\eu)^k$ , so

V_{2k} = \frac{\pi^k}{k!} \sim \frac{(\pi\eu/k)^k}{\sqrt{2\pi k}} = \frac1{\sqrt{\pi n}}\Bigl(\frac{2\pi\eu}n\Bigr)^{n/2} \qquad (n = 2k).

For odd $n$ : $V_{2k+1} = 2W_{2k+1}V_{2k} \leq 2V_{2k}$ , so the same super-geometric decay bounds hold (up to a factor $2$ and a shift of one in the exponent) — for every $q > 0$ , $V_n = o(q^n)$ .

16. $V_2/4 = \pi/4 \approx 0.785$ ; $V_3/8 = \pi/6 \approx 0.524$ ; $V_{10}/2^{10} = \frac{\pi^5}{120\cdot1024} \approx 0.0025$ . In general $\frac{V_n/2^n}{V_{n-2}/2^{n-2}} = \frac{2\pi}{4n} = \frac{\pi}{2n} \to 0$ : the ratio tends to $0$ (super-geometrically). The inscribed ball occupies a vanishing fraction: the cube’s volume migrates to its corners.

17. By question 1 the inner ball of radius $1 - \varepsilon$ has volume $V_n(1-\varepsilon)^n$ , so the outer shell carries the fraction $1 - (1 - \varepsilon)^n \to 1$ . For $n = 100$ , $\varepsilon = 0.01$ : $(0.99)^{100} = \eu^{100\ln0.99} \approx \eu^{-1.005} \approx 0.366$ : about $63\%$ of the ball lies within $1\%$ of its surface.

18. $(W_n)$ decreases, so $W_n \leq W_{n-1} \leq W_{n-2} = \frac{n}{n-1}W_n$ : squeezing, $W_{n-1}/W_n \to 1$ . Multiplying by $W_nW_{n-1} = \frac\pi{2n}$ : $W_n^2 \sim \frac\pi{2n}$ , i.e. $W_n \sim \sqrt{\pi/(2n)}$ . Lower bound: $W_n^2 \geq W_nW_{n+1} = \frac{\pi}{2(n+1)}$ , so $W_n \geq \sqrt{\pi/(2(n+1))}$ for every $n$ .

19. Numerator: $1 - x^2 \leq \eu^{-x^2}$ gives $(1 - x^2)^{\frac{n-1}2} \leq \eu^{-(n-1)x^2/2}$ , and with $a = \frac{n-1}2$ ,

\int_\delta^1(1 - x^2)^{\frac{n-1}2}\dd x \leq \int_\delta^\infty\eu^{-ax^2}\dd x \leq \int_\delta^\infty\frac x\delta\,\eu^{-ax^2}\dd x = \frac{\eu^{-a\delta^2}}{2a\delta} = \frac{\eu^{-(n-1)\delta^2/2}}{(n-1)\delta}.

Denominator: $2W_n \geq \sqrt{2\pi/(n+1)}$ by question 18. Dividing gives the displayed bound. For $\delta = s/\sqrt{n-1}$ it becomes $\sqrt{\tfrac{n+1}{2\pi(n-1)}}\;\eu^{-s^2/2}/s = O\bigl( \eu^{-s^2/2}/s\bigr)$ , uniformly in $n$ : outside the slab $\abs{x_1} \leq s/\sqrt{n-1}$ there is almost no volume, for $s$ moderately large — and by symmetry the same holds for every direction.

20. The two statements coexist because they describe different coordinates of the same point. Almost every point of $B_n(1)$ has norm close to $1$ (question 17: radial concentration near the sphere), yet each of its $n$ coordinates is small, of order $1/\sqrt n$ (question 19), which is consistent since $n$ coordinates of size $1/\sqrt n$ have norm of order $1$ . High-dimensional volume concentrates where all coordinates share the norm budget equally — near the sphere, but far from every coordinate axis pole.

21. Slice $\Delta_n$ at $x_n = t \in \intcc01$ : the slice is $\{x' \in \R^{n-1} : x_i \geq 0,\ \sum x_i \leq 1 - t\} = (1-t)\Delta_{n-1}$ , of volume $(1-t)^{n-1}\operatorname{vol}(\Delta_{n-1})$ by homogeneity. So

\operatorname{vol}(\Delta_n) = \operatorname{vol}(\Delta_{n-1})\int_0^1(1 - t)^{n-1}\dd t = \frac{\operatorname{vol}(\Delta_{n-1})}{n} \quad\Longrightarrow\quad \operatorname{vol}(\Delta_n) = \frac1{n!}\,.

22. The $2^n$ sign-orthants cut $C_n$ into $2^n$ copies of $\Delta_n$ (with negligible overlaps on coordinate hyperplanes): $\operatorname{vol}(C_n) = \frac{2^n}{n!}$ . If $\sum\abs{x_i} \leq 1$ then $\sum x_i^2 \leq \bigl(\sum\abs{x_i}\bigr)^2 \leq 1$ : $C_n \subseteq B_n(1)$ ; and $B_n(1) \subseteq \intcc{-1}1^n$ since $\abs{x_i} \leq \norm x$ . Hence $\frac{2^n}{n!} \leq V_n \leq 2^n$ — consistent with question 15, which places $V_n$ between the factorial and geometric scales.

23. For $(x, y)$ in the unit disk, the slice of $B_4(1)$ is the disk of radius $\sqrt{1 - x^2 - y^2}$ in the $(z, w)$ -plane, of area $\pi(1 - x^2 - y^2)$ . In polar coordinates:

V_4 = \iint_{x^2+y^2\leq1}\pi(1 - x^2 - y^2)\,\dd x\,\dd y = \pi\int_0^{2\pi}\!\!\int_0^1(1 - \rho^2)\rho\, \dd\rho\,\dd\alpha = \pi\cdot2\pi\cdot\frac14 = \frac{\pi^2}2 ,

agreeing with question 6.

24. The probability is the volume ratio $\dfrac{V_{20}}{2^{20}} = \dfrac{\pi^{10}}{10!\cdot2^{20}} \approx \dfrac{0.0258}{1\,048\,576} \approx 2.5\cdot10^{-8}$ . The number of draws until the first hit is of order the inverse, about $4\cdot10^7$ : a rejection sampler that worked beautifully for the disk ( $\pi/4$ of hits) is useless in dimension $20$ — the curse of dimensionality in one line.

25. Route one (Parts I–II) used: Fubini-type slicing of the iterated integral, one-variable substitution in each coordinate (homogeneity), and the Wallis integrals — pure one-variable calculus plus induction. Route two (Part III) used: Fubini for the product structure of $I_n$ , the polar change of variables through the Gauss integral of Example 20.17, and the $\Gamma$ function’s functional equation. They meet in $V_n = \pi^{n/2}/\Gamma(\frac n2 + 1)$ , with Stirling (Theorem 6.13) converting the formula into asymptotics. The Year 3 volume rebuilds all of this on Lebesgue’s integral: there Fubini and the change of variables are theorems for general integrable functions, spherical coordinates exist in every dimension, and the same ball volumes reappear as the worked dividends of the product-measure and Stirling problems — with dominated convergence replacing our hand-made squeezes.

20.1 Line integrals

20.2 Double integrals

20.3 The Green–Riemann theorem

20.4 Triple integrals

20.5 Exercises

20.6 Problem: the volume of the ball in dimension nnn

20.6 Problem: the volume of the ball in dimension $n$