Volume of the n-dimensional unit ball

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(a)

Prove that lim⁡n→∞λn(Bn)=0\lim_{n \rightarrow \infty } \lambda_{n}\left(\mathbf{B}_{n}\right) = 0.

(b)

Find the value of nn that maximizes λn(Bn)\lambda_{n}\left(\mathbf{B}_{n}\right).

For readers familiar with the gamma function Γ\Gamma : Prove that

λn(Bn)=πn/2Γ(n2+1) \lambda _{n}\left(\mathbf{B}_{n}\right) = \frac{\pi ^{n / 2}}{\Gamma \left(\frac{n}{2}+1\right)}

for every positive integer nn.

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Find the volume of the nn-dimensional ball

x12+x22+⋯+xn2⩽a2. x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\leqslant a^{2} .
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Find ∬⋯∫Ωxn2 dx1 dx2⋯dxn\displaystyle \iint \cdots \int_{\Omega } x_{n}^{2}\, dx_{1}\, dx_{2}\cdots dx_{n}, where the region Ω\Omega is determined by the inequalities:

x12+x22+⋯+xn−12⩽a2,−h2⩽xn⩽h2. x_{1}^{2}+x_{2}^{2}+\cdots +x_{n-1}^{2}\leqslant a^{2} , \qquad -\dfrac {h}{2}\leqslant x_{n}\leqslant \dfrac {h}{2} .
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Compute

∬x12+x22+⋯+xn2⩽1⋯∫dx1 dx2⋯dxn1−x12−x22−⋯−xn2. \iint \limits _{x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\leqslant 1}\cdots \int \dfrac {dx_{1}\, dx_{2}\cdots dx_{n}}{\sqrt{1-x_{1}^{2}-x_{2}^{2}-\cdots -x_{n}^{2}}} .
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Reduce the nn-fold integral (n⩾2n\geqslant 2) spread over the region x12+x22+⋯+xn2⩽R2x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}\leqslant R^{2}

∬Ω⋯∫f(x12+x22+⋯+xn2) dx1 dx2⋯dxn \iint \limits _{\Omega }\cdots \int f\left(\sqrt{x_{1}^{2}+x_{2}^{2}+\cdots +x_{n}^{2}}\right)\, dx_{1}\, dx_{2}\cdots dx_{n}

to a single integral, where f(u)f(u) is a continuous function.

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Let

un(a)=∫x12+x22+…+xn2<a2⋯∫dx1 dx2…dxn,vn=un(1). u_{n}(a) = \int \limits _{x_{1}^{2}+x_{2}^{2}+\ldots +x_{n}^{2}<a^{2}}\cdots \int dx_{1} \, dx_{2}\ldots dx_{n}, \qquad v_{n}=u_{n}(1).

Prove the equality un(a)=anvnu_{n}(a)=a^{n}v_{n}.

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Find the volume of the nn-dimensional ball ∑i=1nxi2⩽R2\sum_{i = 1}^{n} x_{i}^{2} \leqslant R^{2}.

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Find the volume of the nn-dimensional cylinder ∑i=1n−1xi2⩽R2,0⩽xn⩽H\sum_{i = 1}^{n-1} x_{i}^{2} \leqslant R^{2}, 0 \leqslant x_{n} \leqslant H.

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Find the volume of the nn-dimensional cone ∑i=1n−1xi2⩽x2xn2,0⩽xn⩽H\sum_{i = 1}^{n-1} x_{i}^{2} \leqslant x^{2} x_{n}^{2}, \quad 0 \leqslant x_{n} \leqslant H.

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Compute the integral ∫Gxn2dx,x=(x1;…;xn)\int_{G} x_{n}^{2} d x, x = \left(x_{1} ; \ldots ; x_{n}\right), where

G={∑i=1n−1xi2⩽R2,0⩽xn⩽H} G = \left\{ \sum _{i = 1}^{n-1} x_{i}^{2} \leqslant R^{2}, \quad 0 \leqslant x_{n} \leqslant H\right\}
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Find the volume of the nn-dimensional ellipsoid ∑i=1nxi2ai2⩽1\sum_{i = 1}^{n} \frac{x_{i}^{2}}{a_{i}^{2}} \leqslant 1.

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Compute the integral ∫Ω(R2−x12−…−xn2)1/2dx\int_{\Omega }\left(R^{2}-x_{1}^{2}-\ldots -x_{n}^{2}\right)^{1 / 2} d x, where Ω\Omega is the ball ∑i=1xi2⩽R2\sum_{i = 1} x_{i}^{2} \leqslant R^{2}.

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Let the function f(r)f(r) be continuous for r⩾0r \geqslant 0. Reduce to a single integral ∫Ωf(x12+…+xn2)dx\int_{\Omega } f\left(\sqrt{x_{1}^{2}+\ldots +x_{n}^{2}}\right) d x, where Ω\Omega is the ball ∑i=1nxi2⩽R2\sum_{i = 1}^{n} x_{i}^{2} \leqslant R^{2}.

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Let VnV_{n} be the volume of the nn-dimensional ball of radius RR. Prove that:

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(1)

there exists cn>0c_{n} > 0 such that for any RR the volume of the ball is Vn=cnRnV_{n} = c_{n} R^{n}

(2)

V2k+1=2(2π)k(2k+1)!!R2k+1,V2k=(2π)k(2k)!!R2k,k∈NV_{2 k+1} = 2 \frac{(2 \pi )^{k}}{(2 k+1) ! !} R^{2 k+1}, V_{2 k} = \frac{(2 \pi )^{k}}{(2 k) ! !} R^{2 k}, k \in N.

Find the volume of a four-dimensional right circular:

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(1)

cylinder with base radius RR and height HH;

(2)

cone with base radius RR and height HH.

Find the Lebesgue measure of the set

{(x1,…,xm)∈Rm∣∑1⩽k⩽m∣xk∣p⩽1}(p>0). \left\{ \left(x_{1}, \ldots , x_{m}\right) \in \mathbb {R}^{m} \mid \sum _{1 \leqslant k \leqslant m}\left|x_{k}\right|^{p} \leqslant 1\right\} \quad (p>0) .
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(a)

Let

Om(r)={(x1,…,xm)∈Rm∣∑1⩽k⩽m∣xk∣⩽r}. O_{m}(r)=\left\{ \left(x_{1}, \ldots , x_{m}\right) \in \mathbb {R}^{m} \mid \sum _{1 \leqslant k \leqslant m}\left|x_{k}\right| \leqslant r\right\} .

Prove that

∫Om(r)h(∑1⩽k⩽m∣xk∣)dx=2m(m−1)!∫0rtm−1h(t)dt \int _{O_{m}(r)} h\left(\sum _{1 \leqslant k \leqslant m}\left|x_{k}\right|\right) d x=\frac{2^{m}}{(m-1)!} \int _{0}^{r} t^{m-1} h(t) d t
(b)

Let a=(a1,…,am)∈R+m,E={x∈R+m∣(x,a)>1}a=\left(a_{1}, \ldots , a_{m}\right) \in \mathbb {R}_{+}^{m}, E=\left\{ x \in \mathbb {R}_{+}^{m} \mid (x, a)>1\right\}. Compute the integral ∫Ee−(a,x)dx\int_{E} e^{-(a, x)} d x.

Prove that for p>0p>0

∫Rmh((∑1⩽k⩽m∣xk∣p)1/p)dx=mCm(p)∫0∞tm−1h(t)dt \int _{\mathbb {R}^{m}} h\left(\left(\sum _{1 \leqslant k \leqslant m}\left|x_{k}\right|^{p}\right)^{1 / p}\right) d x=m C_{m}(p) \int _{0}^{\infty } t^{m-1} h(t) d t

where Cm(p)=λm{(x1,…,xm)∈Rm∣∣x1∣p+…+∣xm∣p⩽1}C_{m}(p)=\lambda_{m}\left\{ \left(x_{1}, \ldots , x_{m}\right) \in \mathbb {R}^{m} \mid \left|x_{1}\right|^{p}+\ldots +\left|x_{m}\right|^{p} \leqslant 1\right\} (for the value of Cm(p)C_{m}(p) see problem 3.36).

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Let E⊂RnE \subset \mathbb {R}^{n} be a bounded set, 0<λn(E)<∞0<\lambda_{n}(E)<\infty, and let M(E,ε)M(E, \varepsilon ) be the maximal number of ε\varepsilon-separated points lying in EE. Prove that

1εnλn(E)αn⩽M(E,ε)⩽2nεnλn(Eε/2)αn, \frac{1}{\varepsilon ^{n}} \frac{\lambda _{n}(E)}{\alpha _{n}} \leqslant M(E, \varepsilon ) \leqslant \frac{2^{n}}{\varepsilon ^{n}} \frac{\lambda _{n}\left(E_{\varepsilon / 2}\right)}{\alpha _{n}},

where Eε/2E_{\varepsilon / 2} is the (ε/2)(\varepsilon / 2)-neighborhood of the set E,αnE, \quad \alpha_{n} is the volume of the nn-dimensional unit ball.

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The thickness of a watermelon's rind equals 120\frac{1}{20} of the radius of the watermelon (regarded as a ball). What percentage of the volume of the watermelon does the rind constitute

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(a)

in three-dimensional space?

(b)

in 100-dimensional space?

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(a)

The ball Bn(r)B^{n}(r) is tangent to all edges of the nn-dimensional cube [−1,1]n\left[-1,1\right]^{n}. Compare, for large nn, the volume of the cube and the volume of the ball.

(b)

The cube [−1,1]n\left[-1,1\right]^{n} is divided by the coordinate hyperplanes into 2n2^{n} cubes. In each of them we inscribe a ball, and then consider the ball Bn(ρ)B^{n}(\rho ) that is externally tangent to all the balls inscribed in the cubes. For which nn does the inclusion Bn(ρ)⊂[−1,1]nB^{n}(\rho ) \subset [-1,1]^{n} hold? Which is larger for large nn: the volume of the cube [−1,1]n\left[-1,1\right]^{n} or the volume of the ball Bn(ρ)B^{n}(\rho )?

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(a)

Prove that the radius of the ball Bn(rn)B^{n}\left(r_{n}\right) whose volume equals one is expressed by the formula

rn=n2πe(1+aln⁡nn+bn+o(1n)), r_{n}=\sqrt{\frac{n}{2 \pi e}}\left(1+a \frac{\ln n}{n}+\frac{b}{n}+o\left(\frac{1}{n}\right)\right),

and find the coefficients aa and bb. What does the Lebesgue measure (of dimension n−1n-1) of the section of this ball by a hyperplane passing through its center tend to?

(b)

Find the asymptotics of the radius ρn\rho_{n} of the ball Bn(ρn)B^{n}\left(\rho_{n}\right) whose Gaussian volume equals 12\frac{1}{2}.

Let AA and BB be convex centrally symmetric (with center at zero) compacta in Rn\mathbb {R}^{n}. Can one assert that λn(A)⩽λn(B)\lambda_{n}(A) \leqslant \lambda_{n}(B), if every central section of the set AA by a hyperplane has Lebesgue measure (of dimension n−1n-1) smaller than the corresponding section of the set BB? (Compare, for large nn, the sections of the unit cube and of the ball having unit volume.)

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Let ff be as in problem 10.3.7. Letting ∥x∥\left\| \boldsymbol {x}\right\| denote the length of x\boldsymbol {x}, show that

∫Rkf(∥x∥) dλˉk(x)=2πk/2Γ(k/2)∫0∞f(r)rk−1 dλˉ(r). \int _{\mathbb {R}^{k}} f\left(\left\| \boldsymbol {x}\right\| \right) \, \mathrm{d}\bar\lambda _{k}(\boldsymbol {x})=\frac{2\pi ^{k/2}}{\Gamma (k/2)} \int _{0}^{\infty } f(r) r^{k-1} \, \mathrm{d}\bar\lambda (r) .
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Remark.
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In problem 10.3.8, let b1=⋯=bk=2b_{1}=\cdots =b_{k}=2 and c1=⋯=ck=1c_{1}=\cdots =c_{k}=1 to obtain

∫(0,+∞)kf(∥x∥2) dλˉk(x)=πk/22kΓ(k/2)∫0∞f(t)tk2−1 dλˉ(t). \int _{(0,+\infty )^{k}} f\left(\left\| \boldsymbol {x}\right\| ^{2}\right) \, \mathrm{d}\bar\lambda _{k}(\boldsymbol {x})=\frac{\pi ^{k/2}}{2^{k}\Gamma (k/2)} \int _{0}^{\infty } f(t) t^{\frac{k}{2}-1} \, \mathrm{d}\bar\lambda (t) .

If ff is the composition of gg and the square root, where gg is nonnegative and L/B∗\mathscr {L} / \mathcal{B}^{*}-measurable, we have

∫(0,+∞)kg(∥x∥) dλˉk(x)=πk/22kΓ(k/2)∫0∞g(t)tk2−1 dλˉ(t)=πk/22kΓ(k/2)∫0∞g(r)rk−2⋅2r dλˉ(r)=πk/22k−1Γ(k/2)∫0∞g(r)rk−1 dλˉ(r). \begin{aligned} \int _{(0,+\infty )^{k}} g\left(\left\| \boldsymbol {x}\right\| \right) \, \mathrm{d}\bar\lambda _{k}(\boldsymbol {x}) & =\frac{\pi ^{k/2}}{2^{k}\Gamma (k/2)} \int _{0}^{\infty } g\left(\sqrt{t}\right) t^{\frac{k}{2}-1} \, \mathrm{d}\bar\lambda (t) \\ & =\frac{\pi ^{k/2}}{2^{k}\Gamma (k/2)} \int _{0}^{\infty } g(r) r^{k-2} \cdot 2r \, \mathrm{d}\bar\lambda (r) \\ & =\frac{\pi ^{k/2}}{2^{k-1}\Gamma (k/2)} \int _{0}^{\infty } g(r) r^{k-1} \, \mathrm{d}\bar\lambda (r) . \end{aligned}

Observe that ∫Rkg(∥x∥) dλˉk(x)=2k∫(0,+∞)kg(∥x∥) dλˉk(x)\int_{\mathbb {R}^{k}} g\left(\left\| \boldsymbol {x}\right\| \right) \, \mathrm{d}\bar\lambda_{k}(\boldsymbol {x})=2^{k} \int_{(0,+\infty )^{k}} g\left(\left\| \boldsymbol {x}\right\| \right) \, \mathrm{d}\bar\lambda_{k}(\boldsymbol {x}) by the rotational symmetry of the integrand. This gives the result with gg replacing ff.

The λˉk\bar\lambda_{k}-measure of a kk-dimensional ball with radius rr is πk/2rkΓ((k/2)+1)\frac{\pi^{k/2} r^{k}}{\Gamma \left(\left(k/2\right)+1\right)}.

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Remark.
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Use problem 10.3.9 on the function   1B(0,r)(x)=  1(0,r)(∥x∥)\; \mathbb {1}_{B(\boldsymbol {0},r)}(\boldsymbol {x})=\; \mathbb {1}_{(0,r)}\left(\left\| \boldsymbol {x}\right\| \right):

λˉk(B(0,r))=∫Rk  1B(0,r)(x) dλˉk(x)=2πk/2Γ(k/2)∫0∞  1(0,r)(t)tk−1 dλˉ(t). \bar\lambda _{k}\left(B(\boldsymbol {0},r)\right)=\int _{\mathbb {R}^{k}} \; \mathbb {1}_{B(\boldsymbol {0},r)}(\boldsymbol {x}) \, \mathrm{d}\bar\lambda _{k}(\boldsymbol {x})=\frac{2\pi ^{k/2}}{\Gamma (k/2)} \int _{0}^{\infty } \; \mathbb {1}_{(0,r)}(t) t^{k-1} \, \mathrm{d}\bar\lambda (t) .

Here B(0,r)B(\boldsymbol {0},r) denotes the ball in Rk\mathbb {R}^{k} with radius rr; we may consider the center to be the origin 0\boldsymbol {0} of Rk\mathbb {R}^{k} by the translation invariance of λˉk\bar\lambda_{k}.

More generally than problem 10.3.10, the kk-dimensional volume of the ellipsoid given by the equation ∑i=1kxi2ci2<1\sum_{i=1}^{k} \frac{x_{i}^{2}}{c_{i}^{2}}<1 is πk/2Γ((k/2)+1)c1⋯cn\frac{\pi^{k/2}}{\Gamma \left(\left(k/2\right)+1\right)} c_{1} \cdots c_{n}. (Here c1,…,cn>0c_{1}, \ldots , c_{n}>0.)

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References

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