Section 9.12

Multiple (Iterated) Integrals

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Problem 3816

Find the multiple integral ∫⋯∫dx1 dx2…dxn\displaystyle \int \cdots \int dx_{1} \, dx_{2}\ldots dx_{n}, taken over the region where all xk>0x_{k}>0 and x1+x2+…+xn⩽1x_{1}+x_{2}+\ldots +x_{n}\leqslant 1.

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Problem 3817

Find the multiple integral ∫⋯∫x1 dx1 dx2…dxn\displaystyle \int \cdots \int x_{1} \, dx_{1} \, dx_{2}\ldots dx_{n}, taken over the region where xk>0x_{k}>0 and x1+x2+…+xn⩽1x_{1}+x_{2}+\ldots +x_{n}\leqslant 1.

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Problem 3818

Find the multiple integral

∫∣x1∣+∣x2∣+…+∣xn∣<a⋯∫dx1 dx2…dxn. \int \limits _{\left|x_{1}\right|+\left|x_{2}\right|+\ldots +\left|x_{n}\right|<a}\cdots \int dx_{1} \, dx_{2}\ldots dx_{n}.
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Problem 3819

Find the multiple integral ∫01∫01⋯∫01(x12+x22+…+xn2) dx1 dx2…dxn\displaystyle \int_{0}^{1}\int_{0}^{1}\cdots \int_{0}^{1}(x_{1}^{2}+x_{2}^{2}+\ldots +x_{n}^{2}) \, dx_{1} \, dx_{2}\ldots dx_{n}.

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Problem 3820

Find the multiple integral ∫01∫01⋯∫01(x1x2+x1x3+…+xn−1xn) dx1 dx2…dxn\displaystyle \int_{0}^{1}\int_{0}^{1}\cdots \int_{0}^{1}(x_{1}x_{2}+x_{1}x_{3}+\ldots +x_{n-1}x_{n}) \, dx_{1} \, dx_{2}\ldots dx_{n}.

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Problem 3821

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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Problem 3822

Using the previous notation (problem 3821: un(a)=∫⋯∫dx1 dx2…dxnu_{n}(a)=\int \cdots \int dx_{1} \, dx_{2}\ldots dx_{n}, taken over the region x12+x22+…+xn2<a2x_{1}^{2}+x_{2}^{2}+\ldots +x_{n}^{2}<a^{2}, and vn=un(1)v_{n}=u_{n}(1)), the following equalities are obvious:

vn=∬x12+x22<1dx1 dx2∫x32+x42+…+xn2<1−x12−x22⋯∫dx3 dx4…dxn= v_{n} = \iint \limits _{x_{1}^{2}+x_{2}^{2}<1} dx_{1} \, dx_{2} \int \limits _{x_{3}^{2}+x_{4}^{2}+\ldots +x_{n}^{2}<1-x_{1}^{2}-x_{2}^{2}}\cdots \int dx_{3} \, dx_{4}\ldots dx_{n} ={} =∬x12+x22<1(1−x12−x22)n2−1vn−2 dx1 dx2. {}= \iint \limits _{x_{1}^{2}+x_{2}^{2}<1} (1-x_{1}^{2}-x_{2}^{2})^{\frac{n}{2}-1}v_{n-2} \, dx_{1} \, dx_{2}.

Using these, prove that

un(a)=πn/2anΓ(n2+1). u_{n}(a) = \frac{\pi ^{n/2}a^{n}}{\Gamma \big(\frac{n}{2}+1\big)}.

In particular, u1=2au_{1}=2a, u2=πa2u_{2}=\pi a^{2}, u3=43πa3u_{3}=\dfrac {4}{3}\pi a^{3}, u4=π22a4,…u_{4}=\dfrac {\pi^{2}}{2}a^{4}, \ldots

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Problem 3823

If

un=∫x1+x2+…+xn<a0<x1,0<x2,…,0<xn⋯∫x1α1−1x2α2−1…xnαn−1 dx1 dx2…dxn, u_{n} = \int \limits _{\substack {x_{1}+x_{2}+\ldots +x_{n}<a \\ 0<x_{1},0<x_{2},\ldots ,0<x_{n}}}\cdots \int x_{1}^{\alpha _{1}-1}x_{2}^{\alpha _{2}-1}\ldots x_{n}^{\alpha _{n}-1} \, dx_{1} \, dx_{2}\ldots dx_{n},

then it is easy to prove that un(a)=aα1+α2+…+αnvnu_{n}(a)=a^{\alpha_{1}+\alpha_{2}+\ldots +\alpha_{n}}v_{n}, where vn=un(1)v_{n}=u_{n}(1). Using this and the equality

vn=∫01x1α1−1 dx1∫x22+x32+…+xn2<1−x120<x2,…,0<xn⋯∫x2α2−1x3α3−1…xnαn−1 dx2 dx3…dxn, v_{n} = \int \limits _{0}^{1} x_{1}^{\alpha _{1}-1} \, dx_{1} \int \limits _{\substack {x_{2}^{2}+x_{3}^{2}+\ldots +x_{n}^{2}<1-x_{1}^{2} \\ 0<x_{2},\ldots ,0<x_{n}}}\cdots \int x_{2}^{\alpha _{2}-1}x_{3}^{\alpha _{3}-1}\ldots x_{n}^{\alpha _{n}-1} \, dx_{2} \, dx_{3}\ldots dx_{n},

prove that

un(a)=Γ(α1)Γ(α2)…Γ(αn)Γ(1+α1+α2+…+αn)aα1+α2+…+αn. u_{n}(a) = \frac{\Gamma (\alpha _{1})\Gamma (\alpha _{2})\ldots \Gamma (\alpha _{n})}{\Gamma (1+\alpha _{1}+\alpha _{2}+\ldots +\alpha _{n})}a^{\alpha _{1}+\alpha _{2}+\ldots +\alpha _{n}}.
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Problem 3824

Prove the equality:

∫0xdx1∫0x1dx2…∫0xn−1f(x) dxn=∫0xf(t)(x−t)n−1(n−1)! dt. \int \limits _{0}^{x} dx_{1} \int \limits _{0}^{x_{1}} dx_{2}\ldots \int \limits _{0}^{x_{n-1}} f(x) \, dx_{n} = \int \limits _{0}^{x} f(t)\frac{(x-t)^{n-1}}{(n-1)!} \, dt.
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Problem 3825

Prove the equality:

∫0xx1 dx1∫0x1x2 dx2…∫0xn−1xn dxn∫0xnf(t) dt= \int \limits _{0}^{x} x_{1} \, dx_{1} \int \limits _{0}^{x_{1}} x_{2} \, dx_{2}\ldots \int \limits _{0}^{x_{n-1}} x_{n} \, dx_{n} \int \limits _{0}^{x_{n}} f(t) \, dt ={} =12⋅4⋅…⋅2n∫0xf(t)(x2−t2)n dt. {}= \frac{1}{2\cdot 4\cdot \ldots \cdot 2n}\int \limits _{0}^{x} f(t)(x^{2}-t^{2})^{n} \, dt.
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Problem 3826

Reduce to a single integral the multiple integral

∬⋯∫x1α1−1x2α2−1…xnαn−1F(x1+x2+…+xn) dx1 dx2…dxn, \iint \cdots \int x_{1}^{\alpha _{1}-1}x_{2}^{\alpha _{2}-1}\ldots x_{n}^{\alpha _{n}-1}F(x_{1}+x_{2}+\ldots +x_{n}) \, dx_{1} \, dx_{2}\ldots dx_{n},

taken over the region defined by the inequalities

0<x1,0<x2,…,0<xn,x1+x2+…+xn<a. 0<x_{1}, \quad 0<x_{2}, \quad \ldots , \quad 0<x_{n}, \quad x_{1}+x_{2}+\ldots +x_{n}<a.
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Remark.
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One approach is a change of variables by the formulas x1+x2+…+xn=ux_{1}+x_{2}+\ldots +x_{n}=u, x1=x1x_{1}=x_{1}, x2=x2x_{2}=x_{2}, x3=x3,…,xn−1=xn−1x_{3}=x_{3},\ldots ,x_{n-1}=x_{n-1}.

Problem 3827

The points M(x,y,z)M(x,y,z) and M1(ξ,η,ζ)M_{1}(\xi ,\eta ,\zeta ) independently of each other range over the entire volume of a body of density p=p(x,y,z)p=p(x,y,z). The self-potential of the body is called the integral

12∫ ⁣ ⁣∫ ⁣ ⁣∫ ⁣ ⁣∫ ⁣ ⁣∫ ⁣ ⁣∫p(x,y,z)p(ξ,η,ζ) dx dy dz dξ dη dζ(x−ξ)2+(y−η)2+(z−ζ)2. \frac{1}{2}\int \! \! \int \! \! \int \! \! \int \! \! \int \! \! \int \frac{p(x,y,z)p(\xi ,\eta ,\zeta ) \, dx \, dy \, dz \, d\xi \, d\eta \, d\zeta }{\sqrt{(x-\xi )^{2}+(y-\eta )^{2}+(z-\zeta )^{2}}}.

Find the value of the self-potential of a homogeneous sphere x2+y2+z2=a2x^{2}+y^{2}+z^{2}=a^{2}. Its value measures the work done in bringing from infinity into a single body particles that attract each other by Newton's law.

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Remark.
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Two approaches quickly lead to the goal: 1) Use the fact that the integral over the variables (ξ,η,ζ)(\xi ,\eta ,\zeta ) gives the potential of the sphere at an interior point (x,y,z)(x,y,z), found in one of the earlier problems (problem 3799). 2) Account for the work dAdA done when the masses coming from infinity increase the radius of the sphere from rr to r+drr+dr, and integrate the result over rr from 00 to aa.