三重积分计算什么时候用驻坐标什么时候用球坐标（三重积分的概念和计算）

各位看官大家好，是我，美⑨(*^▽^*)

长文预警，建议先点赞再看（笑）

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恭喜提问者，你成功的发现了 Ahmed积分 (Ahmed integral) 的计算方法之一(σﾟ∀ﾟ)σ..:*☆

这个式子通过换元不难得出等价于

∫01∫01∫011(1+x2+y2+z2)2dxdydz\int_0^1\int_0^1\int_0^1\frac{1}{(1+x^2+y^2+z^2)^2}\mathrm d x\mathrm d y\mathrm d z

如果我们通过暴力强行积分算掉两个变量（比如 y,zy,z ）

可以强行分别对 y,zy, z 算，也可以作换元{x=rcos⁡ϕy=rsin⁡ϕ⇒dxdy=rdrdϕ\begin{cases} x=r \cos\phi\cr y=r \sin\phi \end{cases}\Rightarrow\mathrm dx\mathrm dy=r\mathrm dr\mathrm d\phi

下面分别演示一下（大嘘）|ू･ω･` )

∫011(k+y2)2dy=12k(k+1)+arccot(k)2k3/2\int_0^1\frac{1}{(k+y^2)^2}\text{d}y=\frac{1}{2 k (k+1)}+\frac{\mathrm{arccot}\left(\sqrt{k}\right)}{2 k^{3/2}}

然后再换一下 k↦z2+tk\mapsto z^2+t , 对 zz 求积分ヽ(￣▽￣)ﾉ

1k(k+1)=1k−1k+1=1z2+t−1z2+t+1\frac{1}{k(k+1)}=\frac{1}{k}-\frac{1}{k+1}=\frac{1}{z^2+t}-\frac{1}{z^2+t+1}

∫01{1z2+t−1z2+t+1}dz=arccot(t)t−arccot(t+1)t+1\int_0^1\left\{\frac{1}{z^2+t}-\frac{1}{z^2+t+1} \right\}\text{d}z=\frac{\mathrm{arccot}\left(\sqrt{t}\right)}{\sqrt{t}}-\frac{\mathrm{arccot}\left(\sqrt{t+1}\right)}{\sqrt{t+1}}

对于 arccot\mathrm{arccot} 那一部分，直接暴力分部即可(ﾟ▽ﾟ*)

arccot(k)k3/2=arccot(z2+t)(z2+t)3/2\frac{\mathrm{arccot}\left(\sqrt{k}\right)}{k^{3/2}}=\frac{\mathrm{arccot}\left(\sqrt{z^2+t}\right)}{(z^2+t)^{3/2}}

∫01arccot(z2+t)(z2+t)3/2dz=∫01arccot(z2+t)d(ztt+z2)=arccot(t+1)tt+1+∫01z2t(t+z2)(t+z2+1)dz=arccot(t+1)tt+1+∫01{t+1t(t+z2+1)−1t+z2}dz=arccot(t+1)t+1−arccot(t)t+2arccot(t+1)tt+1\begin{aligned} \int_0^1\frac{\mathrm{arccot}\left(\sqrt{z^2+t}\right)}{(z^2+t)^{3/2}}\text{d}z&=\int_0^1\mathrm{arccot}\left(\sqrt{z^2+t}\right)\text{d}\left(\frac{z}{t \sqrt{t+z^2}}\right)\cr &=\frac{\mathrm{arccot}\left(\sqrt{t+1}\right)}{t \sqrt{t+1}}+\int_0^1\frac{z^2}{t \left(t+z^2\right) \left(t+z^2+1\right)}\text{d}z\cr &=\frac{\mathrm{arccot}\left(\sqrt{t+1}\right)}{t \sqrt{t+1}}+\int_0^1\left\{\frac{t+1}{t \left(t+z^2+1\right)}-\frac{1}{t+z^2}\right\}\text{d}z\cr &=\frac{\mathrm{arccot}\left(\sqrt{t+1}\right)}{\sqrt{t+1}}-\frac{\mathrm{arccot}\left(\sqrt{t}\right)}{\sqrt{t}}+\frac{2 \mathrm{arccot}\left(\sqrt{t+1}\right)}{t \sqrt{t+1}} \end{aligned}

然后两式相加即可得ヾ(◍°∇°◍)ﾉﾞ

∫01∫011(t+y2+z2)2dydz=arccot(t+1)tt+1\int_0^1\int_0^1\frac{1}{(t+y^2+z^2)^2}\text{d}y\text{d}z=\frac{\mathrm{arccot}\left(\sqrt{t+1}\right)}{t\sqrt{t+1}}

然后代入

t=x2+1t=x^2+1

我们就得出了Ahmed积分的一种形式ヾ(๑╹◡╹)ﾉ"

∫01∫01∫011(1+x2+y2+z2)2dxdydz=∫01arccotx2+2(x2+1)x2+2dx\int_0^1\int_0^1\int_0^1\frac{1}{(1+x^2+y^2+z^2)^2}\mathrm d x\mathrm d y\mathrm d z =\int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x

另一种方法也是可行的而且计算量会小不少，就是做换元

{x=rcos⁡ϕy=rsin⁡ϕ⇒dxdy=rdrdϕ\begin{cases} x=r \cos\phi\cr y=r \sin\phi \end{cases}\Rightarrow\mathrm dx\mathrm dy=r\mathrm dr\mathrm d\phi

来，咱画个图ヾ(^∀^)ﾉ

z↑|x¯x¯x¯x¯x¯x¯x¯x¯x¯x¯x¯x¯|||1||)θ|r=sec⁡ϕ|||−−−−−−−−→y|O1\kern{10pt}\begin{aligned} &z\\ & \begin{aligned} &\kern{-4pt}\uparrow\\[-30pt] &| \kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}}\kern{-2pt}\bar{\color{white}{x}} \kern{-2pt}|\\[-10pt] &|\kern{42.7pt}|\\[-10pt] \end{aligned} \begin{aligned} &\kern{-56pt} \ *** all 1 \end{aligned} \\[-10pt] &\kern{2pt}\cancel{ \begin{aligned} &\kern{-4.7pt}|\kern{40pt}\\[-10pt] &\kern{-4.7pt}|\kern{15pt}\scriptsize ) \theta \ \end{aligned} } \begin{aligned} &\kern{-48.7pt}| \kern{8pt} \scriptsize r=\sec \phi \normalsize\kern{7.8pt}|\\[-10pt] &\kern{-48.7pt}| \kern{42.7pt}|\\[-10pt] &\kern{-50pt}-\kern{-4.3pt}-\kern{-4.3pt}-\kern{-4.3pt}-\kern{-4.3pt}-\kern{-4.3pt}-\kern{-4.3pt}-\kern{-4.3pt}-\kern{-4.3pt}\rightarrow y \end{aligned} \begin{aligned} &\\ &\kern{-22.8pt} | \end{aligned} \begin{aligned} &\\[-10pt] &\\[-10pt] &\ *** all \kern{-79pt}O \end{aligned}\\[-10pt] &\kern{46pt} \ *** all 1 \end{aligned}

上面这个图是用 LATEX\mathrm{\LaTeX} 码的，就凭这个你就得给我点个赞（、(◕ᴗ◕✿)

原式原式∫01∫01(原式)dydz=2∫0π/4∫0sec⁡ϕ(原式)rdrdϕ\int_0^1\int_0^1(原式)\text{d}y\text{d}z =2\int_0^{\pi/4}\int_0^{\sec \phi }(原式)r \text{d}r\text{d}\phi

我们可以得到(ﾉﾟ∀ﾟ)ﾉ

∫01∫011(a+y2+z2)2dydz=2∫0π/4∫0sec⁡ϕrdrdϕ(a+r2)2\int_0^1\int_0^1\frac{1}{(a+y^2+z^2)^2}\mathrm dy\mathrm dz=2\int_0^{\pi/4}\int_0^{\sec \phi } \frac{r \text{d}r\text{d}\phi}{(a+r^2)^2}

然后先对 rr 算一下不定积分（凑微分）ヾ(^Д^*)/

∫r(r2+a)2dr=12∫1(r2+a)2d(r2)=−12(a+r2)\int \frac{r}{(r^2+a)^2}\text{d}r=\frac{1}{2}\int \frac{1}{(r^2+a)^2}\text{d}(r^2)=-\frac{1}{2 \left(a+r^2\right)}

所以定积分(✧◡✧)

∫0sec⁡ϕr(r2+a)2dr=12(a+1)−12(a+sec2⁡ϕ)\int_0^{\sec \phi} \frac{r}{(r^2+a)^2}\text{d}r=\frac{1}{2 (a+1)}-\frac{1}{2 \left(a+\sec ^2\phi\right)}

然后对 ϕ\phi 积分，得到同样的结果(*´・ｖ・)

∫01∫01∫011(1+x2+y2+z2)2dxdydz=∫01arccotx2+2(x2+1)x2+2dx\int_0^1\int_0^1\int_0^1\frac{1}{(1+x^2+y^2+z^2)^2}\mathrm d x\mathrm d y\mathrm d z =\int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x

其实Ahmed积分的一般形式是这个：(⁎˃ᴗ˂⁎)

A(p,q,r)=def∫01B(q(px2+1))(r+1)px2+1dxA(p,q,r)\xlongequal{\text{def}}\int_0^1\frac{B(q(px^2+1))}{(r+1)px^2+1}\text{d}x

其中

B(x)=arctan⁡xx=∑n≥0(−)n2n+1xnB(x)=\frac{\arctan \sqrt{x}}{\sqrt{x}}=\sum_{n\geq0}\frac{(-)^n}{2n+1}x^n

其中比较常见的一类就是Ｏ(≧▽≦)Ｏ

∫01arctan⁡x2+a(x2+1)x2+adx=A(1a,a,a−1)\int_0^1\frac{\arctan\sqrt{x^2+a}}{\left(x^2+1\right) \sqrt{x^2+a}}\text{d}x=A\left( \frac{1}{a},a,a-1 \right)

在这一题里面 a=2a=2 ( 因为

arctan+arccot=π/2\arctan+\text{arccot}=\pi/2 本质是一样的)

问题转化为了这个φ(≧ω≦*)♪

∫01arccotx2+2(x2+1)x2+2dx\int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x

（很明显三角换元什么的常规技巧似乎不太行，那咋办？）Σ (ﾟДﾟ;)

觉得哪一块难处理？(꒪Д꒪)ノ

是不是( • ̀ω•́ )✧

arccotx2+2x2+2\frac{\mathrm{arccot}\sqrt{x^2+2}}{\sqrt{x^2+2}}

这一部分似乎好丑啊，怎么把它整漂亮一点？o(￣▽￣)ｄ

想到

∫011x2+y2dy=1xarctan⁡1x=1xarccotx\int_0^1 \frac{1}{x^2+y^2} \mathrm dy=\frac{1}{x}\arctan\frac{1}{x}=\frac{1}{x}\mathrm{arccot}x

似乎给我们提供了一条路( ･´ω`･ )

arccotx2+2x2+2=∫01dy(x2+2)2+y2=∫01dyx2+y2+2\frac{\mathrm{arccot}\sqrt{x^2+2}}{\sqrt{x^2+2}}=\int_0^1\frac{\mathrm dy}{(\sqrt{x^2+2})^2+y^2}=\int_0^1\frac{\text{d}y}{x^2+y^2+2}

然后回到原式╰(￣▽￣)╭

∫01arccotx2+2(x2+1)x2+2dx=∫01∫011(x2+1)(x2+y2+2)dydx\int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x=\int_0^1\int_0^1\frac{1}{(x^2+1)(x^2+y^2+2)}\text{d}y\text{d}x

似乎有点对称，我们把 x,yx,y 互换(*￣3￣)╭

∫01arccotx2+2(x2+1)x2+2dx=∫01∫011(y2+1)(x2+y2+2)dydx\int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x=\int_0^1\int_0^1\frac{1}{(y^2+1)(x^2+y^2+2)}\text{d}y\text{d}x

不如我们相加除以二吧（ㄟ( ▔, ▔ )ㄏ ）

∫01arccotx2+2(x2+1)x2+2dx=12∫01∫01{1x2+1+1y2+1}dydxx2+y2+2\int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x=\frac12\int_0^1\int_0^1\left\{\frac{1}{x^2+1}+\frac{1}{y^2+1}\right\}\frac{\text{d}y\text{d}x}{x^2+y^2+2}

然后发现后面可以约分？(=ﾟωﾟ)ﾉ

∫01arccotx2+2(x2+1)x2+2dx=12∫01∫01dxdy(x2+1)(y2+1)\int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x=\frac12\int_0^1\int_0^1\frac{\text{d}x\text{d}y}{(x^2+1)(y^2+1)}

然后爽了，我们可以分开来算(灬°ω°灬)

∫01arccotx2+2(x2+1)x2+2dx=12(∫01dxx2+1)⋅(∫01dyy2+1)=12⋅π4⋅π4=π232\begin{aligned} \int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x&=\frac12\left( \int_0^1\frac{\text{d}x}{x^2+1} \right)\cdot\left( \int_0^1\frac{\text{d}y}{y^2+1} \right)\cr &=\frac12\cdot \frac\pi 4\cdot \frac\pi 4=\frac{\pi^2}{32} \end{aligned}

似乎就搞定了（？）●ヽ(ﾟ∀ﾟ)ﾉ●

没错确实搞定了∑d(*ﾟ∀ﾟ*)

我们得出结论（◑▽◐）

∫01∫01∫011(1+x2+y2+z2)2dxdydz=π232\color{red}{\boxed{\large\color{black}{\int_0^1\int_0^1\int_0^1\frac{1}{(1+x^2+y^2+z^2)^2}\mathrm d x\mathrm d y\mathrm d z =\frac{\pi^2}{32}}}}

就这么结束是不是有点仓促（(Ｔ▽Ｔ)）

再写一下推论吧(^o^)／

由于

∫011(x2+1)x2+2dx=x↦tan⁡t∫0π/4dtsec2⁡t+1=∫0π/4cos⁡tdtcos2⁡t+1=∫0π/4dsin⁡t2−sin2⁡t=π6\begin{aligned} \int_0^1\frac{1}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x&\xlongequal{x\mapsto\tan t}\int_0^{\pi/4}\frac{\text{d}t}{\sqrt{\sec^2t+1}}\cr &=\int_0^{\pi/4}\frac{ \cos t\text{d}t}{\sqrt{\cos^2t+1}}\cr &=\int_0^{\pi/4}\frac{ \text{d}\sin t}{\sqrt{2-\sin^2t}}\cr &=\frac{\pi}{6} \end{aligned}

所以

A(12,2,1)=∫01arctan⁡x2+2(x2+1)x2+2dx=∫01π/2−arccotx2+2(x2+1)x2+2dx=π2∫011(x2+1)x2+2dx−∫01arccotx2+2(x2+1)x2+2dx=π2⋅π6−π232=5π296\begin{aligned} A\left( \frac{1}{2},2,1 \right)&=\int_0^1\frac{\arctan\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x\cr &=\int_0^1\frac{\pi/2-\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x\cr &=\frac\pi 2\int_0^1\frac{1}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x-\int_0^1\frac{\mathrm{arccot}\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x\cr &=\frac{\pi}{2}\cdot\frac \pi 6-\frac{\pi^2}{32}\cr &=\frac{5\pi^2}{96} \end{aligned}

A(12,2,1)=∫01arctan⁡x2+2(x2+1)x2+2dx=5π296\color{red}{\boxed{\color{black}{\large A\left( \frac{1}{2},2,1 \right)=\int_0^1\frac{\arctan\sqrt{x^2+2}}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x=\frac{5\pi^2}{96}}}}

理论上来说应该是结束了，但是我还想水一水（ε≡ (๑>₃<)

比如除了这种方法还有其他方法没有？[・_・?]

当然是有的（笑），否则我不会继续水(*๓´╰╯`๓)

我们还可以含参解决，比如我们令

I(a)=∫01arccot(ax2+2)(x2+1)x2+2dx\mathscr{I}(a)=\int_0^1\frac{\mathrm{arccot}\left( a\sqrt{x^2+2} \right)}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x

我们要求的就是

I=I(1)=I(a)=∫01arccot(x2+2)(x2+1)x2+2dx\mathscr{I}=\mathscr{I}(1)=\mathscr{I}(a)=\int_0^1\frac{\mathrm{arccot}\left(\sqrt{x^2+2} \right)}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x

现在我们对 aa 求导

I′(a)=∂∂a∫01arccot(ax2+2)(x2+1)x2+2dx=−∫011(x2+1)(a2(x2+2)+1)dx=∫01{a2(a2+1)(a2x2+2a2+1)−1(a2+1)(x2+1)}dx=a(a2+1)2a2+1arctan⁡(a2a2+1)−π4(a2+1)\begin{aligned} \mathscr{I}^\prime(a)&=\frac{\partial}{\partial a}\int_0^1\frac{\mathrm{arccot}\left( a\sqrt{x^2+2} \right)}{\left(x^2+1\right) \sqrt{x^2+2}}\text{d}x\cr &=-\int_0^1\frac{1}{\left(x^2+1\right) \left(a^2 \left(x^2+2\right)+1\right)}\text{d}x\cr &=\int_0^1\left\{ \frac{a^2}{\left(a^2+1\right) \left(a^2 x^2+2 a^2+1\right)}-\frac{1}{\left(a^2+1\right) \left(x^2+1\right)} \right\}\text{d}x\cr &=\frac{a }{\left(a^2+1\right) \sqrt{2 a^2+1}}\arctan\left(\frac{a}{\sqrt{2 a^2+1}}\right)-\frac{\pi }{4 \left(a^2+1\right)}\cr \end{aligned}

哦哟，有希望，积分回去？

似乎不大行，∑(O_O；)

试试倒代换吧

I(1)=I(∞)−∫1∞[a(a2+1)2a2+1arctan⁡(a2a2+1)−π4(a2+1)]da=π216−∫1∞a(a2+1)2a2+1arctan⁡(a2a2+1)da⏟J\begin{aligned} \mathscr{I}(1)&=\mathscr{I}(\infty)-\int_1^\infty\left[ \frac{a }{\left(a^2+1\right) \sqrt{2 a^2+1}}\arctan\left(\frac{a}{\sqrt{2 a^2+1}}\right)-\frac{\pi }{4 \left(a^2+1\right)} \right]\text{d}a\cr &=\frac{\pi^2}{16}-\underset{\mathscr{J}}{\underbrace{\int_1^\infty\frac{a }{\left(a^2+1\right) \sqrt{2 a^2+1}}\arctan\left(\frac{a}{\sqrt{2 a^2+1}}\right)\text{d}a}} \end{aligned}

然后对 J\mathscr{J} 使用倒代换，然后我们惊奇的发现

J=∫01arccota2+2(a2+1)a2+2da\mathscr{J}=\int_0^1\frac{\mathrm{arccot}\sqrt{a^2+2}}{\left(a^2+1\right) \sqrt{a^2+2}}\text{d}a

Amazing!~~~(@[]@!!)

所以说

I=J=12⋅π216=π232\mathscr{I}=\mathscr{J}=\frac{1}{2}\cdot\frac{\pi^2}{16}=\frac{\pi^2}{32}

我们从另一个角度证明了我们的结果是正确的

啊终于写完了，不点个赞再走嘛？(￣▽￣)~*

双击屏幕支持我哦