一类较复杂的三重积分
[i=s] 本帖最后由 青青子衿 于 2018-11-17 16:48 编辑 [/i]\[ \int_0^c\int_0^b\int_0^a\left(x^2+y^2\right)\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z \]
\begin{align*}
\color{red}{\frac{abc\left(43a^2+43b^2+10c^2\right)}{360}\sqrt{a^{\overset{\,}{2}}+b^2+c^2}+\frac{c^6}{45}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)}\\
\color{red}{-\frac{2a^6}{45}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{2b^6}{45}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)}\\
\color{red}{+\frac{ac\left(27a^4+10a^2c^2-9c^4\right)}{360}\operatorname{arsinh}\left(\frac{b}{\sqrt{a^2+c^2}}\right)}\\
\color{red}{+\frac{bc\left(27b^4+10b^2c^2-9c^4\right)}{360}\operatorname{arsinh}\left(\frac{a}{\sqrt{b^2+c^2}}\right)}\\
\color{red}{+\frac{ab\left(9a^4+10a^2b^2+9b^4\right)}{90}\operatorname{arsinh}\left(\frac{c}{\sqrt{a^2+b^2}}\right)}\\
\end{align*}
\begin{align*}
\frac{abc\left(43a^2+43b^2+10c^2\right)}{360}\sqrt{a^{\overset{\,}{2}}+b^2+c^2}+\frac{c^6}{45}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)\\
-\frac{2a^6}{45}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)-\frac{2b^6}{45}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)\\
+\frac{ac\left(27a^4+10a^2c^2-9c^4\right)}{360}\ln\left(\frac{b}{\sqrt{a^2+c^2}}+\sqrt{1+\frac{b^2}{a^2+c^2}}\,\right)\\
+\frac{bc\left(27b^4+10b^2c^2-9c^4\right)}{360}\ln\left(\frac{a}{\sqrt{b^2+c^2}}+\sqrt{\frac{a^2}{b^2+c^2}+1}\,\right)\\
+\frac{ab\left(9a^4+10a^2b^2+9b^4\right)}{90}\ln\left(\frac{c}{\sqrt{a^2+b^2}}+\sqrt{1+\frac{c^2}{a^2+b^2}}\,\right)\\
\end{align*}
\[ \iiint\limits_{x^\overset{\,}{2}+y^2+z^2\le a^2} \left(x^2+y^2\right)\sqrt{x^2+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z \]
\[ \dfrac{4\pi}{9}a^6 \]
\[ \int_0^c\int_0^b\int_0^a\,z^2\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y{\rm\,d}z \]
\begin{align*}
\color{red}{\frac{abc\left(5a^2+5b^2+38c^2\right)}{360}\sqrt{a^{\overset{\,}{2}}+b^2+c^2}-\frac{c^6}{18}\arctan\left(\frac{ab}{c\sqrt{a^2+b^2+c^2}}\right)}\\
\color{red}{+\frac{a^6}{90}\arctan\left(\frac{bc}{a\sqrt{a^2+b^2+c^2}}\right)+\frac{b^6}{90}\arctan\left(\frac{ac}{b\sqrt{a^2+b^2+c^2}}\right)}\\
\color{red}{+\frac{ac^3\left(5a^2+9c^2\right)}{90}\operatorname{arsinh}\left(\frac{b}{\sqrt{a^2+c^2}}\right)}\\
\color{red}{+\frac{bc^3\left(5b^2+9c^2\right)}{90}\operatorname{arsinh}\left(\frac{a}{\sqrt{b^2+c^2}}\right)}\\
\color{red}{-\frac{ab\left(9a^4+10a^2b^2+9b^4\right)}{360}\operatorname{arsinh}\left(\frac{c}{\sqrt{a^2+b^2}}\right)}\\
\end{align*} [i=s] 本帖最后由 青青子衿 于 2019-1-1 17:17 编辑 [/i]
[b]回复 [url=http://kuing.orzweb.net/redirect.php?goto=findpost&pid=28665&ptid=5708]1#[/url] [i]青青子衿[/i] [/b]
\[ \int_0^b\int_0^a\left(x^2+y^2\right)\sqrt{x^{\overset{\,}{2}}+y^2+z^2}{\rm\,d}x{\rm\,d}y \]
\begin{align*}
\color{red}{\frac{7ab}{120}\left(3a^2+3b^2+2z^2\right)\sqrt{a^{\overset{\,}{2}}+b^2+z^2}}\\
\color{red}{+\frac{2}{15}z^5\arctan\left(\frac{ab}{z\sqrt{a^2+b^2+z^2}}\right)}\\
\color{red}{+\frac{a}{120}\left(9a^4+10a^2z^2-15z^4\right)\operatorname{arsinh}\left(\frac{b}{\sqrt{a^2+z^2}}\right)}\\
\color{red}{+\frac{b}{120}\left(9b^4+10b^2z^2-15z^4\right)\operatorname{arsinh}\left(\frac{a}{\sqrt{b^2+z^2}}\right)}\\
\end{align*}[code]\int_0^a\int_0^{a-x}\int_0^{a-x-y}\sqrt{x^2+y^2+z^2}dzdydx\\
3\int_0^{\frac{\pi}{4}}\int_{\arctan\left(\sec\varphi\right)}^{\frac{\pi}{2}}\int_0^{\frac{a}{\left(\sin\theta\right)\left(\cos\varphi\right)+\left(\sin\theta\right)\left(\sin\varphi\right)+\cos\theta}}r^3\left(\sin\theta\right)drd\theta d\varphi
[/code]...[code]\int_0^{1-x_1-y_1}\sqrt{x_1^2+y_1^2+z_1^2}dz_1
\frac{1}{2}\left(1-x_1-y_1\right)\sqrt{x_1^2+y_1^2+\left(1-x_1-y_1\right)^2}+\frac{1}{2}\left(x_1^2+y_1^2\right)\ln\left(\frac{1-x_1-y_1}{\sqrt{x_1^2+y_1^2}}+\sqrt{\frac{\left(1-x_1-y_1\right)^2}{x_1^2+y_1^2}+1}\right)[/code]{:time:}
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