问题及解答

设 $D=\{(x,y)\in\mathbb{R}^2\mid 1\leqslant x^2+y^2\leqslant 2\}$, 比较下面三个二重积分的大小

\[
M=\iint_{D}\ln(x^2+y^2)\mathrm{d}x\mathrm{d}y,
\]

\[
N=\iint_{D}\bigl[\ln(x^2+y^2)\bigr]^2\mathrm{d}\sigma,
\]

\[
P=\iint_{D}(x^2+y^2-1)\mathrm{d}x\mathrm{d}y.
\]

Answer:   $N < M < P$.

Hint: we have the inequality $\ln(1+t) < t$ for $t > 0$.

如果 $D$ 改为

$D=\{(x,y)\in\mathbb{R}^2\mid 1\leqslant x^2+y^2\leqslant 4\}$, 那就只能手算了. 此时答案是:  $M < N < P$.

具体的, 

\[
\begin{aligned}
M&=\pi(8\ln 2-3),\\
N&=4\pi\bigl[4(\ln 2)^2-4\ln 2+\frac{3}{2}\bigr],\\
P&=\frac{9}{2}\pi.
\end{aligned}
\]

令 $t=x^2+y^2$, 则由条件 $t\in[1,2]$, 推出 $0\leqslant\ln t\leqslant \ln 2 < 1$. 因此

\[\iint_D\bigl[\ln(x^2+y^2)\bigr]^2\mathrm{d}x\mathrm{d}y < \iint_D\ln(x^2+y^2)\mathrm{d}\sigma.\]

另一方面, 由于有不等式 $\ln(1+x) < x$, $\forall\ x > 0$. 因此

\[\ln(1+(x^2+y^2-1)) < x^2+y^2-1,\]

即

\[\iint\ln(x^2+y^2)\mathrm{d}\sigma < \iint_D (x^2+y^2-1)\mathrm{d}x\mathrm{d}y.\]

总结有:  $N < M < P$.

若要详细计算 $M,N,P$ 三个值, 则可使用极坐标法. 令 $x=r\cos\theta$, $y=r\sin\theta$. 根据条件,

\[D=\{(r,\theta)\mid r\in[1,\sqrt{2}],\ \theta\in[0,2\pi]\}\]

\[
\begin{split}
M&=\int_0^{2\pi}\mathrm{d}\theta\int_1^{\sqrt{2}}(\ln r^2)r\mathrm{d}r\\
&=2\pi\cdot\int_1^{\sqrt{2}}2r\ln r\mathrm{d}r\\
&=2\pi\cdot\int_1^{\sqrt{2}}\ln r\mathrm{d}r^2\\
&=2\pi\cdot\biggl[r^2\ln r\biggr|_1^{\sqrt{2}}-\int_1^{\sqrt{2}}r^2\mathrm{d}\ln r\biggr]\\
&=2\pi\cdot\biggl[2\ln\sqrt{2}-\int_1^{\sqrt{2}}r\mathrm{d}r\biggr]\\
&=2\pi(\ln 2-\frac{1}{2})
\end{split}
\]

\[
\begin{split}
N&=\int_0^{2\pi}\mathrm{d}\theta\int_1^{\sqrt{2}}(\ln r^2)^2 r\mathrm{d}r\\
&=2\pi\cdot\int_1^{\sqrt{2}}4r(\ln r)^2\mathrm{d}r\\
&=4\pi\cdot\int_1^{\sqrt{2}}(\ln r)^2\mathrm{d}r^2\\
&=4\pi\cdot\biggl[r^2(\ln r)^2\biggr|_{1}^{\sqrt{2}}-\int_1^{\sqrt{2}}r^2\mathrm{d}(\ln r)^2\biggr]\\
&=4\pi\cdot\biggl[2(\ln\sqrt{2})^2-\int_1^{\sqrt{2}}2r\ln r\mathrm{d}r\biggr]\\
&=4\pi\cdot\biggl[\frac{1}{2}(\ln 2)^2-\int_1^{\sqrt{2}}\ln r\mathrm{d}r^2\biggr]\\
&=4\pi\cdot\biggl[\frac{1}{2}(\ln 2)^2-\biggl(r^2\ln r\biggr|_{1}^{\sqrt{2}}-\int_1^{\sqrt{2}}r^2\mathrm{d}\ln r\biggr)\biggr]\\
&=4\pi\cdot\biggl[\frac{1}{2}(\ln 2)^2-\Bigl(2\ln\sqrt{2}-\frac{1}{2}r^2\biggr|_{1}^{\sqrt{2}}\Bigr)\biggr]\\
&=4\pi\cdot\Bigl(\frac{1}{2}(\ln 2)^2-\ln 2+\frac{1}{2}\Bigr)\\
&=2\pi\cdot(\ln 2-1)^2
\end{split}
\]

\[
\begin{split}
P&=\int_0^{2\pi}\mathrm{d}\theta\int_1^{\sqrt{2}}(r^2-1)r\mathrm{d}r\\
&=2\pi\cdot\int_1^{\sqrt{2}}(r^3-r)\mathrm{d}r\\
&=2\pi\cdot\bigl(\frac{1}{4}r^4-\frac{1}{2}r^2\bigr)\biggr|_{1}^{\sqrt{2}}\\
&=\frac{\pi}{2}
\end{split}
\]

最终计算得

\[
\begin{aligned}
M&=2\pi(\ln 2-\frac{1}{2}),\\
N&=2\pi(1-\ln 2)^2,\\
P&=\frac{\pi}{2}.
\end{aligned}
\]