计算 $$\bex I=\iint_D\frac{\rd x\rd y}{\sqrt{1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}\cdot \sex{x^2+y^2+1-\frac{x^2}{a^2}-\frac{y^2}{b^2}}^{3/2}}, \eex$$ 其中 $a,b>0$, $$\bex D=\sed{(x,y);\frac{x^2}{a^2}+\frac{y^2}{b^2}\leq 1,\ x\geq 0,\ y\geq 0}. \eex$$post
解答: 做极坐标变换 $$\bex x=ar\cos \theta,\quad y=br\sin\theta, \eex$$ 则 $$\beex \bea I&=\int_0^{\pi/2}\rd \theta\int_0^1 \frac{r}{\sqrt{1-r^2}\cdot \sex{a^2r^2\cos^2\theta+b^2r^2\sin^2\theta+1-r^2}^{3/2}}\rd r\\ &=\frac{1}{2}\int_0^{\pi/2}\rd \theta \int_0^1 \frac{\rd s}{\sqrt{1-s}\cdot \sez{1+(a^2\cos^2\theta+b^2\sin^2\theta-1)s}^{3/2}}\quad\sex{r^2=s}. \eea \eeex$$ 为此, 计算 $$\beex \bea J(c)&=\int_0^1 \frac{\rd s} {\sqrt{1-s}\cdot(1+cs)^{3/2}}\\ &=\int_0^1\frac{2t\rd t} {t\sex{1+c-ct^2}^{3/2}}\quad\sex{\sqrt{1-s}=t}\\ &=2\int_0^1\frac{\rd t}{(1+c-ct^2)^{3/2}}. \eea \eeex$$ 若 $c=0$, 则 $J(0)=2$. 若 $c>0$, 则 $$\beex \bea J(c)&=2\int_0^{\arcsin\sqrt{\frac{c}{1+c}}} \frac{1}{(1+c)^{3/2}\cos^3\phi}\cdot \sqrt{\frac{1+c}{c}}\cos\phi\rd \phi\\ &=\frac{2}{\sqrt{c}(1+c)}\tan\sex{\arcsin\sqrt{\frac{c}{1+c}}}\\ &\frac{2}{1+c}. \eea \eeex$$ 若 $c<0$, 则 $$\beex \bea J(c)&=2\int_0^1 \frac{\rd t}{(1+c+|c|t^2)^{3/2}}\\ &=2\int_0^{\arctan\sqrt{\frac{|c|}{1+c}}} \frac{1}{(1+c)^{3/2}\sec^3\phi}\cdot \sqrt{\frac{1+c}{|c|}}\sec^2\phi\rd \phi\\ &=\frac{2}{\sqrt{|c|}(1+c)}\sin\sex{\arctan\sqrt{\frac{|c|}{1+c}}} \quad\sex{\sqrt{|c|}^2+\sqrt{1+c}=1}\\ &=\frac{2}{1+c}. \eea \eeex$$ 综上, $$\bex c>-1\ra J(c)=\int_0^1 \frac{\rd s} {\sqrt{1-s}\cdot(1+cs)^{3/2}}=\frac{2}{1+c}, \eex$$ 咱们如今能够结束这 $I$ 的计算: $$\beex \bea I&=\frac{1}{2}\int_0^{\pi/2} J(a^2\cos^2\theta+b^2\sin^2\theta-1)\rd \theta\\ &=\frac{1}{2}\int_0^{\pi/2}\frac{2}{a^2\cos^2\theta+b^2\sin^2\theta}\rd \theta\\ &=\int_0^{\pi/2} \frac{\cos^2\theta+\sin^2\theta}{a^2\cos^2\theta+b^2\sin^2\theta}\rd \theta\\ &=\int_0^{\pi/2} \frac{\rd \tan \theta}{a^2\tan^2\theta+b^2}\\ &=\frac{1}{ab}\left.\arctan\sex{\frac{a}{b}\tan\theta}\right|_{\theta=0}^{\theta=\pi/2}\\ &=\frac{\pi}{2ab}. \eea \eeex$$blog