四分之一椭圆上二重积分的计算

计算 $$bex I=iint_Dfrac{rd xrd y}{sqrt{1-frac{x^2}{a^2}-frac{y^2}{b^2}}cdot sex{x^2+y^2+1-frac{x^2}{a^2}-frac{y^2}{b^2}}^{3/2}}, eex$$ 其中 $a,b>0$, $$bex D=sed{(x,y);frac{x^2}{a^2}+frac{y^2}{b^2}leq 1, xgeq 0, ygeq 0}. eex$$

解答: 作极坐标变换 $$bex x=arcos theta,quad y=brsintheta, eex$$ 则 $$beex bea I&=int_0^{pi/2}rd thetaint_0^1 frac{r}{sqrt{1-r^2}cdot sex{a^2r^2cos^2theta+b^2r^2sin^2theta+1-r^2}^{3/2}}rd r\ &=frac{1}{2}int_0^{pi/2}rd theta int_0^1 frac{rd s}{sqrt{1-s}cdot sez{1+(a^2cos^2theta+b^2sin^2theta-1)s}^{3/2}}quadsex{r^2=s}. eea eeex$$ 为此, 计算 $$beex bea J(c)&=int_0^1 frac{rd s} {sqrt{1-s}cdot(1+cs)^{3/2}}\ &=int_0^1frac{2trd t} {tsex{1+c-ct^2}^{3/2}}quadsex{sqrt{1-s}=t}\ &=2int_0^1frac{rd t}{(1+c-ct^2)^{3/2}}. eea eeex$$ 若 $c=0$, 则 $J(0)=2$. 若 $c>0$, 则 $$beex bea J(c)&=2int_0^{arcsinsqrt{frac{c}{1+c}}} frac{1}{(1+c)^{3/2}cos^3phi}cdot sqrt{frac{1+c}{c}}cosphird phi\ &=frac{2}{sqrt{c}(1+c)}tansex{arcsinsqrt{frac{c}{1+c}}}\ &frac{2}{1+c}. eea eeex$$ 若 $c<0$, 则 $$beex bea J(c)&=2int_0^1 frac{rd t}{(1+c+|c|t^2)^{3/2}}\ &=2int_0^{arctansqrt{frac{|c|}{1+c}}} frac{1}{(1+c)^{3/2}sec^3phi}cdot sqrt{frac{1+c}{|c|}}sec^2phird phi\ &=frac{2}{sqrt{|c|}(1+c)}sinsex{arctansqrt{frac{|c|}{1+c}}} quadsex{sqrt{|c|}^2+sqrt{1+c}=1}\ &=frac{2}{1+c}. eea eeex$$ 综上, $$bex c>-1ra J(c)=int_0^1 frac{rd s} {sqrt{1-s}cdot(1+cs)^{3/2}}=frac{2}{1+c}, eex$$ 我们现在可以结束这 $I$ 的计算: $$beex bea I&=frac{1}{2}int_0^{pi/2} J(a^2cos^2theta+b^2sin^2theta-1)rd theta\ &=frac{1}{2}int_0^{pi/2}frac{2}{a^2cos^2theta+b^2sin^2theta}rd theta\ &=int_0^{pi/2} frac{cos^2theta+sin^2theta}{a^2cos^2theta+b^2sin^2theta}rd theta\ &=int_0^{pi/2} frac{rd tan theta}{a^2tan^2theta+b^2}\ &=frac{1}{ab}left.arctansex{frac{a}{b}tantheta}right|_{theta=0}^{theta=pi/2}\ &=frac{pi}{2ab}. eea eeex$$