ここで勉強すれば数学検定１級の壁は超えられるか。

D={$(x,y)$|$x^{2}+y^{2}\leqq 1$｝とおくとき、次の二重積分の値を求める。

$\iint _{D}\sqrt {\dfrac {1-x^{2}-y^{2}}{1+x^{2}+y^{2}}}dxdy$

$x=r\cos \theta ,y=r\sin \theta$とおくと、$dxdy=rdtd\theta$

求める積分　I=$\iint _{D}\sqrt {\dfrac {1-x^{2}-y^{2}}{1+x^{2}+y^{2}}}=\int ^{2\pi }_{0}d\theta\int ^{1}_{0}\sqrt {\dfrac {1-r^{2}}{1+r^{2}}}\cdot rdr$

ここで

$r^{2}=t\Rightarrow 2r\cdot dr=dt\Rightarrow dr=\dfrac {dt}{2r}$

$r:0\rightarrow 1\Rightarrow r^{2}:0\rightarrow 1\Rightarrow t:0\rightarrow 1$

なので

$I=2\pi \int ^{1}_{0}\dfrac {r}{2r}\sqrt {\dfrac {1-t}{1+t}}dt=\pi \int ^{1}_{0}\sqrt {\dfrac {1-t}{1+t}}dt$

$=\pi \int ^{1}_{0}\dfrac {1-t}{\sqrt {1-t^{2}}}dt=\pi \left[ \int ^{1}_{0}\left( \dfrac {1}{\sqrt {1-t^{2}}}-\dfrac {t}{\sqrt {1-t^{2}}}\right) dt\right]$

$=\pi \left\{ \int ^{1}_{0}\left( \sin ^{-1}t\right)^{'} dt+\int ^{1}_{0}\left( \sqrt {1-t^{2}}\right) ^{'}dt\right\}$

$=\pi \left\{ \left[ \sin ^{-1}t\right] ^{1}_{0}+\left[ \sqrt {1-t^{2}}\right] ^{1}_{0}\right\}$

$=\pi \left\{ \left( \dfrac {\pi }{2}-0\right) +\left( 0-1\right) \right\} = \dfrac {\pi ^{2}}{2}-\pi$

$\dfrac {\pi ^{2}}{2}-\pi$　・・・答え