x2a2+y2b2=1(a,b>0)\frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \qquad (a,b>0)a2x2+b2y2=1(a,b>0) y=b2(1−x2a2)y=\sqrt{b^2\left(1-\frac{x^2}{a^2}\right)}y=b2(1−a2x2) 4∫0ab2(1−x2a2)dx=4ba∫0aa2−x2 dx=4ba[12(xa2−x2+a2sin−1xa)]0a=4ba⋅a22⋅π2=πab4\int_0^a\sqrt{b^2\left(1-\frac{x^2}{a^2}\right)}dx=\frac{4b}{a}\int_0^a\sqrt{a^2-x^2}\,dx=\frac{4b}{a}\left[\frac{1}{2}\left(x\sqrt{a^2-x^2}+a^2\sin^{-1}\frac{x}{a}\right)\right]_0^a=\frac{4b}{a}\cdot\frac{a^2}{2}\cdot\frac{\pi}{2}=\pi ab4∫0ab2(1−a2x2)dx=a4b∫0aa2−x2dx=a4b[21(xa2−x2+a2sin−1ax)]0a=a4b⋅2a2⋅2π=πab また、x=acostx=a\cos tx=acost 、y=bsinty=b\sin ty=bsint は {b2x2=a2b2cos2t①a2y2=a2b2sin2t②\begin{cases} b^2x^2=a^2b^2\cos^2 t & \text{①} \\ a^2y^2=a^2b^2\sin^2 t & \text{②} \end{cases}{b2x2=a2b2cos2ta2y2=a2b2sin2t①② ①+② b2x2+a2y2=a2b2b^2x^2+a^2y^2=a^2b^2b2x2+a2y2=a2b2 x2a2+y2b2=1\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1a2x2+b2y2=1 から同値になる。