【高校数学Ⅲ】6-1-2 体積｜要点まとめ

直角二等辺三角形\(PQR\)の面積を\(S(x)\)とすると、
\(\displaystyle S(x)=\frac{1}{2}\{x(1-x)\}^2\)
\(\displaystyle =\frac{1}{2}x^2(1-x)^2\)
よって、求める体積\(V\)は
\(\displaystyle V=\int_0^1 S(x)dx\)
\(\displaystyle =\int_0^1 \frac{1}{2}x^2(1-x)^2dx\)
\(\displaystyle =\frac{1}{2}\int_0^1 (x^4-2x^3+x^2)dx\)
\(\displaystyle =\frac{1}{2}\left[\frac{1}{5}x^5-\frac{1}{2}x^4+\frac{1}{3}x^3\right]_0^1\)
\(\displaystyle =\frac{1}{2}\left(\frac{1}{5}-\frac{1}{2}+\frac{1}{3}\right)\)
\(\displaystyle =\frac{1}{60}\)

求める体積\(V\)は
\(\displaystyle V=\pi\int_0^\pi y^2dx\)
\(\displaystyle =\pi\int_0^\pi \sin^2xdx\)
\(\displaystyle =\pi\int_0^\pi \frac{1-\cos2x}{2}dx\)
\(\displaystyle =\frac{\pi}{2}\left[x-\frac{1}{2}\sin2x\right]_0^\pi\)
\(\displaystyle =\frac{\pi^2}{2}\)

\(x^2+(y-2)^2=1\)より、\(y=2\pm\sqrt{1-x^2}\)
\(y_1=2+\sqrt{1-x^2}\)の回転体の体積を\(V_1\)
\(y_2=2-\sqrt{1-x^2}\)の回転体の体積を\(V_2\)
求める体積\(V\)は
\(\displaystyle V=V_1-V_2\)
\(\displaystyle =\pi\int_{-1}^1 (2+\sqrt{1-x^2})^2dx\)
\(\displaystyle \ \ \ -\pi\int_{-1}^1 (2-\sqrt{1-x^2})^2dx\)
\(\displaystyle =\pi\int_{-1}^1 4・2\sqrt{1-x^2}dx\)
\(\displaystyle =16\pi\int_0^1 \sqrt{1-x^2}dx\)
\(x=\sin\theta\)とおくと、\(dx=\cos\theta d\theta\)なので、
\(\displaystyle =16\pi\int_0^{\frac{\pi}{2}} \cos\theta・\cos\theta d\theta\)
\(\displaystyle =16\pi\int_0^{\frac{\pi}{2}} \frac{1+\cos2\theta}{2} d\theta\)
\(\displaystyle =8\pi\left[\theta+\frac{1}{2}\sin2\theta\right]_0^{\frac{\pi}{2}}\)
\(=4\pi^2\)