問
\(区間\;0\leq x \leq 1 において y=xとy=x^2で囲まれた領域をy=xを軸として回転させた時の体積\)
\(y=x,\;y=x^2\)を\(-\frac{\pi}{4}\)回転させる
求めたい\(-\frac{\pi}{4}\)回転させた側を\((x, y)\)とすると，それを\(\frac{\pi}{4}\)回転させた結果として\((X,Y)\)として考える．
$$\begin{eqnarray}
\begin{bmatrix}
X \\
Y
\end{bmatrix}
&=&
\begin{bmatrix}
\cos{\left(\frac{\pi}{4}\right)} & -\sin{\left(\frac{\pi}{4}\right)} \\
\sin{\left(\frac{\pi}{4}\right)} & \cos{\left(\frac{\pi}{4}\right)}
\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}
\\&=&
\begin{bmatrix}
\frac{\sqrt{2}}{2} & -\frac{\sqrt{2}}{2} \\
\frac{\sqrt{2}}{2} & \frac{\sqrt{2}}{2}
\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}
\\&=&
\frac{\sqrt{2}}{2}
\begin{bmatrix}
1 & -1 \\
1 & 1
\end{bmatrix}\begin{bmatrix}x \\ y\end{bmatrix}
\end{eqnarray}$$
$$\left\{\begin{eqnarray}
X&=&\frac{\sqrt{2}}{2}\left(x-y\right)
\\Y&=&\frac{\sqrt{2}}{2}\left(x+y\right)
\end{eqnarray}\right.$$
$$\left\{\begin{eqnarray}
X&=&\alpha\left(x-y\right)
\\Y&=&\alpha\left(x+y\right)
\end{eqnarray}\right.\;\cdots\;\alpha=\frac{\sqrt{2}}{2}$$
$$\begin{eqnarray}
f&=&Y-X
\\&=&\alpha\left(x+y\right)-\left\{\alpha\left(x-y\right)\right\}
\\&=&\alpha\left(x+y-x+y\right)
\\&=&\alpha2y
\\&=&\sqrt{2}y
\\f(y)&=&0
\\\sqrt{2}y&=&0
\\y&=&\frac{0}{\sqrt{2}}=0\;\cdots\;X軸
\\
\\g&=&Y-X^2
\\&=&\alpha\left(x+y\right)-\left\{\alpha\left(x-y\right)\right\}^2
\\&=&\alpha\left(x+y\right)-\alpha^2\left(x^2-2xy+y^2\right)
\\&=&\alpha x+\alpha y-\alpha^2x^2+2\alpha^2xy-\alpha^2y^2
\\&=&-\alpha^2y^2+\left(\alpha+2\alpha^2x\right)y+\left(\alpha x-\alpha^2 x^2\right)
\\g(y)&=&0
\\y&=&\frac{-\left(\alpha+2\alpha^2x\right)\pm\sqrt{\left(\alpha+2\alpha^2x\right)^2-4\cdot\left(-\alpha^2\right)\cdot\left(\alpha x-\alpha^2 x^2\right)}}{2\left(-\alpha^2\right)}
\\&=&\frac{-\alpha\left(1+2\alpha x\right)\pm\sqrt{\alpha^2+4\alpha^3x\cancel{+4\alpha^4x^2}+4\alpha^3x\cancel{-4\alpha^3 x}}}{-2\alpha^2}
\\&=&\frac{-\alpha\left(1+2\alpha x\right)\pm\sqrt{\alpha^2+8\alpha^3 x}}{-2\alpha^2}
\\&=&\frac{-\alpha\left(1+2\alpha x\right)\pm\sqrt{\alpha^2\left(1+8\alpha x\right)}}{-2\alpha^2}
\\&=&\frac{-\alpha\left(1+2\alpha x\right)\pm\alpha\sqrt{1+8\alpha x}}{-2\alpha^2}
\\&=&\frac{\left(1+2\alpha x\right)\pm\sqrt{1+8\alpha x}}{2\alpha}
\\&=&\frac{1+2\alpha x\pm\sqrt{1+8\alpha x}}{2\alpha}
\\&=&\frac{1}{2\alpha}+\frac{2\alpha x}{2\alpha}\pm\frac{\sqrt{1+8\alpha x}}{2\alpha}
\\&=&\frac{1}{2\alpha}+x\pm\frac{\sqrt{1+8\alpha x}}{2\alpha}
\\&=&\frac{1}{2\frac{\sqrt{2}}{2}}+x\pm\frac{\sqrt{1+8\frac{\sqrt{2}}{2} x}}{2\frac{\sqrt{2}}{2}}
\\&=&\frac{1}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}+x\pm\frac{\sqrt{1+4\sqrt{2} x}}{\sqrt{2}}\frac{\sqrt{2}}{\sqrt{2}}
\\&=&\frac{\sqrt{2}}{2}+x\pm\frac{\sqrt{2}\sqrt{1+4\sqrt{2} x}}{2}
\\&=&\frac{\sqrt{2}}{2}+x\pm\frac{\sqrt{2\left(1+4\sqrt{2} x\right)}}{2}
\\&=&\frac{\sqrt{2}}{2}+x\pm\frac{\sqrt{2+8\sqrt{2} x}}{2}
\end{eqnarray}$$
\(区間\;0\leq x \leq 1における回転軸y=xの長さは\sqrt{2}．\)
\(これより，-\frac{\pi}{4}回転させた後は，区間0から\sqrt{2}を扱うことになる．\)
\(x=0\)のとき\(+\)側の式は
$$\begin{eqnarray}
\left.\frac{\sqrt{2}}{2}+x+\frac{\sqrt{2+8\sqrt{2} x}}{2}\right|_{x=0}&=&\frac{\sqrt{2}}{2}+0+\frac{\sqrt{2+8\sqrt{2}\cdot0}}{2}
\\&=&\frac{\sqrt{2}}{2}+\frac{\sqrt{2}}{2}
\\&=&\sqrt{2}\gt0
\end{eqnarray}$$
\(-\)側の式は
$$\begin{eqnarray}
\left.\frac{\sqrt{2}}{2}+x-\frac{\sqrt{2+8\sqrt{2} x}}{2}\right|_{x=0}&=&\frac{\sqrt{2}}{2}+0-\frac{\sqrt{2+8\sqrt{2}\cdot0}}{2}
\\&=&\frac{\sqrt{2}}{2}-\frac{\sqrt{2}}{2}
\\&=&0
\end{eqnarray}$$
であり,\(x=0\)で\(y=0\)となる，原点を通る\(-\)側の式をここでは用いる．
ちなみに，第三項の分子の平方根内の値が非負となる区間は
$$\begin{eqnarray}
2+8\sqrt{2} x &\geq&0
\\8\sqrt{2} x&\geq&-2
\\x&\geq&-\frac{2}{8\sqrt{2} }
\\x&\geq&-\frac{1}{4\sqrt{2}}
\end{eqnarray}$$
\( x\geq-\frac{1}{4\sqrt{2}}\)となり，特に問題ない．
\(f-g\)となる領域をX軸まわりに回転させた体積を求める
$$\begin{eqnarray}
V&=&2\pi\int_0^{\sqrt{2}} \left[0 - \left(x+\frac{\sqrt{2}}{2}-\frac{\sqrt{2+8\sqrt{2} x}}{2}\right)\right]^2\mathrm{d}x
\\&=&2\pi\int_0^{\sqrt{2}}\left(-x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2} x}}{2}\right)^2\mathrm{d}x
\\&=&2\pi\int_0^{\sqrt{2}}\left\{
-x\left(-x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2} x}}{2}\right)
+\frac{-\sqrt{2}}{2}\left(-x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2} x}}{2}\right)
+\frac{\sqrt{2+8\sqrt{2} x}}{2}\left(-x-\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2} x}}{2}\right)
\right\}\mathrm{d}x
\\&=&2\pi\int_0^{\sqrt{2}}\left\{
x^2+x\frac{\sqrt{2}}{2}-x\frac{\sqrt{2+8\sqrt{2} x}}{2}
+\frac{\sqrt{2}}{2}x+\frac{\sqrt{2}}{2}\frac{\sqrt{2}}{2}+\frac{-\sqrt{2}}{2}\frac{\sqrt{2+8\sqrt{2} x}}{2}
-\frac{\sqrt{2+8\sqrt{2} x}}{2}x-\frac{\sqrt{2+8\sqrt{2} x}}{2}\frac{\sqrt{2}}{2}+\frac{\sqrt{2+8\sqrt{2} x}}{2}\frac{\sqrt{2+8\sqrt{2} x}}{2}
\right\}\mathrm{d}x
\\&=&2\pi\int_0^{\sqrt{2}}\left\{
x^2
-2x\frac{\sqrt{2+8\sqrt{2} x}}{2}
-2\frac{\sqrt{2}}{2}\frac{\sqrt{2+8\sqrt{2} x}}{2}
+\left(\frac{\sqrt{2+8\sqrt{2} x}}{2}\right)^2
+2\frac{\sqrt{2}}{2}x
+\left(\frac{\sqrt{2}}{2}\right)^2
\right\}\mathrm{d}x
\\&=&2\pi\int_0^{\sqrt{2}}\left\{
x^2
-x\sqrt{2+8\sqrt{2} x}
-\frac{\sqrt{2}}{2}\sqrt{2+8\sqrt{2}} x
+\frac{2+8\sqrt{2} x}{4}
+\sqrt{2}x
+\frac{2}{4}
\right\}\mathrm{d}x
\\&=&2\pi\int_0^{\sqrt{2}}\left\{
x^2
-x\sqrt{2+8\sqrt{2} x}
-\frac{\sqrt{2}}{2}\sqrt{2+8\sqrt{2}} x
+\frac{2}{4}
+\frac{8\sqrt{2} x}{4}
+\sqrt{2}x
+\frac{1}{2}
\right\}\mathrm{d}x
\\&=&2\pi\int_0^{\sqrt{2}}\left\{
x^2
-x\sqrt{2+8\sqrt{2} x}
-\frac{\sqrt{2}}{2}\sqrt{2+8\sqrt{2}} x
+2\sqrt{2} x
+\sqrt{2}x
+\frac{1}{2}
+\frac{1}{2}
\right\}\mathrm{d}x
\\&=&2\pi\int_0^{\sqrt{2}}\left\{
x^2
-x\sqrt{2+8\sqrt{2} x}
-\frac{\sqrt{2}}{2}\sqrt{2+8\sqrt{2}} x
+3\sqrt{2} x
+1
\right\}\mathrm{d}x
\\&=&2\pi\left[
\int_0^{\sqrt{2}}x^2\mathrm{d}x
-\int_0^{\sqrt{2}}x\sqrt{2+8\sqrt{2} x}\mathrm{d}x
-\frac{\sqrt{2}}{2}\int_0^{\sqrt{2}}\sqrt{2+8\sqrt{2} x}\mathrm{d}x
+3\sqrt{2} \int_0^{\sqrt{2}}x\mathrm{d}x
+\int_0^{\sqrt{2}}\mathrm{d}x
\right]
\\&=&2\pi\left(
\frac{2}{3}\sqrt{2}
-\frac{149}{60}\sqrt{2}
-\frac{\sqrt{2}}{2}\cdot\frac{13}{3}
+3\sqrt{2}\cdot1
+\sqrt{2}
\right)
\\&=&2\pi\left(\frac{2}{3}-\frac{149}{60}-\frac{13}{6}+4\right)\sqrt{2}
\\&=&2\pi\frac{40-149-130+240}{60}\sqrt{2}
\\&=&2\pi\frac{\sqrt{2}}{60}
\\&=&\frac{\pi\sqrt{2}}{30}
\end{eqnarray}$$
個々の積分について
$$\begin{eqnarray}
\int_0^{\sqrt{2}}\mathrm{d}x&=&\left[x\right]_0^{\sqrt{2}}
\\&=&\left[\sqrt{2}-0 \right]
\\&=&\sqrt{2}
\end{eqnarray}$$
$$\begin{eqnarray}
\int_0^{\sqrt{2}}x\mathrm{d}x&=&\left[\frac{1}{2}x^2\right]_0^{\sqrt{2}}
\\&=&\left[\frac{1}{2}\sqrt{2}^2-\frac{1}{2}0^2 \right]
\\&=&\left[1-0\right]
\\&=&1
\end{eqnarray}$$
$$\begin{eqnarray}
\int_0^{\sqrt{2}}\sqrt{2+8\sqrt{2} x}\mathrm{d}x
\\&=&\frac{1}{8\sqrt{2}}\int_2^{18}\sqrt{u}\mathrm{d}u\;\cdots\;u=2+8\sqrt{2} x,\;\frac{\mathrm{d}u}{\mathrm{d}x}=8\sqrt{2},\;\mathrm{d}x=\frac{1}{8\sqrt{2}}\mathrm{d}u,\;x=0\rightarrow t=2,\;x=\sqrt{2}\rightarrow t=18
\\&=&\frac{1}{8\sqrt{2}}\left[\frac{1}{\frac{1}{2}+1}u^{\frac{1}{2}+1}\right]_2^{18}
\\&=&\frac{1}{8\sqrt{2}}\left[\frac{1}{\frac{3}{2}}u^{\frac{3}{2}}\right]_2^{18}
\\&=&\frac{1}{8\sqrt{2}}\left[\frac{2}{3}u^{\frac{3}{2}}\right]_2^{18}
\\&=&\frac{1}{8\sqrt{2}}\frac{2}{3}\left[u^{\frac{3}{2}}\right]_2^{18}
\\&=&\frac{1}{8\sqrt{2}}\frac{2}{3}\left[18^{\frac{3}{2}}-2^{\frac{3}{2}}\right]
\\&=&\frac{1}{8\sqrt{2}}\frac{2}{3}\left(18\cdot18^{\frac{1}{2}}-2\cdot2^{\frac{1}{2}}\right)
\\&=&\frac{1}{8\sqrt{2}}\frac{2}{3}\left(18\cdot\left(3^2\cdot2\right)^{\frac{1}{2}}-2\cdot2^{\frac{1}{2}}\right)
\\&=&\frac{1}{8\sqrt{2}}\frac{2}{3}\left(18\cdot3\cdot2^{\frac{1}{2}}-2\cdot2^{\frac{1}{2}}\right)
\\&=&\frac{1}{8\sqrt{2}}\frac{2}{3}\left(54\cdot2^{\frac{1}{2}}-2\cdot2^{\frac{1}{2}}\right)
\\&=&\frac{1}{8\sqrt{2}}\frac{2}{3}\left(54-2\right)\cdot2^{\frac{1}{2}}
\\&=&\frac{1}{8\sqrt{2}}\frac{2\cdot52}{3}\sqrt{2}
\\&=&\frac{1}{8\sqrt{2}}\frac{104}{3}\sqrt{2}\;\cdots\;\int_2^{18}\sqrt{u}\mathrm{d}u=\frac{104}{3}\sqrt{2}
\\&=&\frac{104}{24}
\\&=&\frac{13}{3}
\end{eqnarray}$$
$$\begin{eqnarray}
\int_0^{\sqrt{2}}x\sqrt{2+8\sqrt{2} x}\mathrm{d}x
\\&=&\frac{1}{8\sqrt{2}}\int_2^{18}\frac{u-2}{8\sqrt{2}}\sqrt{u}\mathrm{d}u\;\cdots\;u=2+8\sqrt{2} x,\;\frac{\mathrm{d}u}{\mathrm{d}x}=8\sqrt{2},\;\mathrm{d}x=\frac{1}{8\sqrt{2}}\mathrm{d}u,\;x=0\rightarrow t=2,\;x=\sqrt{2}\rightarrow t=18,\;x=\frac{u-2}{8\sqrt{2}}
\\&=&\frac{1}{8\sqrt{2}}\frac{1}{8\sqrt{2}}\int_2^{18}\left(u-2\right)\sqrt{u}\mathrm{d}u
\\&=&\frac{1}{128}\int_2^{18}\left(u\sqrt{u}-2\sqrt{u}\right)\mathrm{d}u
\\&=&\frac{1}{128}\int_2^{18}\left(u^{\frac{3}{2}}-2u^{\frac{1}{2}}\right)\mathrm{d}u
\\&=&\frac{1}{128}\int_2^{18}u^{\frac{3}{2}}\mathrm{d}u-\frac{2}{128}\int_2^{18}u^{\frac{1}{2}}\mathrm{d}u
\\&=&\frac{1}{128}\left[\frac{1}{\frac{3}{2}+1}u^{\frac{3}{2}+1}\right]_2^{18}
-\frac{1}{64}\frac{104}{3}\sqrt{2}\;\cdots\;\int_2^{18}\sqrt{u}\mathrm{d}u=\frac{104}{3}\sqrt{2}, 前述の積分結果より
\\&=&\frac{1}{128}\left[\frac{1}{\frac{5}{2}}u^{\frac{5}{2}}\right]_2^{18}
-\frac{13}{24}\sqrt{2}
\\&=&\frac{1}{128}\frac{2}{5}\left[u^{\frac{5}{2}}\right]_2^{18}
-\frac{13}{24}\sqrt{2}
\\&=&\frac{1}{128}\frac{2}{5}\left[18^{\frac{5}{2}}-2^{\frac{5}{2}}\right]-\frac{13}{24}\sqrt{2}
\\&=&\frac{1}{128}\frac{2}{5}\left[18^2\cdot18^{\frac{1}{2}}-2^2\cdot2^{\frac{1}{2}}\right]-\frac{13}{24}\sqrt{2}
\\&=&\frac{1}{128}\frac{2}{5}\left[18^2\cdot(3^2\cdot2)^{\frac{1}{2}}-2^2\cdot2^{\frac{1}{2}}\right]-\frac{13}{24}\sqrt{2}
\\&=&\frac{1}{128}\frac{2}{5}\left[18^2\cdot3\cdot2^{\frac{1}{2}}-2^2\cdot2^{\frac{1}{2}}\right]-\frac{13}{24}\sqrt{2}
\\&=&\frac{1}{128}\frac{2}{5}\left[(18^2\cdot3-2^2)\cdot2^{\frac{1}{2}}\right]-\frac{13}{24}\sqrt{2}
\\&=&\frac{1}{128}\frac{2}{5}968\sqrt{2}-\frac{13}{24}\sqrt{2}
\\&=&\frac{121}{40}\sqrt{2}-\frac{13}{24}\sqrt{2}
\\&=&\frac{121\cdot3-13\cdot5}{120}\sqrt{2}
\\&=&\frac{363-65}{120}\sqrt{2}
\\&=&\frac{298}{120}\sqrt{2}
\\&=&\frac{149}{60}\sqrt{2}
\end{eqnarray}$$
$$\begin{eqnarray}
\int_0^{\sqrt{2}}x^2\mathrm{d}x&=&\left[\frac{1}{3}x^3\right]_0^{\sqrt{2}}
\\&=&\left[\frac{1}{3}\sqrt{2}^3-\frac{1}{3}0^3 \right]
\\&=&\left[\frac{1}{3}\sqrt{2}^3-0\right]
\\&=&\frac{1}{3}\sqrt{2}^3
\\&=&\frac{2}{3}\sqrt{2}
\end{eqnarray}$$
