Ответ:
Формула объёма тела вращения около оси ОХ:
[tex]\displaystyle \bf V_{ox}=\pi \int\limits_{a}^{b}\, y^2(x)\, dx[/tex]
[tex]\displaystyle \bf y=cos2x\ ,\ x=0\ ,\ y=0\ \ ,\ \ 0\leq x\leq \dfrac{\pi }{4}\\\\V_{ox}=\pi \int\limits_{0}^{\pi /4}\, cos^22x\, dx=\pi \int\limits_{0}^{\pi /4}\, \frac{1+cos4x}{2}\, dx=\frac{\pi }{2}\int\limits_{0}^{\pi /4}\, (1+cos4x)\, dx=\\\\\\=\frac{\pi }{2}\cdot \Big(x+\frac{1}{4}\, sin4x\Big)\, \Big|_0^{\pi /4}=\frac{\pi }{2}\cdot \Big(\frac{\pi }{4}+\frac{1}{4}\, \underbrace{\bf sin\pi }_{0}\Big)=\frac{\pi ^2}{8}[/tex]
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Ответ:
Формула объёма тела вращения около оси ОХ:
[tex]\displaystyle \bf V_{ox}=\pi \int\limits_{a}^{b}\, y^2(x)\, dx[/tex]
[tex]\displaystyle \bf y=cos2x\ ,\ x=0\ ,\ y=0\ \ ,\ \ 0\leq x\leq \dfrac{\pi }{4}\\\\V_{ox}=\pi \int\limits_{0}^{\pi /4}\, cos^22x\, dx=\pi \int\limits_{0}^{\pi /4}\, \frac{1+cos4x}{2}\, dx=\frac{\pi }{2}\int\limits_{0}^{\pi /4}\, (1+cos4x)\, dx=\\\\\\=\frac{\pi }{2}\cdot \Big(x+\frac{1}{4}\, sin4x\Big)\, \Big|_0^{\pi /4}=\frac{\pi }{2}\cdot \Big(\frac{\pi }{4}+\frac{1}{4}\, \underbrace{\bf sin\pi }_{0}\Big)=\frac{\pi ^2}{8}[/tex]