Tính thể tích khối tròn xoay tạo thành khi quay quanh trục Ox hình phẳng giới hạn bởi các đường \(x=\frac{\pi}{2};x=\pi;y=0;y=\sqrt{1+\cos^4x+\sin^4x}\) ?
\(\frac{7}{8}\pi\) \(\frac{7}{8}\pi^2\) \(\frac{7}{8}\pi^3\) \(\frac{5}{8}\pi^2\) Hướng dẫn giải:\(V=\pi\int\limits^{\pi}_{\frac{\pi}{2}}\left[\left(\sqrt{1+\cos^4x+\sin^4x}\right)^2-0^2\right]\text{dx}\)
\(=\pi\int\limits^{\pi}_{\frac{\pi}{2}}\left(1+\cos^4x+\sin^4x\right)\text{dx}\)
\(=\pi\int\limits^{\pi}_{\frac{\pi}{2}}\left[1+\left(\cos^2x+\sin^2x\right)^2-2\cos^2x\sin^2x\right]\text{dx}\)
\(=\pi\int\limits^{\pi}_{\frac{\pi}{2}}\left[1+1^2-\frac{1}{2}\sin^22x\right]\text{dx}\)
\(=\pi\int\limits^{\pi}_{\frac{\pi}{2}}\left[2-\frac{1}{2}\frac{1-\cos4x}{2}\right]\)
\(=\pi\int\limits^{\pi}_{\frac{\pi}{2}}\left(\frac{7}{4}+\frac{1}{4}\cos4x\right)\text{dx}\)
\(=\pi\left(\frac{7}{4}+\frac{1}{16}\sin4x\right)|^{\pi}_{\frac{\pi}{2}}\)
\(=\frac{7\pi^2}{8}\)
