Respuesta :
Answer:
Required centroid [tex](\bar{x}, \bar{y})=(\frac{M_y}{M},\frac{M_x}{M})=(-\frac{3+2\sqrt{3}}{4},\frac{4\pi}{1-\sqrt{3}})[/tex].
Step-by-step explanation:
Given functions are,
[tex]f(x)=y=6\sin (4x), g(x)=y=6\cos (4x)[/tex] and [tex][a,b]=[0, \frac{\pi}{6}][/tex]
To find centroid of the region bounded by the above curves,
Mass=[tex]\int_{0}^{\frac{\pi}{6}}(f9x)-g(x))dx[/tex]
[tex]=\int_{0}^{\frac{\pi}{6}}(6\sin(4x)-6\cos(4x))dx[/tex]
[tex]=6\Big[\frac{-\cos (4x)}{4}\Big]_{0}^{\frac{\pi}{6}}-6\Big[\frac{\sin (4x)}{4}\Big]_{0}^{\frac{\pi}{6}}[/tex]
[tex]=\frac{3}{2}\Big[-\cos (\frac{2\pi}{3})+\cos 0-\sin (\frac{2\pi}{3})+\sin (0)\Big][/tex]
[tex]=\frac{3}{2}\Big[\frac{1}{2}-\frac{\sqrt{3}}{2}\Big]=\frac{3}{4}(1-\sqrt{3})[/tex]
Moments are,
- [tex]M_x=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}((f(x))^2-(g(x))^2)dx[/tex]
[tex]=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}(36\sin^2 (4x)+36\cos^2(4x))dx[/tex]
[tex]=\frac{1}{2}\int_{0}^{\frac{\pi}{6}}36dx[/tex]
[tex]=\frac{1}{2}\times 36\times \frac{\pi}{6}=3\pi[/tex]
and,
- [tex]M_y=6\int_{a}^{b}x(f(x)-g(x))dx[/tex]
[tex]=6\int_{0}^{\frac{\pi}{6}}x(\sin (4x)-\cos (4x))dx[/tex]
[tex]=6\int_{0}^{\frac{\pi}{6}}x\sin (4x)dx-6\int_{0}^{\frac{\pi}{6}}x\cos (4x)dx[/tex]
[tex]=6\Big[-\frac{x\cos (4x)}{4}\Big]_{0}^{\frac{\pi}{6}}+\frac{3}{2}\Big[\frac{\sin (4x)}{4}\Big]_{0}^{\frac{\pi}{6}}-6\Big[\frac{x\sin (4x)}{4}\Big]_{0}^{\frac{\pi}{6}}+\frac{3}{2}\Big[-\frac{\cos (4x)}{4}\Big][/tex]
[tex]=(-\frac{6}{4}\times \frac{\pi}{6}\times (-\frac{1}{2}))+(\frac{3}{2}\times \frac{1}{4}\times \frac{\sqrt{3}}{2})-((-6)\times \frac{\pi}{6}\times \frac{\sqrt{3}}{2})-(\frac{3}{8}\times (-\frac{3}{2}))[/tex]
[tex]=\frac{3\sqrt{3}}{16}(\sqrt{3}+1)[/tex]
Thus, centroid,
[tex](\bar{x}, \bar{y})=(\frac{M_y}{M},\frac{M_x}{M})=(-\frac{3+2\sqrt{3}}{4},\frac{4\pi}{1-\sqrt{3}})[/tex]
