Question

Find the centroid of the region bounded by the given curves.

Answer 1

\[y = 6\sin3x, y = 6\cos3x, x = 0, x = \pi/12\]

Answer 2

\[A = \int\limits_{}^{}(6\cos3x-6\sin3x)dx = 6\int\limits_{}^{}(\cos3x-\sin3x)dx \]\[= 6[\frac{ 1 }{ 3 }\sin3x-(-\frac{ 1 }{ 3 }\cos3x)] = 2[\sin3x+\cos3x]_{0}^{\pi/12}\]\[2[(\frac{ \sqrt{2} }{ 2 }+\frac{ \sqrt{2} }{ 2 })-(0+1)] = 2[\sqrt{2}-1]=2\sqrt{2}-2\]

Answer 3

\[x = \frac{ 1 }{ A }\int\limits_{a}^{b}x[f(x)-g(x)]dx=\frac{ 1 }{ 2\sqrt{2}-2 }\int\limits_{a}^{b}x)6\cos3x-6\sin3x)dx\]\[\frac{ 1 }{ 2\sqrt{2}-2 }[6\int\limits_{}^{}xcos3x*dx-6\int\limits_{}^{}xsin3x*dx]\]\[u = x, du = dx, dv = \cos3xdx, v = \frac{ 1 }{ 3 }\sin3x\]\[u = x, du =dx, dv = \sin3xdx, v = -\frac{ 1 }{ 3 }\cos3x\]\[\frac{ 1 }{ 2\sqrt{2}-2 }[6[\frac{ 1 }{ 3 }xsin3x-\int\limits_{}^{}\frac{ 1 }{ 3 }\sin3xdx] - 6[-\frac{ 1 }{ 3 }xcos3x+\int\limits_{}^{}\frac{ 1 }{ 3 }\cos3xdx] ]\]\[\frac{ 1 }{ 2\sqrt{2}-2 }*6[\frac{ x }{ 3 }\sin3x-\frac{ 1 }{ 3 }[-\frac{ 1 }{ 3 }\cos3x] ]-[-\frac{ x }{ 3 }\cos3x+\frac{ 1 }{ 3 }[\frac{ 1 }{ 3 }\sin3x] ] ]\]\[\frac{ 6 }{ 2\sqrt{2}-2 }[\frac{ x }{ 3 }\sin3x+\frac{ 1 }{ 9 }\cos3x+\frac{ x }{ 3 }\cos3x-\frac{ 1 }{ 9 }\sin3x]_{0}^{\pi/12}\]