Sketch the region bounded on the left by $x=1$, bounded above by $y=\frac{1}{x^3}$, and bounded below by $y=-\frac{1}{x^3}$.
a) find the centroid of the region for $1 \leq x \leq 6$
b) find the centroid of the region for $1 \leq x \leq b$
c) Where is the centroid as $ b\to \infty$
I have calculated part a to be (48/13, 0) but I cannot get past part b
On
HINT: Due to symmetry about the x-axis of the region bounded above by the curve $y=\frac{1}{x^3}$, below by the curve $y=\frac{-1}{x^3}$ & on the left of the line $x=1$ , the centroid of the bounded region will lie on the x-axis at some point $(\bar{x}, 0)$ given as $$\bar{x}=\frac{\int xdA}{\int dA}$$
Where, $dA=\text{area of the bounded region}=\int ydx$, hence we get $$\bar{x}=\frac{\int xydx}{\int y dx}$$
$$A=\int_{1}^{b}2ydx=\int_{1}^{b}\frac{2}{x^3}dx=1-\frac{1}{b^2}$$ when the $b=6$ $$A=1-\frac{1}{36}=\frac{35}{36}$$
now we will calculate the moment about y-axis to find the x center $$M_y=\int_{1}^{b}2y(x-1)dx=\int_{1}^{b}\frac{2}{x^3}(x-1)dx=\frac{1-2b}{b^2}+1$$ when $b=6$ $$M_y=\frac{25}{36}$$
$$x_c=\frac{M_y}{A}=(25/36)/(35/36)=\frac{25}{36}=\frac{5}{7}$$ then you can conclude the answer when the $b=infinity$