Question

1.Find the centroid of the region bounded by the curve 𝑦 = 𝑥3 and 𝑦 = 4𝑥 in the first quadrant. 

2. Find the centroid of the solid of revolution generated by revolving the region bounded by 𝑦 = 4𝑥 − 𝑥2 and the x-axis about the y-axis.

Answer 1

1

$Mx=\frac{1}{2}\int_a^b(f(x)^2-g(x)^2)dx=\frac{1}{2}\int_0^1[(4x)^2-(x^3)]$

$=2.59523$

$My=\int_a^bx[f(x)-g(x)]dx=\int_0^1[x[(4x)^2-(x^3)dx]$

$=\:3.85714$

$M=\int_a^b[f(x)-g(x)]dx=\int_0^1[4x-x^3]$

$=1.75$

$\bar{x},\bar{y}=(\frac{My}{M},\frac{Mx}{M})=(2.20408,1.48289)$

2

$y=4x-x^2$

$4x-x^2=x$

$x=0,3$

we are interested in the point in the first quadrant, the transition point is

$x_t,y_t=3,3$

$\bar{y}=\int\frac{ydv}{v}$ where v = volume

$V=\int dV=\int \pi x^2dy$

$V=\int_0^{y_t}\pi^2dy+\int_{y_t}^4 \pi(\sqrt{4-y)}dy$

$V=\pi[(\frac{1}{3}y^3)^{y_t}_0+(4y-\frac{1}{2}y^2)]^4_{y_t}$

$V=\frac{\pi}{12}(121-17\sqrt{17)}$

$\int ydV=\int y(\pi x^2 dy)$

$\int ydV=\int_0^{y_t}\pi^3dy+\int_{y_t}^4\pi y(4-y)dy$

$\int ydV=\pi [(\frac{1}{4}y^4)_0^{y_t}+(2y^2-\frac{1}{3}y^3)_{y_t}^4]$

$\int ydV=\frac{x}{24}(135+17\sqrt{17)}$

finally we find the the centroid:

$\bar{y}=\frac{\frac{x}{24}(135+17\sqrt{17)}}{\frac{x}{12}(121-17\sqrt{17)}}$

$\bar{y}=\frac{83+17\sqrt{17}}{76}\approx 2.014379$