Centroid of volume formed by rotating about the y-axis the area bounded by 3 lines

Using Vertical Strip
$V = \Sigma \left[ 2\pi {\displaystyle \int_{x_1}^{x_2}} xy \, d_x \right]$

$V = 2\pi {\displaystyle \int_0^2} x (3x - x) \, d_x + 2\pi {\displaystyle \int_2^4} x [ \, (8 - x) - x \, ] \, d_x$

$V = 32\pi ~ \text{unit}^3$
 

By symmetry
$X_G = 0$
 

Solving for Y_G
$V \, Y_G = \Sigma \left[ 2\pi {\displaystyle \int_{x_1}^{x_2}} y_c xy \, d_x \right]$

$V \, Y_G = \Sigma \left[ 2\pi {\displaystyle \int_{x_1}^{x_2}} \frac{1}{2}(y_U + y_L) x (y_U - y_L) \, d_x \right]$

$V \, Y_G = \Sigma \left[ \pi {\displaystyle \int_{x_1}^{x_2}} x({y_U}^2 - {y_L}^2) \, d_x \right]$

$32\pi \, Y_G = \pi {\displaystyle \int_0^2} x(9x^2 - x^2) \, d_x + \pi {\displaystyle \int_2^4} x [ \, (8 - x)^2 - x^2 \, ] \, d_x$

$32\pi \, Y_G = \frac{352}{3}\pi$

$Y_G = \frac{11}{3}$
 

Centroid of the solid is at (0, 11/3)
 

Using Horizontal Strip
$V = \Sigma \left[ \pi {\displaystyle \int_{x_1}^{x_2}} ({x_R}^2 - {x_L}^2) \, dy \right]$

$V = \pi {\displaystyle \int_0^4} (y^2 - \frac{1}{9}y^2) \, dy + \pi {\displaystyle \int_4^6} [ \, (8 - y)^2 - \frac{1}{9}y^2) \, ] \, dy$

$V = 32\pi ~ \text{unit}^3$
 

$V \, Y_G = \Sigma \left[ \pi {\displaystyle \int_{x_1}^{x_2}} y_c ({x_R}^2 - {x_L}^2) \, dy \right]$

$32\pi \, Y_G = \pi {\displaystyle \int_0^4} y(y^2 - \frac{1}{9}y^2) \, dy + \pi {\displaystyle \int_4^6} y[ \, (8 - y)^2 - \frac{1}{9}y^2) \, ] \, dy$

$32\pi \, Y_G = \frac{352}{3}\pi$

$Y_G = \frac{11}{3}$
 

Centroid of the solid is at (0, 11/3)