708 Centroid and area of spandrel by integration

$y = kx^n$
 

At (b, h)
$h = kb^n$

$k = \dfrac{h}{b^n}$
 

Thus,
$y = \dfrac{h}{b^n}x^n$
 

Differential area
$dA = y \, dx$

$dA = \dfrac{h}{b^n}x^n \, dx$
 

Area by integration
$\displaystyle A = \dfrac{h}{b^n}\int_0^b x^n \, dx$

$A = \dfrac{h}{b^n} \left[ \dfrac{x^{n + 1}}{n + 1} \right]_0^b$

$A = \dfrac{h}{(n + 1)b^n} \left[ (b^{n + 1}) - (0^{n + 1}) \right]$

$A = \dfrac{1}{(n + 1)}bh$           answer
 

 

x-coordinate of the centroid
$\displaystyle A \, x_G = \int_a^b x_c \, dA$

$\displaystyle \dfrac{1}{(n + 1)}bh \, x_G = \int_0^b x \left( \dfrac{h}{b^n}x^n \, dx \right)$

$\displaystyle \dfrac{1}{(n + 1)}bh \, x_G = \dfrac{h}{b^n} \int_0^b x^{n + 1} \, dx$

$\dfrac{1}{(n + 1)}bh \, x_G = \dfrac{h}{b^n} \left[ \dfrac{x^{n + 2}}{n + 2} \right]_0^b$

$\dfrac{1}{(n + 1)}bh \, x_G = \dfrac{h}{(n + 2)b^n} \left[ (b^{n + 2}) - (0^{n + 2}) \right]$

$\dfrac{1}{(n + 1)}bh \, x_G = \dfrac{h}{(n + 2)b^n} \left[ b^{n + 2} \right]$

$\dfrac{1}{(n + 1)}bh \, x_G = \dfrac{1}{n + 2} b^2h$

$x_G = \dfrac{n + 1}{n + 2}b$
 

Location of centroid from the line x = b
$\bar{x} = b - x_G$

$\bar{x} = b - \dfrac{n + 1}{n + 2}b$

$\bar{x} = \dfrac{(n + 2) - (n + 1)}{n + 2}b$

$\bar{x} = \dfrac{1}{n + 2}b$           answer
 

y-coordinate of the centroid
$\displaystyle A \, \bar{y} = \int_a^b y_c \, dA$

$\displaystyle \dfrac{1}{(n + 1)}bh \, \bar{y} = \int_0^b (\frac{1}{2}y)(y \, dx)$

$\displaystyle \dfrac{1}{(n + 1)}bh \, \bar{y} = \frac{1}{2} \int_0^b y^2 \, dx$

$\displaystyle \dfrac{1}{(n + 1)}bh \, \bar{y} = \frac{1}{2} \int_0^b \left( \dfrac{h}{b^n}x^n \right)^2 \, dx$

$\displaystyle \dfrac{1}{(n + 1)}bh \, \bar{y} = \dfrac{h^2}{2b^{2n}} \int_0^b x^{2n} \, dx$

$\dfrac{1}{(n + 1)}bh \, \bar{y} = \dfrac{h^2}{2b^{2n}} \left[ \dfrac{x^{2n + 1}}{2n + 1} \right]_0^b$

$\dfrac{1}{(n + 1)}bh \, \bar{y} = \dfrac{h^2}{2(2n + 1)b^{2n}} \left[ (b^{2n + 1}) - (0^{2n + 1}) \right]$

$\dfrac{1}{(n + 1)}bh \, \bar{y} = \dfrac{h^2}{2(2n + 1)b^{2n}} \left[ b^{2n + 1} \right]$

$\dfrac{1}{(n + 1)}bh \, \bar{y} = \dfrac{1}{2(2n + 1)}bh^2$

$\bar{y} = \dfrac{n + 1}{2(2n + 1)}h$

$\bar{y} = \dfrac{n + 1}{4n + 2}h$           answer