Question

Find the centroid of the area enclosed by y = 1 + sin(x) and y = 1 − sin(x) on

the interval [0, π]

Answer 1

$f(x) = 1 + sin(x)$ $g(x) = 1-sin(x)$ $a = 0 \space\space\space\space b = \large\frac{\pi}{4}$

We apply the formulae that the coordinates of the centroid (= center of mass assuming constant density) of the region with top f(x), bottom g(x), left-hand side x = a, and right-hand side x = b are

$x = \large\frac{\int_a^bx|f(x)-g(x)|dx}{\int_a^b|f(x)-g(x)|dx},$ $y= \large\frac{\int_a^b\frac{1}{2}|f(x)^2-g(x)^2|dx}{\int_a^b|f(x)-g(x)|dx}$

$\int_a^b|f(x)-g(x)|dx = \int_0^{\frac{\pi}{4}}|1+sin(x)-1+sin(x)|=\int_0^{\frac{\pi}{4}}2sin(x) = |-2cos(x)|_0^\frac{\pi}{4} = -2cos\frac{\pi}{4}+2cos0 = 2-\sqrt{2}$

$\int_a^bx|f(x)-g(x)|dx = \int_0^{\frac{\pi}{4}}x|1+sin(x)-1+sin(x)|=\int_0^{\frac{\pi}{4}}2xsin(x) = 2|sin(x)-xcos(x)|_0^\frac{\pi}{4} = 2(sin\frac{\pi}{4}-\frac{\pi}{4}cos(\frac{\pi}{4}))-2(sin0-0*cos0) = \sqrt{2}(1-\frac{\pi}{4})$

$\int_a^b\frac{1}{2}|f(x)^2-g(x)^2|dx = \int_0^{\frac{\pi}{4}}|1+2sin(x)+sin^2(x)-1+2sin(x)-sin^2(x)|=\int_0^{\frac{\pi}{4}}4sin(x) = |-4cos(x)|_0^\frac{\pi}{4} = -4cos\frac{\pi}{4}+4cos0 = 4-2\sqrt{2}$

$x = \large\frac{\int_a^bx|f(x)-g(x)|dx}{\int_a^b|f(x)-g(x)|dx} = \frac{\sqrt{2}(1-\frac{\pi}{4})}{2-\sqrt{2}} = \frac{4-\pi}{4\sqrt{2}-1} ,$ $y= \large\frac{\int_a^b\frac{1}{2}|f(x)^2-g(x)^2|dx}{\int_a^b|f(x)-g(x)|dx} = \frac{ 4-2\sqrt{2}}{2-\sqrt{2}}$ $=2$