Answer to Question #169125 in Calculus for Sean Cody

Question #169125

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

the interval [0, π]


1
Expert's answer
2021-03-07T17:34:55-0500

f(x)=1+sin(x)f(x) = 1 + sin(x) g(x)=1sin(x)g(x) = 1-sin(x) a=0    b=π4a = 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=abxf(x)g(x)dxabf(x)g(x)dx,x = \large\frac{\int_a^bx|f(x)-g(x)|dx}{\int_a^b|f(x)-g(x)|dx}, y=ab12f(x)2g(x)2dxabf(x)g(x)dxy= \large\frac{\int_a^b\frac{1}{2}|f(x)^2-g(x)^2|dx}{\int_a^b|f(x)-g(x)|dx}


abf(x)g(x)dx=0π41+sin(x)1+sin(x)=0π42sin(x)=2cos(x)0π4=2cosπ4+2cos0=22\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}


abxf(x)g(x)dx=0π4x1+sin(x)1+sin(x)=0π42xsin(x)=2sin(x)xcos(x)0π4=2(sinπ4π4cos(π4))2(sin00cos0)=2(1π4)\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})


ab12f(x)2g(x)2dx=0π41+2sin(x)+sin2(x)1+2sin(x)sin2(x)=0π44sin(x)=4cos(x)0π4=4cosπ4+4cos0=422\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=abxf(x)g(x)dxabf(x)g(x)dx=2(1π4)22=4π421,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=ab12f(x)2g(x)2dxabf(x)g(x)dx=42222y= \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=2


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