Question #200170

Find the centroid of the region bounded by y=x² and y=2


Expert's answer

Given the region is bounded by y=x2andy=2.Area,A=∫−222−x2dx=[2x−x33]−22=3.771Let (X, Y) be the centroid.X=∫−22x(2−x2)dxA=[x2−x44]−223.771=0Y=12∫−22(22−(x2)2)dxA=12∫−22(4−x4)dx3.771=[4x−x55]−223.771=12×9.053.771=1.2Centroid is(0,1.2).\text{Given} \text{ the region is bounded by }y=x^2and y=2.\newline \text{Area},A=\int_{-\sqrt2}^{\sqrt2 }2-{x^2} dx\newline = [2x-\frac{x^3}{3}]_{-\sqrt2}^{\sqrt2 } \newline =3.771\newline \text{Let (X, Y) be the centroid.}\newline X=\frac{\int_{-\sqrt2}^{\sqrt2 }x(2-x^2) dx}{A}\newline =\frac{[x^2-\frac{x^4}{4}]_{-\sqrt2}^{\sqrt2 }}{3.771}\newline =0 \newline Y=\frac{\frac{1}{2}\int_{-\sqrt2}^{\sqrt2 }(2^2-(x^2)^2) dx}{A}\newline =\frac{\frac{1}{2}\int_{-\sqrt2}^{\sqrt2 }(4-x^4) dx}{3.771}\newline =\frac{[4x-\frac{x^5}{5}]_{-\sqrt2}^{\sqrt2 }}{3.771}\newline =\frac{\frac{1}{2}×9.05}{3.771}\newline =1.2\newline \text{Centroid is} (0, 1.2).


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