Answer to Question #200170 in Calculus for Melvin

Question

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

Answer 1

$\text{Given} \text{ the region is bounded by }y=x^2and y=2.\newline \text{Area},A=\int_{-\sqrt2}^{\sqrt2 }2-{x^2} dx\newline = [2x-\frac{x^3}{3}]_{-\sqrt2}^{\sqrt2 } \newline =3.771\newline \text{Let (X, Y) be the centroid.}\newline X=\frac{\int_{-\sqrt2}^{\sqrt2 }x(2-x^2) dx}{A}\newline =\frac{[x^2-\frac{x^4}{4}]_{-\sqrt2}^{\sqrt2 }}{3.771}\newline =0 \newline Y=\frac{\frac{1}{2}\int_{-\sqrt2}^{\sqrt2 }(2^2-(x^2)^2) dx}{A}\newline =\frac{\frac{1}{2}\int_{-\sqrt2}^{\sqrt2 }(4-x^4) dx}{3.771}\newline =\frac{[4x-\frac{x^5}{5}]_{-\sqrt2}^{\sqrt2 }}{3.771}\newline =\frac{\frac{1}{2}×9.05}{3.771}\newline =1.2\newline \text{Centroid is} (0, 1.2).$