Answer to Question #347485 in Calculus for kiking

Question

find the centroid of the region with the indicated boundaries y = 4 - x^2 and the x-axis

Answer 1

The coordinates of the centroid of a region bouded by curves $y=f(x)$ and $y=g(x)$

($f(x) \geq g(x)$ ):

$\overline{x}=\frac{1}{A}\int_a^b x[f(x)-g(x)]dx,$

$\overline{y}=\frac{1}{2A}\int_a^b [(f(x))^2-(g(x))^2]dx,$

where a and b the endpoints and A the area of the region.

We have:

$A=\int_{-2}^2 (4-x^2)dx= (4x-\frac{x^3}{3})|_{-2}^2=$

$= (4\cdot 2-\frac{2^3}{3})|_{-2}^2 - (4\cdot (-2)-\frac{(-2)^3}{3})|_{-2}^2=$

$=8-\frac{8}{3}+8-\frac{8}{3}=16-\frac{16}{3}=16-5\frac{1}{3}=10\frac{2}{3}.$

$\overline{x}=\frac{3}{32}\int_{-2}^2 x[4-x^2]dx= \frac{3}{32}\int_{-2}^2 (4x-x^3)dx=$

$=\frac{3}{32}(4\frac{x^2}{2}-\frac{x^4}{4})|_{-2}^{2}=\frac{3}{32}(2x^2-\frac{x^4}{4})|_{-2}^{2}=$

$=\frac{3}{32}[(2\cdot 2^2-\frac{2^4}{4})-(2\cdot (-2)^2-\frac{(-2)^4}{4})]|_{-2}^{2}=0.$

$\overline{y}=\frac{3}{64}\int_{-2}^2 (4-x^2)^2dx=\frac{3}{64}\int_{-2}^2 (16-8x^2+x^4)dx=$

$=\frac{3}{64}\int_{-2}^2 (16-8x^2+x^4)dx=\frac{3}{64} (16x-8\frac{x^3}{3}+\frac{x^5}{5})|_{-2}^{2}=$

$=\frac{3}{64} [(16\cdot2-\frac{8\cdot 2^3}{3}+\frac{2^5}{5})-(16\cdot(-2)-\frac{8\cdot (-2)^3}{3}+\frac{(-2)^5}{5})]=$

$=\frac{3}{64}[(32-\frac{64}{3}+\frac{32}{5})-(-32+\frac{64}{3}-\frac{32}{5})]=$

$=\frac{3}{64}(64-42\frac{2}{3}+12\frac{4}{5})=\frac{3}{64}(21\frac{1}{3}+12\frac{4}{5})=$

$=\frac{3}{64}(21\frac{5}{15}+12\frac{12}{15})=\frac{3}{64}\cdot34\frac{2}{15}=$

$=\frac{3}{64}\cdot \frac{512}{15}=\frac{8}{5}=1\frac{3}{5}.$

Answer: $(0, 1\frac{3}{5}).$