Question

Centroid of Plane Area

Locate the centroid of the region 𝑅 bounded by the given curves. Sketch the graph.

1.) 𝑅:𝑦=𝑥^2, 𝑦=2𝑥^2 −3𝑥

2.) 𝑅:𝑦^2=4𝑥, 𝑦=4, 𝑦−𝑎𝑥𝑖𝑠

3.) 𝑅:𝑥=2𝑦−𝑦^2, 𝑦−𝑎𝑥𝑖𝑠

4.) 𝑅:𝑦=2𝑥+1, 𝑥+𝑦=7, 𝑥=8

Answer 1

1)$x=\frac{\iint\limits_D xdxdy}{\iint\limits_D dxdy}=\frac{\int\limits_0^3\int\limits_{2x^2-3x}^{x^2}xdydx}{\int\limits_0^3\int\limits_{2x^2-3x}^{x^2}dydx}=\frac{\int\limits_0^3(x^2-2x^2+3x)xdx}{\int\limits_0^3(x^2-2x^2+3x)dx}=\frac{6.75}{4.5}=1.5\\ y=\frac{\iint\limits_D ydxdy}{\iint\limits_D dxdy}=\frac{0.5*\int_0^3 ((x^2)^2-(2x^2-3x)^2)dx}{4.5}=\frac{0.5*16.2}{4.5}=1.8$

2

$x=\frac{\iint\limits_D xdxdy}{\iint\limits_D dxdy}=\frac{\int\limits_0^44x\sqrt{x}dx}{\int\limits_0^44\sqrt{x}dx}=\frac{256/5}{64/3}=2.4$

due to symmetry about the x-axis, y = 0

3

$x=\frac{\iint\limits_D xdxdy}{\iint\limits_D dxdy}=\frac{\int_0^1 2*x*\sqrt{1-x}dx}{\int_0^1 2*\sqrt{1-x}dx}=\frac{8/15}{4/3}=0.4$

due to symmetry with respect to y = 1, y = 1

4

this is a triangle

$x=(1/3)*(2+8+8)=6\\ y=(1/3)*(5 -1+17)=7$