Find centroid of region of two curves

Question

Find the coordinates (to three decimal places) of the centroid of $y=2^x$, and $y=x^2$.

EDIT: $(0\le x\le2)$

I understand this with a triangle, not with curves.

Answer 1

The computation of the centroid in $R^2$, of a region bounded by two continuous functions, goes, by definition, as follows. (Note that, over $[0,2]$, $x^2 \le 2^x$.)

First, one has to calculate the area, $\mathscr a$, of the region $$A=\{x,y\ ;\ 0\le x\le 2, \ x^2\le y\le 2^x\}.$$ $ \color{white}{bbbbbbbbbbb}$ $$\text{The region with the centroid to be calculated below.}$$ $$\color{white}{nnn}$$ We have for the area: $$\mathscr a= \iint_Adydx=\int_0^2\ \left[\int_{x^2}^{2^x}dy\right]\ dx=\int_0^2 2^xdx-\int_0^2x^2dx.$$

Then, for the coordinates of the centroid:
$$\overline x=\frac{1}{a}\int_0^2 x(2^x-x^2)\ dx=\frac{1}{\mathscr a}\left(\int_0^2x2^xdx-\int_0^2x^3dx\right),$$ $$\overline y=\frac{1}{2\mathscr a}\int_0^2 (2^x+x^2)(2^x-x^2)\ dx=\frac{1}{2\mathscr a}\left(\int_0^22^{2x}dx-\int_0^2x^4dx\right).$$ Now, we calculate the following integrals

$$\int_0^2 2^xdx=\frac{3}{\ln2},$$ $$\int_0^2x^2dx=\frac{8}{3},$$ $$\int_0^2 x2^x\ dx=\frac{8}{\ln2}-\frac{3}{(\ln2)^2},$$ $$\int_0^2x^3\ dx=4.$$ $$\int_0^22^{2x}\ dx=\frac{15}{2\ln2},$$ $$\int_0^2x^4\ dx=\frac{32}{5}.$$

Having done all this, one can easily put together the pieces: $$a=\frac{3}{ln2}-\frac{8}{3}=1.6614,$$ $$\overline x=\frac{1}{\mathscr a}\left(\frac{8}{\ln2}-\frac{3}{(\ln2)^2}-4\right)=0.7809,$$ $$\overline y=\frac{1}{2\mathscr a}\left(\frac{15}{2\ln2}-\frac{32}{5}\right)=1.3303.$$

Answer 2

Here is the method of finding the centroid according to Wikipedia of the region bounded by the graphs of $f(x)$ from above, $g(x)$ from below, $x=a$ on the left, and $x=b$ on the right. (We assume that $a\le b$ and $f(x)\ge g(x)$ for $a\le x\le b$.)

1) Find the area between the two curves in your given domain with

$$A=\int_a^b [f(x)-g(x)]\,dx$$

In your case, that is

$$A=\int_0^2 [2^x-x^2]\,dx$$

2) Find $\bar x$, the $x$-coordinate of the centroid, with

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

In your case, that is

$$\bar x=\frac 1A\int_0^2 x[2^x-x^2]\,dx$$

3) Find $\bar y$, the $y$-coordinate of the centroid, with

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

In your case, that is

$$\bar y=\frac 1A\int_0^2 \left[\frac{2^x+x^2}2\right][2^x-x^2]\,dx$$