Find the centroid of the region bounded by the graphs of y arcsin x x 0 and y 2

Finding the centroid of the region

Finding the centroid of a region can be a difficult task, but with a little knowledge of calculus it can be done relatively easily. In this section we will discuss how to find the centroid of the region bounded by the graphs of y= arcsin(x) and y= 2.

Bounding the region

In order to calculate the centroid of a region, we first need to bound the region. The most common way to do this is with a minimum bounding rectangle (MBR), which is simply the smallest rectangle that contains the entire region.

To find the MBR of a region, we first need to find the minimum and maximum x-coordinates and y-coordinates of all the points in the region. Once we have these four values, we can define our MBR as follows:

Finding the centroid

There are a few methods for finding the centroid of a region, but the most common is the “center of mass” method. This involves taking the average of all the points in the region. So, if you have a region that is defined by a set of points (x1,y1), (x2,y2), …, (xn,yn), then the centroid would be:

(x1+x2+…+xn)/n, (y1+y2+…+yn)/n

Another common method is the “geometric center” method. This involves taking the mean of all the coordinates of the points in the region. So, using the same example above, the geometric center would be:

(x1+x2+…+xn)/n, (y1+y2+…+yn)/n

Examples

We will be finding the centroid of the region that is bounded by the graphs of y = arcsin(x), x = 0, and y = 2. This region is shown in the picture below.

Example 1

To find the centroid of the region bounded by the graphs of y=arcsin(x) and y=2, we’ll need to use the formula for the centroid of a region:

centroid = (1/A) integral[ydx] from a to b

In this case, a=0 and b=1, so our integral becomes:

centroid = (1/A) integral[ydx] from 0 to 1

Since we’re taking the integral of a function that’s a product of two functions (y and dx), we’ll need to use the product rule:

centroid = (1/A) integral[ydx] from 0 to 1
= (1/A) integral[(arcsin(x))(1/cos^2(x))] from 0 to 1

Example 2

To find the centroid of the region bounded by the graphs of y=arcsin x and y= 2x, we will use the formula below.

$$ \bar{x}= \frac{\int_{a}^{b}xf(x)dx}{\int_{a}^{b}f(x)dx} $$

In this case, we are looking for the centroid of the shaded region in the graph.

First, we will calculate the integrals on the numerator and denominator separately. For the numerator, we will use the substitution x=sin u:

$$ \int_{a}^{b}xf(x)dx = \int_{0}^{\pi/2} (\sin u)(2\sin u) du = 2\int_{0}^{\pi/2}\sin^2u du = [u-\sin u \cos u]_0^{\pi/2} = \pi/2 – 1 $$

For the denominator, we will use the substitution x=tan v:

$$ \int_{a}^{b}f(x)dx = \int_{0}^{\pi/4}\frac{1}{\cos^2v}\frac{1}{1+v^2}\ dv = \int_{0}^{\pi/4}\frac{1}{1+v^2}\ dv = [v-\tan v]_0^{\pi/4} = \pi/4 – 1 $$

Now that we have both integrals, we can plug them into our formula for $\bar{x}$:

$$ \bar{x}= \frac{\int_{a}^{b}xf(x)dx}{\int_{a]^{b]}f(x)dx)}=\frac{\pi/2-1}{\pi/4-1}\approx 1.61904 $$

jerichorosas