Finding a centriod (center of mass)

[thename1000]

Studying for a test and I can't really grasp a full example problem. I would also like help on a couple further example problems that are not done for me.

Homework Statement

Find the exact coordinates of the centriod:

a. y=4-x^2, y=0
b. 3x+2y=6, y=0, x=0
c. y=e^x, y=0, x=0, x=1
d. y=1/x, y=0, x=1, x=2

Homework Equations

equation 1: xbar=(1/A) of integral from a to b of xf(x) dx
equation 2: ybar=(1/A) of integral from a to b of (1/2)* [f(x)]^2 dx

A= area
f(x)=the function

The Attempt at a Solution

a.) I have this entire problem worked out for me. I understand why xbar=0, but not how to solve for ybar. In my notes it goes from the line ybar=equation 2 (with f(x) filled in)
to A= integral of -2 to 2 of (4-x^2)dx
to 2 * integral from 0 to 2 of (4-x^2)dx
I don't know why the limits changed or where the 2 came from.

b-d
Could someone run down how to do these. Don't complete the problem or anything but I'm not sure about:

*What to do with the second conditions. (y=1, x=0, y=0 etc.
*To find x intercepts for integration do I just solve for x?

It looks to me like you follow these steps:

1. Find out if either xbar or ybar are zero (odd). Which saves time.
2. Find area by integrating the function. <<Not clear to me how exactly to do this.
3. Plug A into given equation
4. Integrate

Homework Statement

thanks

 

[lanedance]

first think about the shape, in this case an negtive paraboloid bounded by the x axis(y=0)

so xbar is zero as the shape is symmetric about the y axis

to compute ybar use the follwing steps, first compute the whole area by either an intergal over x or y, as we are trying to find y bar, use an integral over y (though its a little trickier to set-up)

now for a given y, is what is the area, dA(y), of a sliver of width dy? rearranging for x

$$x = \sqrt{4-y}$$
then counting the contribution on each side of the y-axis gives:
$$dA(y) = 2.x.dy = 2.sqrt(4-y).dy$$
to find the area integrate from the x-axis to the turning point
$$A = \int dA(y) = \int_0^4 2.sqrt(4-y).dy$$
note you could set this up in termes of x and it would be a bit easier, though we need it again later

to work out the centre of mass now sum up the contribution of each area element using an integral
$$\bar{y}.A = \int y.dA(y) = \int_0^4 2.y.sqrt(4-y).dy$$

the the average centre of mass is given by
$$\bar{y} = \frac{\int y.dA(y)}{\int dA(y)} = \frac{\bar{y}.A }{A}$$