# Average of an integral question

1. Dec 15, 2007

### FatCat0

1. The problem statement, all variables and given/known data
This is a very simplified physics problem, just need help with the calc part:

What x value represents the average of the area for the semicircle with the equation y = +- (r^2 - x^2)^(1/2)?

2. Relevant equations
I called the integral A(x) because it represents area:
A(x) = +- ( (.5) ( x (r^2 - x^2)^(1/2) + r^2 arcsin (x/r) ) )
r > 0

3. The attempt at a solution
I bounded the region to 0 < x < r (so it represents a semi-circle of radius r in the positive x region).

A(0) = 0
A(r) = +- ( (.5) ( r (r^2 - r^2)0^(1/2) + r^2 arcsin (r/r) ) )
A(r) = +- ( (.5) ( (0) + (r^2 * 90) )
A(r) = +- (45 * r^2)

Then the average formula is (I think) ( A(0) + A(r) ) / (0 + r), but when I do that my average is +- (45 * r)...which can't be right, since the average x value has to be somewhere between 0 and r

Maybe you just...can't do the average for something that's split into two formulas like this? If you can't, is there any way you can?

2. Dec 15, 2007

### FatCat0

As an aside, even disregarding the +- thing, this doesn't work for a quarter-circle either, since the average couldn't be located 44 times farther from the origin than the shape
was.

3. Dec 15, 2007

### FatCat0

Another aside, maybe calling it the average was bad? I need to find the center of mass, which will be along the x-axis. Everything to the right of that point must equal everything to the left, in the system of course. Maybe I'm doing THAT wrong too.

4. Dec 15, 2007

### Feldoh

Well $$y = +/- \sqrt{(r^2 - x^2)}$$ makes an entire circle. But it's a complicated derivation to find the center of mass of a semi-circle.

Essentially your y-coordinate turn out to be: $$\frac{4R}{3\pi}$$

Last edited: Dec 15, 2007
5. Dec 15, 2007

### FatCat0

How did you get the that answer? I couldn't think of a way beyond taking $$y = +- \sqrt{r^{2} - x^{2}}$$, bounding it to a region of x that formed a semi-circle, integrating that, then trying to find the average.

Could you elucidate me as to the method you used? Or possibly point out where I went wrong in mine?

6. Dec 15, 2007

### HallsofIvy

Staff Emeritus
First you said "average area for a semicircle" which doesn't make sense. A semicircle doesn't have an "average" area, it has an area! Now you are saying you are looking for the "center of mass" (technically, the centroid). That doesn't have anything to do with an "average" value! The y coordinate of the centroid of a region R is given by
$$\frac{\int_R\int y dA}{\int\int dA}$$
For a semicircle of radius R, that is
$$\frac{\int_{x=0}^R \int_{y=0}^{\sqrt{R^2- x^2}}y dy dx}{\frac{1}{2}\pi R^2}$$

Of course it might be simpler to do the integral in polar coordinates:
$$\frac{\int_{r=0}^R\int_{\theta= 0}^\pi [r sin(\theta)][r dr d\theta]}{\frac{1}{2}\pi R^2}$$

Either of those will give the result Feldoh cites.

7. Dec 15, 2007

### Feldoh

Edit: Halls beat me to it :)

8. Dec 15, 2007

### FatCat0

How would you word the average thing? What I meant was sort of what I was saying in post 3; I was looking for an x-value where the area inside the semicircle to the left of it would be equal to the area in the semicircle to the right of it.

Both of those equations you posted, Ivy, are a little bit beyond me. We literally started integrals yesterday in calc, and I know a bit from physics already.

If it's not too much trouble, walk me through either of those methods just a bit? I think I see what you're doing, but don't quite get the integration part. Would the numerator of the first one become 0/(1/2*pi*R^2) and (R^2 - x^2)^(1/2) / (1/2 * pi * R^2) ? Where does the x come in, or does it at all?

Sorry, like I said...integral noob....

9. Dec 15, 2007

### Feldoh

The average value of an integral is an extension of the mean value theorem which says:

Avg Value = $$\frac{1}{b-a} \int_{a}^{b} f(x) dx$$ This gives you the average value of a function over the interval.

As I said it's sort of complicated but here's a link to get you started: http://mathworld.wolfram.com/GeometricCentroid.html

10. Dec 16, 2007

### FatCat0

Ahh, I'll work on it later and come back if I get even more confused.

Danke =)