# Centroid of parabola

## Homework Statement

y = 16 -x^2 find centroid bounded by x axis

## Homework Equations

x = (1/A) ∫ x(f(x)) dx and (1/A) (1/2)(f(x))^2 dx = y

## The Attempt at a Solution

I just applied it. It is a weird because I would of thought that x would of been at 0. But I didn't get that I x = 9/16 and y = 8/5. Y might be OK but wouldn't x have to be at 0?

I calculated the area I got 256/3

So x = (3/256) ∫(x(16-x^2) dx = (3/256)∫ 16x - x^3 dx = (3/256) (8x^2 -(x^4/4)) between [-4,4] = 9/16

For y = (3/256) ∫ (1/2)(16-x^2)^2 dx = (3/256)∫ 256 - 32x^2 + x^4 dx = (3/512)(256x = (32/2)x^3 - (x^5/5)) between [-4,4] = 8/5

Like I said it seems that x would be at 0 but I'm not sure I think I did everthing OK.

Mark44
Mentor

## Homework Statement

y = 16 -x^2 find centroid bounded by x axis

## Homework Equations

x = (1/A) ∫ x(f(x)) dx and (1/A) (1/2)(f(x))^2 dx = y

## The Attempt at a Solution

I just applied it. It is a weird because I would of thought that x would of been at 0. But I didn't get that I x = 9/16 and y = 8/5. Y might be OK but wouldn't x have to be at 0?
Your intuition is correct, but your work isn't. ##\bar{x}## = 0, which is pretty obvious from the symmetry of the region.
I calculated the area I got 256/3

So x = (3/256) ∫(x(16-x^2) dx = (3/256)∫ 16x - x^3 dx = (3/256) (8x^2 -(x^4/4)) between [-4,4] = 9/16

For y = (3/256) ∫ (1/2)(16-x^2)^2 dx = (3/256)∫ 256 - 32x^2 + x^4 dx = (3/512)(256x = (32/2)x^3 - (x^5/5)) between [-4,4] = 8/5

Like I said it seems that x would be at 0 but I'm not sure I think I did everthing OK.

So, Even though it is the long way if I did it the way I did it should workout to be 0. Why do I get what I get ? did you look at y ? What did you think of y?

I found my mistake for x. Do you think y looks OK I went over it again seems OK to me.

Mark44
Mentor
Your integral doesn't look right to me. It should be
$$\frac 1 A \int_0^{16} y * (x_{right} - x_{left}) dy$$

The integrand represents the moment of a horizontal strip across the parabola, and the moment in this case is the length of the lever arm (y) times the area of the horizontal strip ΔA = (xright - xleft)Δy.

Your function is y = 16 - x2, or equivalently, x = ±√(16 - y).

Part of your problem seems to be treating x and y cavalierly - you can't do that and come up with the right answer.

With all of these applications of integration, it's extremely helpful to have a good drawing of the situation. Just memorizing some formulas and using them blindly isn't a good strategy.

##\frac 1 A \int_0^{16} (1/2)[f(x)]^2 dx##

This is what my book gives for calculation of y. I don't understand your formula. Well I do but why would I need to deviate from what the book says for this problem. I don't see why?

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Mark44
Mentor
##\frac 1 A \int_0^{16} (1/2)[f(x)]^2 dx##

This is what my book gives for calculation of y. I don't understand your formula. Well I do but why would I need to deviate from what the book says for this problem. I don't see why?

Your formula should be ##\frac 1 A \int_{-4}^{4} (1/2)[f(x)]^2 dx##, with the difference being the limits of integration. You're integrating with respect to x, and x is in the interval [-4, 4].

With that correction, your formula and mine should agree. The difference between them is that my formula uses horizontal strips that extend across the parabola. For each of these the lever arm is the y value on the strip. Each of my horizontal strips is very narrow, so all the strips have about the same y value. In my formula, the moment is y * <width of strip> * Δy.

Your formula uses vertical strips of height f(x). The lever arm is at the middle of the strip, so is (1/2)f(x). The moment is f(x) * (1/2)f(x) * Δx.

I meant [-4,4] I forgot to change it. Yeah I recalculated y with my formula and I got 32/5.

In yours what is x left and right? Would it be after subtracting.

##\frac 1 A \int_0^{16} y * (2\sqrt(16-y)) dy##

Dick
Homework Helper
Your integral doesn't look right to me. It should be
$$\frac 1 A \int_0^{16} y * (x_{right} - x_{left}) dy$$

The books formula is right. It's an equivalent formulation. To show they are the same you use Green's theorem.

Dick
Homework Helper
Your formula uses vertical strips of height f(x). The lever arm is at the middle of the strip, so is (1/2)f(x). The moment is f(x) * (1/2)f(x) * Δx.

Yes, or you can explain it that way instead of using Green's theorem. Actually somewhat nicer.

Pretty out there though for me.

What class out greens theorem come up/

Would*

Dick