## Center of Mass Example Problems Using the Center of Mass Equation Integral

Today, we’re going to be going through some center of mass example problems. So, we’re going to sketch the region bounded by the curves and then find the exact coordinates of the centroid of that region. And the curves that we have in this first example are:

$$y=4-x^2, \ and$$ $$y=0$$

If you would prefer to watch a video of this problem, you can do so here:

## Centroid vs Center of Mass

Before we jump into this first example, I do want to just point out one thing. And that is the difference between a centroid vs center of mass. The reason I wanna bring that up is because, you can see this problem is asking us to find the coordinates of the centroid. But I said I was going to show you center of mass example problems. Well, they’re pretty much the same thing in most contexts. In this context specifically, they’re going to be the same.

Really, the only difference when you’re looking at a centroid versus a center of mass is that a center of mass is referring to the center point of some actual physical object that actually has a mass. The context that you’d usually see for that is when you have a thin plate, which is described by some functions or some region, and you want to find the center of mass of that plate. Whereas, a centroid usually comes into play where you are just described some region that exists bound between two curves on an x-y-plane.

Like this case here for example. We were just given these functions and we are looking at the region that is trapped between those functions. Since we’re looking at some region, “centroid” would be the proper terminology there. But if you’re ever given some uniform density object like a thin plate, for example, the centroid and the center of mass are actually going to be the same thing. So you would get the same point whether you were thinking of it in either context. They are going to use the same formulas to figure those out.

So that brings me to the center of mass equation integrals, which are two equations that are on my calculus 2 study guide. If you haven’t checked that out you can click here to learn more about that. You can go download that right away. It’s only a few bucks, it’s pretty affordable. And I highly recommend you grab yourself a copy of that. It’s available right away, so you can go start using that today. Before I show you how to apply the formulas on my study guide though, let’s go ahead and start with graphing the given curves.

## Sketch the Given Curves and the Bounded Region

First we will start with $y=4-x^2$. This is just going to be a downward facing parabola with the vertex at the point (0, 4). Then, the line $y=0$ is going to be a horizontal line on the x-axis. Therefore, we would get a region like the one below.

So once you’ve sketched your region and kind of given yourself a visualization of what we’re trying to do here, the best place to go from there is to just go straight into the center of mass equation integrals.

## Center of Mass Equation Calculus

There’s a separate equation for the x-coordinate of the centroid and for the y-coordinate of the center of mass. Those equations, from my calculus 2 study guide are:

$$\bar{x} = \frac{1}{A} \int_a^b x f(x) \ dx$$ $$\bar{y} = \frac{1}{A} \int_a^b \frac{1}{2} \Big[ f(x) \Big]^2 \ dx$$

Where the centroid of the region will be at the point $( \bar{x}, \ \bar{y} )$.

Before I show you how to use these, I do just wanna point out the different pieces of these equations. First of all, we have A in both of these equations. A is just the area of the region whose centroid, or center of mass, we’re trying to find. We would first need to figure out the area between these two curves. And that’s what A would be. I’m not going to show you how to do that here, but you can see more about that by clicking here.

Then we have the bounds of our integrals, the a and b. Those bounds of the integrals are just going to be the x values that are the left and right edge of our region. So in this case, a will be -2. Meanwhile, b will be 2. And that’ll be true for both of those integrals.

Finally, we have f(x). That is just going to be the function that creates this region. These functions assume that the lower bound of the region will be formed by the line $y=0$. Based on that, f(x) is just going to be the top function, $y=4-x^2$.

You can simply use those pieces in the formulas above, and I’ll show you how to do that shortly. However, before we do that I do want to point out one thing.

## Finding the Center of Mass of a Symmetrical Region

This is an interesting example, because you can see this region whose centroid we are looking for, is actually symmetrical in the x direction. This region is symmetrical to the right and to the left of the y-axis. Whenever you have symmetry in your region that actually saves you some work. Since we have symmetry in the x direction, that tells us that the x coordinate of our centroid must be on that line of symmetry, which is the line $x=0$. As a result, we already know that we’ll have $\bar{x} = 0$.

Since we already know $\bar{x}=0$, all we need to do is use the above formula to calculate $\bar{y}$.

## How to Apply the Center of Mass Formula

If you use that method shown here, you would figure out that the area of this region is $\frac{32}{3}$. And like I said earlier, the bounds of the integral are the left edge and the right edge of this area. So that gives us $a=-2$ and $b=2$. I also mentioned above that we will know that $f(x)=4-x^2$. Plugging this into the $\bar{y}$ equation tells us:

$$\bar{y} = \frac{1}{32/3} \int_{-2}^{2} \frac{1}{2} \Big[ 4-x^2 \Big]^2 \ dx$$

I do want to point out something important about the above equation. When you are applying these equations you put the f(x) all in brackets or parentheses, because we need make sure to square this whole function. From here we can simplify things a bit. When you have a fraction in the denominator of another fraction like we do here, you want to keep in mind that dividing is the same as multiplying by its reciprocal. Therefore, $\frac{1}{32/3}$ can be rewritten as $\frac{3}{32}$.

Then we can expand and simplify the rest of the integral.

$$\bar{y} = \frac{3}{32} \cdot \frac{1}{2} \int_{-2}^{2} (4-x^2)(4-x^2) \ dx$$

$$\bar{y} = \frac{3}{64} \int_{-2}^{2} 16-8x^2+x^4 \ dx$$

$$\bar{y} = \frac{3}{64} \Bigg[ 16x \ – \frac{8}{3}x^3+ \frac{x^5}{5} \Bigg]_{-2}^2$$

$$\bar{y} = \frac{3}{64} \Bigg[ \Bigg( 16(2) \ – \frac{8}{3}(2)^3+ \frac{(2)^5}{5} \Bigg) – \Bigg( 16(-2) \ – \frac{8}{3}(-2)^3+ \frac{(-2)^5}{5} \Bigg) \Bigg]$$

$$\bar{y} = \frac{3}{64} \cdot \frac{512}{15}$$

$$\bar{y} = \frac{8}{5}$$

That tells us that the y-coordinate of our centroid of this region is $\bar{y}=\frac{8}{5}$. And we already figured out that the x-coordinate of our centroid was $\bar{x}=0$. Therefore, the centroid of this region between these two given functions is going to be $(0, \ \frac{8}{5})$.

Again, the center of mass equations are on my calculus two study guide. You can click here to check that out and get your copy today!