How would I find the center of mass if the density is not uniform?

For a two-dimensional region R with density function p(x,y), which may or may not be uniform: x-coordinate of center of mass = double integral on R of x*p(x,y)dA / double integral on R of p(x,y)dA = double integral on R of x*p(x,y)dA / mass y-coordinate of center of mass = double integral on R of y*p(x,y)dA / double integral on R of p(x,y)dA = double integral on R of y*p(x,y)dA / mass Note that each coordinate of the center of mass can be thought of as a weighted average of that coordinate over the region, with the values of the density function used as the weights. Similar formulas can be used for regions besides two-dimensional regions. The concept is the same.

The example of density using the area are quite confusing. logically it is not acceptable as mass is product of density and volume. please Find some other example to explain.

How about this -- the height of the plate is 1, and the density doesn't depend on h. That is, at any given point if you move up and down through the "plate," the density doesn't change. Then we have ρ(x,y,h) = (sin(π x) + 1) y, where the variable h is "useless." So we don't need to account for it in our integral: m = ∫∫∫ρ dh dx dy= ∫∫ (∫ρ dh) dx dy = ∫∫ (ρ ∫dh) dx dy= ∫∫ (ρ h) dx dy = ∫∫ρ (1) dx dy = ∫∫ρ dx dy. We can take ρ out of the first integral because it doesn't depend on h. And since h is 1, it disappears nicely. So above we turned the density back into our good ol' three dimensional ρ. However, just because we live in three dimensions doesn't mean that is the only number of dimensions in which density exists. It is perfectly fine for an infinitesimally thin plane to have 2D density, σ = m/A = (a number) kg/㎡. Or 1D density on a line, λ = m/𝓁 = (a number) kg/m. They just describe the mass per area or length. You get these a lot in multivariable calculus and physics. You can even have 4D density, though I don't know what the symbol for it is. It would have units kg/m^4. Neither is mass the only thing that can have density. Charge density is almost as common in physics, σ = q/A = (a number) C/㎡, where q is charge, and C is coulombs. So in conclusion, you can make the density 3D density in this example to help you understand it, but really density doesn't have to be 3D. Then the metal plate is infinitesimally thin.

Why do we divide the answer by the area pi/2?

I had the same question initially, but I resolved it for myself by thinking of it as an average value. Recall the idea of the average value of a function f(x) on some interval [a,b]. It is calculated by taking the area under f(x) and dividing by the length of that interval (which is b-a). The b-a term is analogous to the pi/2 area term here.

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Multivariable calculus

Course: Multivariable calculus > Unit 4

Lesson 6: Double integrals (articles)

Double integrals beyond volume

Google Classroom

Double integrals do more than find volume under three-dimensional graphs. Here we cover other uses, a more general notation for double integrals, and explain the "feel" of double integration.

Background

Double integrals over non-rectangular regions

What we're building to

Double integrals are used anytime you get that feeling where you want to chop up a two-dimensional region into infinitely many infinitely small areas, multiply each one by some value, then add them up.
The more general notation for a double integral is
$\begin{aligned} \iint_\blueE{R} f\,\redE{dA} \end{aligned}$
where
start color #0c7f99, R, end color #0c7f99 is the region that you are integrating over.
start color #bc2612, d, A, end color #bc2612 signifies a "tiny chunk of area", which typically means d, x, d, y or d, y, d, x, unless another coordinate system is being used.
f, left parenthesis, x, comma, y, right parenthesis is a two-variable function.

Example 1: Mass of a plate

Imagine a metal plate 3 meters wide and 2 meters tall. Our goal will be to find its mass based on its density, but the catch is that the density is not constant over the plate.

To be able to describe this variable density with a function, start by situating the plate on the x, y-plane:

Its lower left corner at the origin, and its long side resting on the x-axis.

Let's say that the density of this plate, in start text, k, g, end text, slash, start text, m, end text, squared, is expressed with the following function.

sigma, left parenthesis, x, comma, y, right parenthesis, equals, left parenthesis, sine, left parenthesis, pi, x, right parenthesis, plus, 1, right parenthesis, y

(sigma is a typical variable name for two-dimensional density). Density is mass per unit area, so it might seem strange to define it using a function which takes in individual points. After all, what does it mean for a single point like left parenthesis, 1, comma, 2, right parenthesis to have density sigma, left parenthesis, 1, comma, 2, right parenthesis? If you prefer, you can interpret this function as giving the density within tiny region around each point.

To find the mass of the plate, you can imagine chopping it up into many tiny pieces, each one a rectangle, then adding up their masses.

Think of each of these rectangles as having a tiny width, d, x, and a tiny height, d, y.

Think about a specific rectangle, perhaps the one containing the point left parenthesis, 1, comma, 2, right parenthesis. Since this rectangle is really small, the density within it will pretty much equal the constant sigma, left parenthesis, 1, comma, 2, right parenthesis. The more finely you cut things, and the smaller the rectangles, the closer it is to being true that the density of each rectangle is constant.

This means we can find the mass of each such rectangle. For example,

\begin{aligned} \underbrace{ \sigma(1, 2) }_{\text{density}} \,\underbrace{ dx\,dy }_{\text{tiny area}} = (\sin(\pi)+1)(2)\,dx\,dy = 2\,dx\,dy \end{aligned}

To get the total mass of the plate, we integrate all of these tiny masses together. Since we are integrating over a two-dimensional region, we use a double integral. Caution: the order of your integrals depends on whether you express the tiny area of each rectangle as d, x, d, y or d, y, d, x

Concept check: Which of the following double integrals represents the mass of our metal plate, which is 3 meters wide and 2 meters tall:

(Choice A)
$\begin{aligned} \int_0^3 \int_0^2 \sigma(x, y)\,dx\,dy \end{aligned}$
(Choice B)
$\begin{aligned} \int_0^2 \int_0^3 \sigma(x, y)\,dx\,dy \end{aligned}$

Concept check: Using the function sigma, left parenthesis, x, comma, y, right parenthesis, equals, left parenthesis, sine, left parenthesis, pi, x, right parenthesis, plus, 1, right parenthesis, y, evaluate this double integral. (If you are unsure about how to do this, consider reviewing the article introducing double integrals)

\begin{aligned} \int_0^2 \int_0^3 (\sin(\pi x) + 1)y \,dx\,dy = \end{aligned}

Thinking about tiny areas

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When I first introduced double integrals, it was in the context of computing the volume under a graph. The thought process went something like this:

First cut the volume into infinitely many slices. Each slice represents a constant value for one of the variables, for example x, equals, 0, point, 78.
Find the area of each of those slices. (This is what the inner integral does).
Make each slice an infinitesimal volume by giving it a little depth. Mathematically, this means multiplying the area of each slice by either d, x or d, y, whichever one represents a tiny step perpendicular to the slice.
Integrate those infinitesimal volumes together to get the volume of the solid as a whole. (This is what the outer integral does).

By contrast, the example from the previous section finding the mass of the plate has a different look and feel to it. We start by thinking about tiny areas, then we multiply each one by a constant (the density) and try to add all of them together at once.

Of course, both these perspectives are equivalent. And when it comes to the computation, nothing will look different. You will always set up one integral inside another, compute the inner integral, then compute the outer integral.

Nevertheless, in terms of visualization and conceptual understanding, framing a double integral in terms of tiny areas is distinct from framing it as one linear integral inside another. For example, if you thought about computing the volume under a graph by initially breaking your region of the x, y-plane down into tiny areas, you might imagine adding together the volume of thin columns above those tiny areas.

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General notation for double integrals

When we think about a double integral with respect to tiny areas, it's common to write it abstractly like this:

\begin{aligned} \iint_\blueE{R} f\,\redE{dA} \end{aligned}

start color #0c7f99, R, end color #0c7f99

start color #0c7f99, R, end color #0c7f99 represents the region that we are integrating over. The reason for writing it like this is that while you are setting things up, or reasoning about double integrals in general, you typically don't want to get bogged down with the specific (and potentially complicated) definition of your region while you scribble things down.

When it comes time to compute the integral, we replace this \iint, start subscript, start color #0c7f99, R, end color #0c7f99, end subscript with an actual pair of single-integrals with bounds that can be computed. When start color #0c7f99, R, end color #0c7f99 is a rectangle, those bounds will be constants:

integral, start subscript, y, start subscript, 1, end subscript, end subscript, start superscript, y, start subscript, 2, end subscript, end superscript, integral, start subscript, x, start subscript, 1, end subscript, end subscript, start superscript, x, start subscript, 2, end subscript, end superscript, dots

More generally, when start color #0c7f99, R, end color #0c7f99 is defined in terms of some curves in the x, y-plane, the bounds of the inner integral are expressed as functions of the outer variable:

integral, start subscript, y, start subscript, 1, end subscript, end subscript, start superscript, y, start subscript, 2, end subscript, end superscript, integral, start subscript, x, start subscript, 1, end subscript, left parenthesis, y, right parenthesis, end subscript, start superscript, x, start subscript, 2, end subscript, left parenthesis, y, right parenthesis, end superscript, dots

(See the last article for practice with this idea.)

start color #bc2612, d, A, end color #bc2612

start color #bc2612, d, A, end color #bc2612 represents a tiny area, in the same way that d, x represents a tiny length in an ordinary integral.

You will typically imagine chopping up the region start color #0c7f99, R, end color #0c7f99 into many tiny pieces, and this term represents the area of one of those pieces. Once you get down to computing the double integral, you will replace this with d, x, d, y, or d, y, d, x. In other coordinate systems, there are different ways to break down start color #bc2612, d, A, end color #bc2612, but I'll leave that for the next article.

f, left parenthesis, x, comma, y, right parenthesis

f, left parenthesis, x, comma, y, right parenthesis is some two-variable function. When you chop up your region into many tiny pieces, each piece typically represents some value that you are hoping to add up. Perhaps this value is a tiny bit of mass, or the tiny volume of a slim column under a graph.

Hopefully, you are able to express this tiny amount as something times the area of your tiny piece. For example, the mass of a piece is its density times its area; and the volume of a column above a piece is the height of the column times the area.

In these examples, f, left parenthesis, x, comma, y, right parenthesis represents density, or height. In general, it is the thing that needs to be multiplied by the area start color #bc2612, d, A, end color #bc2612 of a tiny piece, and it generally depends on the position of that tiny piece, expressed with left parenthesis, x, comma, y, right parenthesis-coordinates.

"What if I cannot express the tiny value that I want to add up as something times d, A?"

Well, in that case my friend, double integrals are not the tool for you. Although I cannot think of any examples where that comes up...

There are two benefits to this abstract notation:

Simplicity: When you're starting to set something up, or if you want to quickly reference the idea of a certain double integral without getting into the implementation details, it's nice to be able to write something quickly. Also, many of the theorems and tools coming up in multivariable calculus are expressed abstractly in this notation.
Generality: Writing your integral as \iint, start subscript, start color #0c7f99, R, end color #0c7f99, end subscript, f, start color #bc2612, d, A, end color #bc2612 gives you options as you sit down to compute it. For example, in the next article, we will cover double integrals in polar coordinates, in which case the way you expand start color #bc2612, d, A, end color #bc2612 and the way you put bounds on the two integrals are different than they are for cartesian coordinates.

Example 2: Center of mass

What is the center of mass of a half-disk?

For simplicity, let's say the radius of the disk is 1, and let's orient it such that the diameter rests on the y-axis. Also, assume the disk has uniform density everywhere.

This is a pretty interesting problem, don't you think? You can guess that the answer is something slightly to the left of left parenthesis, 0, point, 5, comma, 0, right parenthesis, but it's not obvious what the specific answer should be, is it?

By the vertical symmetry of this half-disk, you can know that the center of mass will lie on the x-axis. In a sense, what we're looking for is the "average x-value" of points in the disk.

Concept check: If we let H represent this half-disk, with vertical bar, H, vertical bar representing its area, which of the following abstractly written integrals represents the x-value for the center of mass of H?

(Choice A)
$\dfrac{1}{|H|} \begin{aligned} \iint_H x \,dA \end{aligned}$
(Choice B)
$\dfrac{1}{|H|} \begin{aligned} \iint_H \sqrt{1-x^2} \,dA \end{aligned}$

Concept check: What is the area of the half-disk H?

vertical bar, H, vertical bar, equals

Concept check: Which of the following represents the right way to expand the integral \iint, start subscript, H, end subscript, x, d, A into a computable form?

(Choice A)
$\begin{aligned} \int_{-1}^1 \int_0^{\sqrt{1 - y^2}} x\,dx\,dy \end{aligned}$
(Choice B)
$\begin{aligned} \int_{0}^1 \int_{-\sqrt{1-x^2}}^{\sqrt{1 - x^2}} x\,dy\,dx \end{aligned}$

Bring it on home: Solve this integral, and use it to find the center of mass of H.

x-coordinate of center of mass:

Summary

Double integrals are used anytime you get that feeling where you want to chop up a two-dimensional region into infinitely many infinitely small areas, multiply each one by some value, then add them up.

The more general notation for a double integral is

\begin{aligned} \iint_\blueE{R} f\,\redE{dA} \end{aligned}

where

start color #0c7f99, R, end color #0c7f99 is the region that you are integrating over.
start color #bc2612, d, A, end color #bc2612 signifies a "tiny chunk of area", which typically means d, x, d, y or d, y, d, x, unless another coordinate system is being used.
f, left parenthesis, x, comma, y, right parenthesis is a two-variable function.

From this point forward, double integrals will be inextricably tied to most of the new topics in multivariable calculus. And in almost all cases, it helps to think about what's happening inside each "tiny area" of a given region, rather than thinking about integrating something along a line then integrating again in the perpendicular direction.

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David
Posted 7 years ago. Direct link to David 's post “The bounds for the integr...”
The bounds for the integral of the last problem for dy should be from sqrt(1-x^2) to -sqrt(1-x^2)
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- aryadityabasisth
  Posted 4 months ago. Direct link to aryadityabasisth's post “No, After computing the i...”
  No, After computing the inner integral, you get the function which tells you the area of a slither whose base is the horizontal component on the input region... That slither of the input region then varies from y=+1 to y=-1 hence they are the bounds. Another intuition to see this is that when you enter the bounds you suggest, we would get our answer as a function of x which we shouldn't get.
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Minhaj Turjo
Posted 3 years ago. Direct link to Minhaj Turjo's post “What did he mean by "aver...”
What did he mean by "average x-value"? And how come dividing the integral by the total mass of the semi-circle gives this "average x-value"?
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Lucas
Posted 4 years ago. Direct link to Lucas's post “How would I find the cent...”
How would I find the center of mass if the density is not uniform?
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(2 votes)
Answer
- Ian Pulizzotto
  Posted 4 years ago. Direct link to Ian Pulizzotto's post “For a two-dimensional reg...”
  For a two-dimensional region R with density function p(x,y), which may or may not be uniform:
  
  x-coordinate of center of mass =
  double integral on R of x*p(x,y)dA / double integral on R of p(x,y)dA
  = double integral on R of x*p(x,y)dA / mass
  
  y-coordinate of center of mass =
  double integral on R of y*p(x,y)dA / double integral on R of p(x,y)dA
  = double integral on R of y*p(x,y)dA / mass
  
  Note that each coordinate of the center of mass can be thought of as a weighted average of that coordinate over the region, with the values of the density function used as the weights.
  
  Similar formulas can be used for regions besides two-dimensional regions. The concept is the same.
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  (4 votes)
Vinod Rathore
Posted 7 years ago. Direct link to Vinod Rathore's post “The example of density us...”
The example of density using the area are quite confusing. logically it is not acceptable as mass is product of density and volume. please Find some other example to explain.
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(1 vote)
Answer
- Alexander Wu
  Posted 7 years ago. Direct link to Alexander Wu's post “How about this -- the hei...”
  How about this -- the height of the plate is 1, and the density doesn't depend on h. That is, at any given point if you move up and down through the "plate," the density doesn't change. Then we have ρ(x,y,h) = (sin(π x) + 1) y, where the variable h is "useless." So we don't need to account for it in our integral:
  
  m = ∫∫∫ρ dh dx dy= ∫∫ (∫ρ dh) dx dy = ∫∫ (ρ ∫dh) dx dy= ∫∫ (ρ h) dx dy = ∫∫ρ (1) dx dy = ∫∫ρ dx dy.
  
  We can take ρ out of the first integral because it doesn't depend on h. And since h is 1, it disappears nicely.
  
  So above we turned the density back into our good ol' three dimensional ρ. However, just because we live in three dimensions doesn't mean that is the only number of dimensions in which density exists. It is perfectly fine for an infinitesimally thin plane to have 2D density, σ = m/A = (a number) kg/㎡. Or 1D density on a line, λ = m/𝓁 = (a number) kg/m. They just describe the mass per area or length. You get these a lot in multivariable calculus and physics. You can even have 4D density, though I don't know what the symbol for it is. It would have units kg/m^4.
  
  Neither is mass the only thing that can have density. Charge density is almost as common in physics, σ = q/A = (a number) C/㎡, where q is charge, and C is coulombs.
  
  So in conclusion, you can make the density 3D density in this example to help you understand it, but really density doesn't have to be 3D. Then the metal plate is infinitesimally thin.
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  (3 votes)
Michael Robertson
Posted 7 years ago. Direct link to Michael Robertson's post “In the final answer, the ...”
In the final answer, the page won't accept (4/3)pi as an answer, even though this is the answer given in the explanation.
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Aiman
Posted 3 years ago. Direct link to Aiman's post “Why do we divide the answ...”
Why do we divide the answer by the area pi/2?
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(1 vote)
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- 𝜏 Is Better Than 𝝅
  Posted 3 years ago. Direct link to 𝜏 Is Better Than 𝝅's post “I had the same question i...”
  I had the same question initially, but I resolved it for myself by thinking of it as an average value. Recall the idea of the average value of a function f(x) on some interval [a,b]. It is calculated by taking the area under f(x) and dividing by the length of that interval (which is b-a). The b-a term is analogous to the pi/2 area term here.
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  (2 votes)
davorkuharic1997
Posted 6 years ago. Direct link to davorkuharic1997's post “In the example 1 how did ...”
In the example 1 how did we get in a second row (-cos(pi*x))/pi out of first row shouldn't it be -cos(pi*x)*(pi*x^2)/2
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(1 vote)
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- jjklg
  Posted 6 years ago. Direct link to jjklg's post “If I understand what you ...”
  If I understand what you are asking, it seems as though you are trying to use the chain rule for integration, but the chain rule only works for differentiation. To integrate sin (pi*x), you have to think of the chain rule backward. We know that the anti derivative of sin is -cos, and the derivative of pi*x is pi, so the anti derivative of sin (pi*x) is (-cos (pi*x))/pi. To check this answer, just use the chain rule on (-cos(pi*x))/pi to see if you get sin (pi*x), which you do because the derivative of what is inside the -cos () (pi in this case) gets canceled by the pi in the denominator, and the derivative of -cos is sin.
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  (1 vote)