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Let $A$ be the area and $\bar{x}$ the $x$ -coordinate of the centroid of a region $R$ that is bounded by a piecewise smooth, simple closed curve $C$ in the $x y$ -plane. Show that $$\frac{1}{2} \oint_{C} x^{2} d y=-\oint_{C} x y d x=\frac{1}{3} \oint_{C} x^{2} d y-x y d x=A \bar{x}.$$

$$=A \bar{x}$$

Calculus 3

Chapter 16

Integrals and Vector Fields

Section 4

Green’s Theorem in the Plane

Vector Functions

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03:04

In mathematics, a function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. An example is the function that relates each real number x to its square x. The input of a function is called the argument and the output is called the value. The set of all permitted inputs is called the domain of the function. Similarly, the set of all permissible outputs is called the codomain. The most common symbols used to represent functions in mathematics are f and g. The set of all possible values of a function is called the image of the function, while the set of all functions from a set "A" to a set "B" is called the set of "B"-valued functions or the function space "B"["A"].

08:32

In mathematics, vector calculus is an important part of differential geometry, together with differential topology and differential geometry. It is also a tool used in many parts of physics. It is a collection of techniques to describe and study the properties of vector fields. It is a broad and deep subject that involves many different mathematical techniques.

04:26

Let $A$ be the area and …

01:29

Area and the centroid Let …

05:22

02:58

Let $D$ be a region bounde…

Yeah, in this problem, we are practicing an application of greens. The're, um and specifically we're going to be looking at line in the girls and how we can informally prove a specific result, given our knowledge of the the're, um, and the properties of line integral. So the first that we should dio it's let's define the X coordinate of the Centro oId in two dimensions. So we'll call that X bar and then we'll multiply by m y over em. So that would be the double integral with respect to our region R of X in D A over the double integral again with respect to this region are in d A. And then we should also note here that Delta X y Z equals one. Now, when we simplify this, we would get one over a times a double integral with respect to this region R of X n d. A. And now I have to use some properties of the fear. Um so, by green serum for M equal to one half X squared and an equal to zero, we can deduce this information so we'll start with a double integral with respect toe are of x India. We can simplify this using the fear. Um, and we'll get the double integral with respect to our of the partial derivative with respect to X of one half X squared, minus the partial derivative with respect to Y of zero in de y DX. We can simplify that to get the line in the girl. Now, with respect to our curve C of zero d x plus one half x squared in D Y, we would get one half times the line integral X squared in D. Y. And now we can simplify that. We can say a X bar, Remember, our X coordinate from before equals the in a row. We just found the one half our line in a row with respect to this curve of X squared in D Y. Now we'll do the same process now for M equal to negative X y and an equal to zero again. We'll have to use green serum so we'll start with a double integral with respect to this region R of X, d y d x, and then we can simplify that using the the're, um will have a double integral with respect to that region are of the partial derivative of zero with respect to X minus the partial derivative of negative Y X. With respect to why in de y t X, you would get the negative line in a girl with respect to our curve C of X y in D X, and we can call that a X bar equals negative, the line integral with respect to our curve C of x t y dx. So now we can to do some information that we were given in the problem. So we're going to get one third time's the line, integral with respect to our curve of X squared y minus X y DX, we would simplify that weaken split up that line integral because that's a property of integration. We would get one third time's the line integral again with respect to our curve of X squared in D Y. Minus one third times are in line integral of X, y and D X. But remember, we defined these line into girls with respect to our X coordinate, so we would get using that knowledge one third times, too. A X bar plus one third a X bar. We could simplify that to get three a X bar over three are three's would cancel and we would get a X bar, and that's what we were trying to prove. So I hope that this problem helped to understand how we can use Green. The're, um, in a different setting, just using our knowledge of the line integral in its relationship in green serum, how to informally prove a result that were given.

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