Basic Line-Integral: Just trying to know what is being asked

In summary, the problem involves finding the line integral of a vector function A, given by A = x2ˆx + y2ˆy + z2ˆz, over a specified path (x, √x) from (0,0) to (2, √2). The path must be specified in order to integrate the function. The vector element ds is the derivative of the position vector s, which is represented as s = xˆx + √xˆy. The dot product of A and ds is taken and then integrated over the specified path. A can be any arbitrary vector function and need not have any physical significance.
  • #1
sgholami
7
1
Hello. I'm new to physics, and the problem I have seems so basic, mathematically speaking. I'm just failing to grasp exactly what is being asked. If I can find that, I believe I can find the answer. Here it is:

1. Homework Statement

Let A = x2ˆx + y2ˆy + z2ˆz

Consider the parabolic path y2 = x between the points (0, 0) and (2, √2).

By integrating over x, compute the line integral ∫(A ⋅ ds)

Homework Equations


  • ds = (dx/dt)
  • dy/dx = (1/2)x

The Attempt at a Solution


Ok, so we're given a function, y = √(x), and asked to compute a line-integral "over x" under this curve. My questions at this point are:
  • Does "over x" mean with respect to x? (This may just be a problem of semantics.)
  • But, what is ds? Above is my guess at what it should be. Is it just a unit of change along the function y?
I would very much appreciate any help you may be able to provide.
 
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  • #2
You are not integrating a function y(x). The thing to be integrated is the vector function A. This is a function of the position vector (x, y, z). When integrating such a beast, a path or area or volume has to be specified to integrate it over. In this case, you are given the path (x, √x) from (0,0) to (2, √2).
ds is a vector element along that line, (dx, dy). It is not dx/dt.
What does not make sense is that A is specified as a 3D vector, but everything else is as though we are working in 2D.
 
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  • #3
Thank you, haruspex.

I was able to find a solution to this problem.

Like you said, the goal is to integrate a vector function (A = x2ˆx + y2ˆy + z2ˆz) dotted with the derivative of the position vector (ds), over a specified path [i.e. from (0, 0) to (2, √2)]. Even though the vector function A is given in three dimensions, we can just ignore the z-dimension, because the function we're integrating over has only two dimensions.

But what is ds? It's just the derivative of s, which is a position vector, or a vector whose coordinates point to any position along our function (y2 = x...or...y = √x). Because it's a vector, it can be represented as s = (some explanation of the behavior of x)ˆx + (some explanation of the behavior of y)ˆy. Again, the z-dimension can be ignored. It is important to get both of those relationships in terms of x. The x-coordinate is obvious (why?) and the manner in which the y-coordinate behaves is easily explained by the given function. So, we have:

s = xˆx + √xˆy

To find ds, which is an infinitesimal change along the function's path, we just find the derivative of s in terms of x, as below. Please forgive the crude notation.

ds = dxˆx + (½x)ˆy → ds = dxˆx + (½x)ˆy * dx​

Once we know all that, we just "pull the lever" and do the calculations we're asked to. First we take the dot product of those two vectors (A and ds), then integrate the result, and evaluate it in terms of our given x- and y-coordinates.

(Important side-note: How does A relate to the given function? Of that, I'm not sure, and it's really necessary information to explain all of what's happening.)
 
  • #4
sgholami said:
How does A relate to the given function?
I do not understand the question. ##x^2\hat x+y^2\hat y+z^2\hat z ## is just some arbtrary vector function of position. A is the name the question setter has given it.
 
  • #5
haruspex said:
I do not understand the question. ##x^2\hat x+y^2\hat y+z^2\hat z ## is just some arbtrary vector function of position. A is the name the question setter has given it.

Right; but I'm not sure how A relates to the function, ##y = \sqrt x##. It seems that the function gives us our path, but then what is A used for?
 
  • #6
sgholami said:
Right; but I'm not sure how A relates to the function, ##y = \sqrt x##. It seems that the function gives us our path, but then what is A used for?
It is just the function to be integrated. It need not have any physical significance, and there need be no relationship between A and the path function.
Here's a scalar example which has physical meaning: suppose the coefficient of kinetic friction of a level floor varies from place to place and is described by f(x,y). The integral of Mgf along some path p(x,y)=0 gives the work done in dragging an object weight Mg along that path.
 
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  • #7
haruspex said:
It is just the function to be integrated. It need not have any physical significance, and there need be no relationship between A and the path function.
Here's a scalar example which has physical meaning: suppose the coefficient of kinetic friction of a level floor varies from place to place and is described by f(x,y). The integral of Mgf along some path p(x,y)=0 gives the work done in dragging an object weight Mg along that path.

Ah! I see! I just find it counter-intuitive that A need not have any relationship to ##y = \sqrt x##. I now understand that my original question should have been how the integral we're asked to evaluate, ##\int (A \cdot ds)##, is relevant to the function ##y##. They just seemed like separate ideas, but I now see that ##y## tells us both about the bounds of this integral and about ds.

Thank you very much!
 

What is a line integral?

A line integral is a type of integral in calculus that is used to calculate the total value of a function along a given curve or path. It takes into account both the length of the curve and the values of the function along that curve.

How is a line integral calculated?

A line integral is calculated by breaking down the curve into smaller segments and taking the sum of the function values at each point along the curve. This sum is then multiplied by the length of each segment and added together to get the total value of the line integral.

What is the purpose of a line integral?

Line integrals are used in various fields of science, including physics and engineering, to calculate quantities such as work, displacement, and electric potential. They are also used in vector calculus to calculate the flux of a vector field across a given curve.

What is the difference between a line integral and a regular integral?

The main difference between a line integral and a regular integral is that a line integral is calculated along a specific curve or path, while a regular integral is calculated over an interval of values. Line integrals also take into account the direction of the curve, while regular integrals do not.

What are some real-world applications of line integrals?

Line integrals have various real-world applications, such as calculating the amount of work done by a force along a specific path, finding the electric potential along a wire or circuit, and calculating the total fluid flow through a curved pipe. They are also used in computer graphics to create smooth, curved lines and shapes.

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