Finding integral of dot product of F and dl

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SUMMARY

The discussion focuses on calculating the line integral of the vector field F = yz ˆx + zx ˆy + xy ˆz from the origin (0, 0, 0) to the point (1, 2, 3) using three different methods. For part (a), the user attempted to evaluate the integral along a straight line path, but the final expression was incorrect as it did not yield a numerical value. In part (b), the user correctly computed the integral along three separate paths, yielding the expression yz + 2zx + 3xy. The discussion emphasizes the importance of parameterizing the path for accurate evaluation and hints at the nature of the vector field F, suggesting it may be conservative, which would allow for path independence in part (c).

PREREQUISITES
  • Understanding of vector fields and line integrals
  • Familiarity with parameterization of paths in calculus
  • Knowledge of dot products in vector calculus
  • Concept of conservative vector fields and path independence
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  • Learn how to parameterize paths for line integrals in vector calculus
  • Study the properties of conservative vector fields and how to identify them
  • Explore numerical evaluation techniques for line integrals
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Students studying vector calculus, particularly those working on line integrals and vector fields, as well as educators seeking to clarify concepts related to path independence and parameterization.

r3dxp
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Homework Statement


1. For the vector field
F = yz ˆx + zx ˆy + xy ˆz
(^x means the unit vector of x)
find the integral of F • dl from (0, 0, 0) to (1, 2, 3) in Cartesian coordinates
in each of the following ways:
(a) along a straight line path from (0, 0, 0,) to (1, 2, 3)
(b) along straight line paths from (0, 0, 0,) to (1, 0, 0), then from (1, 0, 0)
to (1, 2, 0), then from (1, 2, 0) to (1, 2, 3)
(c) without choosing a particular path


for part a), i used the dl as (^x + 2 ^y + 3 ^z)dx since vector r= ^x + 2 ^y + 3 ^z
thus, integral of F dot dl = yzx + z(x^2) + 3/2 (x^2)y evaluated from x=0 to x=1, and get yz+z+3/2y

for part b), i added the three different solutions from path 1, path 2, and path 3.
path1: using dl1=dx ^x, integral of path 1 of F*dl1 = yz
path2: using dl2=dy ^y, integral of path 2 of F*dl2 = 2zx
path3: using dl3=dz ^z, integral of path 3 of F*dl3 = 3xy
so my solution to part b) is yz+2zx+3xy

i was wondering, are my solutions to part a) and part b) correct? If not, please guide me through on how to get the solutions to part a,b, and c. thanks a lot!
 
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r3dxp said:
for part a), i used the dl as (^x + 2 ^y + 3 ^z)dx since vector r= ^x + 2 ^y + 3 ^z thus, integral of F dot dl = yzx + z(x^2) + 3/2 (x^2)y evaluated from x=0 to x=1, and get yz+z+3/2y
No, dl is always \left(\begin{array}{c}dx\\dy\\dz \end{array} \right ). Your final answer is supposed to be a numerical value anyway. A routine method would be to parametrise the path taken to evaluate the line integral along the straight line path. However, in this case since the path taken is relatively simple, we can perform the line integral without the parametric method. So firstly find a way to express x,y,z in terms of each other along that straight line path. The vector path representation of that straight line is \mathbf{r}(t) = t \left(\begin{array}{c}1\\2\\3 \end{array} \right ) as can be seen just by visualising the path.

So from the above you can deduce that along the straight line path, x=t, y=2t,z=3t and from this, x=1/2y=1/3z. Now that you've expressed x,y,z in terms of each other you can perform the line integral.

So we have \int_{(0,0,0)}^{(1,2,3)} \left(\begin{array}{c}yz\\zx\\xy \end{array} \right ) \cdot \left(\begin{array}{c}dx\\dy\\dz \end{array} \right ). So now, you just need to evaluate each of the dot products, perform the integral for each dx,dy,dz, taking care to express the Fx,Fy,Fz components in terms of x when integrating with respect to dx, in terms of y when integrating wrt to dy and so on.

for part b), i added the three different solutions from path 1, path 2, and path 3.
path1: using dl1=dx ^x, integral of path 1 of F*dl1 = yz
path2: using dl2=dy ^y, integral of path 2 of F*dl2 = 2zx
path3: using dl3=dz ^z, integral of path 3 of F*dl3 = 3xy
so my solution to part b) is yz+2zx+3xy

i was wondering, are my solutions to part a) and part b) correct? If not, please guide me through on how to get the solutions to part a,b, and c. thanks a lot!
As above, your answer here is incorrect because you're supposed to get a numerical answer. For b) it's just performing multiple line integrals by the method explained above. For c) the way the question is phrased serves as a strong hint as to how one can perform a line integral without specifying any path at all. What kind of vector field do you think F is such that the path taken for line integrals doesn't matter? How can you verify that?
 

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