Finding integral of dot product of F and dl

[r3dxp]

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 alot!

 

[Defennder]

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 F_x,F_y,F_z 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 alot!

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?

 

[piercas]

Your solutions for part a) and part b) are correct. For part c), you can use the fundamental theorem of calculus to evaluate the integral without choosing a particular path. The fundamental theorem of calculus states that if F is a continuous vector field and C is a smooth curve from point A to point B, then the line integral of F along C is equal to the difference of the scalar field F evaluated at points A and B:

∫F•dl = F(B) - F(A)

In this case, point A is (0,0,0) and point B is (1,2,3). So, we can simply evaluate the scalar field F at these two points and take the difference to get the line integral.

F(A) = 0yz + 0zx + 0xy = 0
F(B) = 2yz + 3zx + 6xy = 2yz + 3zx + 6xy

Therefore, the line integral of F along any path from (0,0,0) to (1,2,3) is equal to 2yz + 3zx + 6xy. This is the solution for part c).

I hope this helps clarify the solutions for all three parts. Let me know if you have any further questions.