Vector calculus problem

In summary, the particle travels from (0,0,0) to (2,1,3) along the segments (0,0,0) ->(0,1,0)->(2,1,0)->(2,1,3). The integral is F dot product with dl. I can't figure this out. I tried doing it the way the book says to but I get a different answer than what the book gives.f
  • #1
177
0
f = 2xy in the x dir, (x^2 - z^2) in the y dir, -3xz^2 in the z dir.
particle travles from (0,0,0) to (2,1,3) along the segments (0,0,0) ->(0,1,0)->(2,1,0)->(2,1,3)

the integral is F dot product with dl.
i can't figure this out.

do i dot product each part individually and evaluate? i mean dot the x component of F and dl and evaluate them on an integral?
 
  • #2
You have to cut your integral into 3 parts for the three line segments:

[tex]\int_{(0,0,0)}^{(0,1,0)}\vec F \cdot d\vec l+\int_{(0,1,0)}^{(2,1,0)}\vec F \cdot d\vec l+\int_{(2,1,0)}^{(2,1,3)}\vec F \cdot d\vec l[/tex]

Since each of the the line segments are parallel to either the x,y or x-axis, getting [itex]d\vec l[/itex] is very simple.
 
  • #3
Galileo said:
You have to cut your integral into 3 parts for the three line segments:

[tex]\int_{(0,0,0)}^{(0,1,0)}\vec F \cdot d\vec l+\int_{(0,1,0)}^{(2,1,0)}\vec F \cdot d\vec l+\int_{(2,1,0)}^{(2,1,3)}\vec F \cdot d\vec l[/tex]

Since each of the the line segments are parallel to either the x,y or x-axis, getting [itex]d\vec l[/itex] is very simple.


so the first integral would be from (0,0,0) to (0,1,0), and i would be evaluating 2xy*dx? that's the dot product of Fx and dl sub x. what do i do with the y?
i tried it this way earlier and i get a final answer after evaluating the three integrals of -22. the book says -50.
 
  • #4
Note the direction. If you go from (0,0,0) to (0,1,0) in a straight line, then you only have the y-component to worry about (not the x!)
So the first integral is:

[tex]\int_0^1 F_y(0,y,0)dy=0[/tex]
since the y component of [itex]\vec f[/itex] is zero along this line.
Do the same for the 2nd and 3rd integral. The second yields 4, the third is -54.
 
  • #5
ive got another problem, my electro-magnetics prof didnt get a chance to do any examples on this stuff. D = 2 Ro z^2 in the Ro dir, and Ro cos^2 theta in the z dir.
find closed surface integral of D dot Ds.
0<Ro<5
-1<z<1
0<theta<2pi

is this the same thing or is it different?
 
  • #6
this is the same problem but this time they want me to evaulate it on the straight line from (0,0,0) to (2,1,3).
everything i can find on the net wants me to parametrize it, and the book doesn't say anything about that. so I am sure there's an easier way to do it.

the book gives the answer of -39.5
 
  • #7
if i make z = 3/2 x and y = 1/2 x and substitute those into the integral of F dot dl i end up with an answer of -27.66667. the way I am doing it makes sense from the things I've seen on the net. but the book gives an answer of -39.5. ??
 

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