Nonstandard integral

  • Thread starter tronter
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  • #1
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1. Find the amount of work done by the constant force field [itex] dx + 3dy - dz [/itex] as it moves a particle along the intersection of the planes [itex] x+y+z = 1 [/itex] and [itex] x-2y = -2 [/itex] from where it intersects the [itex] y,z [/itex] plane ([itex] x = 0 [/itex]) to where it intersects the [itex] z,x [/itex] plane ([itex] y = 0 [/itex]).

So the intersection is [tex] z = 0 [/tex] and [tex] z = 3 [/tex]. Then what would I do?
 

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  • #2
HallsofIvy
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I have absolutely no idea what you mean by "the intersection is z= 0 and z= 3" or how you got that! The intersection of two planes is a line, not two points. You are to integrate along that line.

If x+ y+ z= 1, then x= 1- y- zo. If x- 2y= -2, then x= 2y- 2 so we have 2y- 2= 1- y- z.
z= 1- 3y. In other words, the intersection of x+ y+ z= 1 and x- 2y= -2 is given by x= 2t- 2, y= t, z= 1- 3t. That crosses the yz plane when x= 2t- 2= 0 or t= 1. It crosses the xz plane when y= t= 0.

Integrate dx+ 3dy- dz over that line from t= 1 to t= 0.

(I also have no idea why you consider that integral "nonstandard". It's about as "standard" as there is!)
 
  • #3
I sure don't know about this little beast like question my friend.

It sure is a cancerous question.
sorry for my bad english

C to the T to the remBath

much love
xxx
 

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