Line Integral - Work Done

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Homework Statement



A curve is formed by the intersection of y^2/9 + z^2/4 = 1 and the plane x-2y-3z = 0. The particle moving along the curve goes from (6,0,2) to (-6,0,-2). Find the work done on it by the vector field F(x,y,z) = -yi + xj + yzk.

Homework Equations



I'm going to need to find the integral of F dr which is actually going to be the integral of Fr(t) dot with r'(t) dt.

The Attempt at a Solution



I just need to set the integral up correctly, because I'm going to be letting the computer do the integration. So...

My main stumping point is coming up with the correct parameterization for r(t). In order to find a curve formed by the intersection, I used 3costj + 2sintk and then, for my i component, I noted that the plane equation gave me x = 2y + 3z. I then stuck in the y and z components I had (3cost and 2 sint, respectively) and came up with this curve: (6cost+6sint)i + 3costj + 2sintk. That's where I come to an abrupt stop...because I need r(t) going from (6,0,2) to (-6,0,-2), and I'm not quite sure how to do this. More than likely simple, but I'm stumped.

I know that once I get an r(t), I need to stick it into my function F, and then dot itself with the derivative of r(t). I don't think that will be that difficult, especially if I can simplify some (for instance, I see I could simplify the i component of my hpothetical curve to 6(cost+sint)i.

Any assistance would be much appreciated!
 

Answers and Replies

  • #2
HallsofIvy
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Homework Statement



A curve is formed by the intersection of y^2/9 + z^2/4 = 1 and the plane x-2y-3z = 0. The particle moving along the curve goes from (6,0,2) to (-6,0,-2). Find the work done on it by the vector field F(x,y,z) = -yi + xj + yzk.

Homework Equations



I'm going to need to find the integral of F dr which is actually going to be the integral of Fr(t) dot with r'(t) dt.

The Attempt at a Solution



I just need to set the integral up correctly, because I'm going to be letting the computer do the integration. So...

My main stumping point is coming up with the correct parameterization for r(t). In order to find a curve formed by the intersection, I used 3costj + 2sintk and then, for my i component, I noted that the plane equation gave me x = 2y + 3z. I then stuck in the y and z components I had (3cost and 2 sint, respectively) and came up with this curve: (6cost+6sint)i + 3costj + 2sintk.
Good! That's exactly right.

That's where I come to an abrupt stop...because I need r(t) going from (6,0,2) to (-6,0,-2), and I'm not quite sure how to do this. More than likely simple, but I'm stumped.
So you need t so that x= 6 cos t+ 6 sin t= 6, y= 3 cos t= 0, and z= 2 sin t= 2. Hmm, that last equation is the same as sin t= 1 and certainly [itex]t= \pi/2[/itex] satisfies that. [itex]3 cos(\pi/2)= 3(0)= 0[/itex], and it follow from that that [itex]6 cos(\pi/2)+ 6 sin(\pi/2)= 6(0)+ 6(1)= 6[/itex]. Your integral will have lower limit [itex]\pi/2[/itex].
For the upper limit you want x= 6 cos t+ 6 sin t= -6, y= 3 cos t= 0, and z= 2 sin t= -2. I'll leave it to you to figure out what t is here. Be careful about the DIRECTION in which you go between the two points on this closed curve.

I know that once I get an r(t), I need to stick it into my function F, and then dot itself with the derivative of r(t). I don't think that will be that difficult, especially if I can simplify some (for instance, I see I could simplify the i component of my hpothetical curve to 6(cost+sint)i.

Any assistance would be much appreciated!
 
  • #3
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Thank you very much for the help! That really helps me out, and I appreciate it.

Here's what I've done....

For the lower limit, I noticed that z = sint = -1 is satisfied by 3 pie / 2. It follows then that this also satisfies y = 3cost = 0 and x = 6cost + 6sint = -6. Thus I believe my lower limit will be 3 pie / 2.

I proceeded to find the derivative of r(t) and found that r'(t) = (-6sint + 6cost)i -3costj + 2costk. I also found F(r(t)) by sticking the components of r(t) into those of F. I found that F(r(t)) is 3costi + (-6sint+6cost)j - 6cos^2tk.

My integral will take the dot product of F(r(t)) and r'(t). Lots of stuff cancels! I wound up with -18costsint + 18cos^2t + 18costsint - 18cos^2t - 12cos^3t. I'm left with -12cos^3t if I did my math correctly.

My integral is going in the clockwise direction instead of the counterclockwise, so I will place a negative in front of it. I believe my answer is thus the negative integral from 3pie/2 to pie/2 of -12cos^3t dt.

Am I on the right track here?
 

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