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Parametrizing and Line integrals (of a line, parabola, curve.)

  • Thread starter Breedlove
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  • #1
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Homework Statement



In each part, evaluate the integral [tex]\int(3x+2y)dx+(2x-y)dy[/tex]
(A) The line segment from (0,0) to (1,1).
(b) The parabolic arc y=x^2
(c) The curve y=sin(pi(x)/2) from (0,0) to (1,1)
(D) The curve x=y^3 from (0,0) to (1,1).

Homework Equations



[tex]\int f(x,y,z)dz[/tex]=[tex]\int^{a}_{b}f(x(t),y(t),z(t))z'(t)dt[/tex]


The Attempt at a Solution


Okay, so I'm not very good with the latex thing, but basically, from my understanding, you just have to parametrize each equation and then integrate from 0 to 1 for all of them right? The book says I should be getting 3 for all parts, but I;m getting things like 4.5 for a and 17/6 for b.

For a, I said that x=t and y=t (I stink at parametrizing, is this right???) which gives an integral of 9t from 0 to 1 which integrates to 4.5.

For b, I said that x=t and y=t^2 (how does one parametrize any given equation? These basic steps have really been killing me, like, how does one figure out the parametric equations for a circle? Everywhere I looked it just looks like it's given. What if i wanted to parametrize an ellipse or something? I think I must have missed a big chunk of my calc sequence or something.) Anyway, i ended up with an integral of 5t+t^2 from 0 to 1, which gives 17/6 if i'm right.

Okay, secondly, I've tried the advanced searches and I swear I can never get the search thing to give me what I want, so I always end wasting 30 minutes writing up the whole question and stuff and then when i post it, it says something like "related posts" at the bottom which are pretty much exactly what would have helped me. I'm so exasperated!! Please help!! help of any kind!!!
 
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Answers and Replies

  • #2
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Okay, I know how to that my "basically, from my understanding bit" is way off, this question is more about parametrizing than line integrals, i know how to do the line integral, i just don't know how to get it, I know that in part b i forgot about the y'(t)dy
 
  • #3
Dick
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Just start with the first one. There is nothing wrong with your parametrization. But I get integral 6t*dt. How do you get 9t*dt? x=t, y=t, it couldn't get much simpler than that??
 
  • #4
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Bah! Yeah I misunderstood my handwriting. I thought my two was a five. Okay. But for the second one I still get an integrand of 5t+t^2
 
  • #5
Dick
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y=t^2. dy=2t*dt. (2x-y)=2t-t^2. (2x-y)*dy=(2t-t^2)*2*t*dt. You should be integrating a cubic function of t, not a quadratic. Again, nothing wrong with the parametrization, but what's with the integral?
 
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  • #6
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Finally! Excellent. So... how would I go about parametrizing the curve y=sin(pi(x)/2)? I'm thinking that if the parabola is just x=t and y=t^2, then maybe we can just say x=t in this case too?
 
  • #7
Dick
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Why not? That makes y(0)=0 and y(1)=1. Works for me. Right?
 

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