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

1. May 5, 2009

### Breedlove

1. The problem statement, all variables and given/known data

In each part, evaluate the integral $$\int(3x+2y)dx+(2x-y)dy$$
(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).

2. Relevant equations

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

3. 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!!!

Last edited: May 5, 2009
2. May 5, 2009

### Breedlove

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. May 5, 2009

### Dick

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. May 5, 2009

### Breedlove

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. May 5, 2009

### Dick

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?

Last edited: May 5, 2009
6. May 5, 2009

### Breedlove

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. May 5, 2009

### Dick

Why not? That makes y(0)=0 and y(1)=1. Works for me. Right?

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