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

[Breedlove]

Homework Statement

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).

Homework Equations

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

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!

 

[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

 

[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??

 

[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

 

[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?

 

[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?

 

[Dick]

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