How to Evaluate Line Integrals in Cal 4 Without Attending Lectures?

[hottytoddy]

*Please don't just read the first post and leave. I'm adding to the thread with each attempt. I'm not asking someone to do it for me!*

I guess this is a lousy first post, and I apologize, but I am desperate and would appreciate all the help I can get more than anyone could imagine! This is the first problem of the section. I've been sick and haven't made it to any lectures covering this particular material and my final is TOMORROW! Please please help me if you can!

Homework Statement

Evaluate S_c (x + y) ds where C is the straight line segment x=t, y=(1-t), z=0, from (0,1,1) to (1,0,0).

Homework Equations

The Attempt at a Solution

Kind of both of those together. My text say to integrate a continuous function f(x,y,z) over a curve C:

1. Find a smooth parametrization of C
r(t) = g(t)i + h(t)j + k(t)k, t in the interval [a,b]

2. Evaluate the integral as
S_c f(x,y,z) ds = _bS^a f (g(t),h(t),k(t))|v(t)| dt

Can someone at least help me get started and try to figure it out myself?

The italicized S's are supposed to be integral signs!

 

[hottytoddy]

Okay, I know from the back of the book the answer is root2. This is what I did, and it got me the right answer, but if someone could check to make sure I did it right, that'd be awesome. Thank you!

I set the interval from [0,1] because the lowest x and lowest y are 0 and the highest of each is 1. So I have ₀S¹ (x + y) ds

C says x is equal to t and y = 1-t so I plugged those in and got t + (1 - t) which is just 1.

For r(t), I said it was ti + tj because there are no powers to the t's and that's as simple as it gets. Then, |v(t)| = |i + j| which is root2

I wound up with ₀S¹ f(t,t) root2 dt

₀S¹ (t + t) (root2) dt

Pull root2 out (constant) and get root2 ₀S¹ 2t dt

Then integrate and get root2 [ t² ] evaluated from 1 to 0

root2 [1² - 0²] is root2 * 1 = root2

and that's what I was supposed to get. Did I do it right though?

 

[hottytoddy]

The next problem is evaluate _CS (xy + y + z) ds along the curve r(t)=2ti+tj+(2-2t)k, t in [0,1].

I got |v(t)| = |2i + j - 2k| = 3

₀S¹ 2t + t + 2 - 2t 3 dt

₀S¹ t + 2 3 dt

3 ₀S¹ t + 2 dt

3[ (t2)/2 + 2t] eval. from 1 to 0

3[ 1/2 + 2] = 3[ 5/2] = 15/2

I did something wrong though, I think, because the book says the solution is 13/2

 

[hottytoddy]

I tried that last problem again and substituted x,y,z with 2t, t, and (2-2t), respectively. My integral was then

₀S¹ 2t² - t + 2 3 dt

3 ₀S¹ 2t² - t + 2 dt

3[ (2t³)/3 -t²/2 +2t ] from 0 to 1

3[ 2/3 - 1/2 + 2 -0] = 13/2

I got it right, but again, did I do it right?

 

[hottytoddy]

Next problem.

Find the line integral of f(x,y,z) = x + y + z over the straight line segment from (1,2,3) to (0,-1,1). Solution is 3*root14

I got a root14 by finding the distance between the two points and I set my t interval from 0 to root14, but I'm thinking that was wrong.

I had ₀S^root14 t + t + t root3 dt

I set r(t) = ti + tj+ tk and |v(t)| = root3

I worked it out and got (21 times root3)/2

... what gives?

 

[Defennder]

The next problem is evaluate _CS (xy + y + z) ds along the curve r(t)=2ti+tj+(2-2t)k, t in [0,1].

I got |v(t)| = |2i + j - 2k| = 3
₀S¹ 2t + t + 2 - 2t 3 dt

Your integrand expression is incorrect.

I tried that last problem again and substituted x,y,z with 2t, t, and (2-2t), respectively. My integral was then

₀S¹ 2t² - t + 2 3 dt

Now it is right.

 

[Defennder]

I got a root14 by finding the distance between the two points and I set my t interval from 0 to root14, but I'm thinking that was wrong.

That root14 corresponds to $$\left| \frac{d\textbf{r}}{dt} \right|$$ in the formula for a scalar line integral.

I had ₀S^root14 t + t + t root3 dt

That's not correct. What is the formula for evaluating a scalar line integral?

I set r(t) = ti + tj+ tk and |v(t)| = root3

The expression for r(t) as you have given would correspond to a point passing through the origin. Does the line through (1,2,3) to (0,-1,1) pass through the origin?

 

[hottytoddy]

That root14 corresponds to $$\left| \frac{d\textbf{r}}{dt} \right|$$ in the formula for a scalar line integral.

That's not correct. What is the formula for evaluating a scalar line integral?

The expression for r(t) as you have given would correspond to a point passing through the origin. Does the line through (1,2,3) to (0,-1,1) pass through the origin?

Honestly, I have no clue. I wish you were still online, though. I've pretty much given up and I was planning on not studying anymore for this exam since I need a 95 on it to get a C. Your reply has given me hope!

I can't see what you posted after "that root14 corresponds to"... I get a red x.

The line does not go through the origin. I still don't understand how to get the interval. I'm teaching myself over here, and quickly losing faith in my abilities.

The formula for evaluating a scalar line integral... again, no clue.

I know I need A LOT of help over here, but I'm willing to pull an all nighter if someone (or several someones) will help me. My exam is scheduled for 8.5 hours from now.

 

[hottytoddy]

Okay. I have a theory on that last one. At least, I get the right answer. Again, is my technique correct?

f(x,y,z)= x+y+z

r(t) = ti + 3tj + 2tk (the distance between corresponding coordinates in the points given.)

v(t) = i + 3j + 2k |v(t)| = root14 (YAY)

I'm assuming t between 0 and 1 because it works. Not sure why those numbers, but they work so I used them.

₀S¹ t + 3t + 2t root14 dt

root14 ₀S¹ 6t dt

root14 [3t²] from 0 to 1, which comes out to 3root14

What I need to know though, is why 0 to 1? The test problems are taking directly from the book, so I essentially just need to memorize the problems, but working through something is so much easier when you understand it! This is the third problem in the first section of new material. There are 28 more problems that I need to get through tonight. And they keep getting harder!

 

[Defennder]

What do your notes and textbook say about evaluating scalar line integrals? Keep in mind that you're looking for the part on line integrals of scalar functions. Look it up in the book index and read that relevant section if necessary. I've done last minute studying before and I would say it's always better to do last minute preparation than none at all, although it's undoubtedly better to be prepared beforehand.

 

[Defennder]

Okay. I have a theory on that last one. At least, I get the right answer. Again, is my technique correct?

f(x,y,z)= x+y+z

r(t) = ti + 3tj + 2tk (the distance between corresponding coordinates in the points given.)

v(t) = i + 3j + 2k |v(t)| = root14 (YAY)

I'm assuming t between 0 and 1 because it works. Not sure why those numbers, but they work so I used them.

₀S¹ t + 3t + 2t root14 dt

root14 ₀S¹ 6t dt

root14 [3t²] from 0 to 1, which comes out to 3root14

You did get the right answer, but I don't think your expression for r(t) is correct. Bear in mind that r(t) is the vector equation of the line through those 2 points specified by the question. What is the vector equation of the line through any 2 given points?

 

[hottytoddy]

All is says that makes any sense whatsoever is the snippet I put in the first post.

1. Find a smooth parametrization of C
r(t) = g(t)i + h(t)j + k(t)k, t in the interval [a,b]

2. Evaluate the integral as
Sc f(x,y,z) ds = bSa f (g(t),h(t),k(t))|v(t)| dt

 

[hottytoddy]

Is it as simple as (x1-x0)ti + (y1-y0)ti + (z1-z0)ti? So, (0-1) for x, (-1-2) for y, and (1-3) for z? Or am I backwards? I've been awake for nearly 24 hours already. My brain isn't functioning fully

 

[hottytoddy]

Logically, I want to say r(t) = ti + 2tj + 3tk because one of the points is (1,2,3), but there's no way it's that simple... It has to have 1,2, and 3 for coefficients, because that gives root14.

 

[hottytoddy]

The next one is even worse. I'm going to post it before I work on it to give you some time to look at it also. You are helping me so much! I wish there was some way I could repay you!

Integrate f(x,y,z) = (x + y + z)/(x² + y² + z²) over the path r(t) = ti + tj + tk, 0< a<= t <= b.

 

[hottytoddy]

Okay, that last one wasn't so bad once I got into it.

r(t) is given. From that |v(t)| = root3

Interval is given: [a,b]

Sub t into each x, y, and z. Reduce the equation and you get:

root3 _aS^b (t^-1) dt

root3 [ln b - ln a] = root3 ln(b/a) which is the answer.

Is my technique right?

 

[hottytoddy]

Next problem. Three more and then next section.

Integrate f over the given curve:

f (x,y) = x³/y
C: y = x²/2
0<= x <= 2

I don't even know where to start! There's no example like this in the text. I have nothing to follow.

 

[Defennder]

Is it as simple as (x1-x0)ti + (y1-y0)ti + (z1-z0)ti? So, (0-1) for x, (-1-2) for y, and (1-3) for z? Or am I backwards? I've been awake for nearly 24 hours already. My brain isn't functioning fully

Logically, I want to say r(t) = ti + 2tj + 3tk because one of the points is (1,2,3), but there's no way it's that simple... It has to have 1,2, and 3 for coefficients, because that gives root14.

No, as said earlier, the parametrisation you need is the vector equation of a line. What does a path paramatrisation represent? It maps values of a parameter, say t to a vector which extends from the origin to that point of the path it corresponds to. So far example, the line y=2x+c may be paramatrised as $$\textbf{r}(t) = t\textbf{i} + (2t+c) \textbf{j}$$. There are special parametric representations as well without using either x or y as the varying parameter. To take an example, the parametrisation of a circle x^2 + y^2 = r^2 would could be $$\textbf{r}(t) = r \cos t \textbf{i} + r \sin t \textbf{j} \ \mbox{where} \ t \in [0,2\pi]$$

The next one is even worse. I'm going to post it before I work on it to give you some time to look at it also. You are helping me so much! I wish there was some way I could repay you!

Integrate f(x,y,z) = (x + y + z)/(x2 + y2 + z2) over the path r(t) = ti + tj + tk, 0< a<= t <= b.
...
Okay, that last one wasn't so bad once I got into it.

r(t) is given. From that |v(t)| = root3

Interval is given: [a,b]

Sub t into each x, y, and z. Reduce the equation and you get:

root3 aSb (t-1) dt

root3 [ln b - ln a] = root3 ln(b/a) which is the answer.

Is my technique right?

Yes it looks correct. In this case it appears you know how to evaluate a line integral. Your problem it appears thus far is understanding how to do a parametrisation of a path. Just look that up in any textbook.

Next problem. Three more and then next section.

Integrate f over the given curve:

f (x,y) = x3/y
C: y = x2/2
0<= x <= 2

I don't even know where to start! There's no example like this in the text. I have nothing to follow.

Well, as before start by first coming up with a parametrisation of the curve y=x^2/2. It should be a vector function of the form r(t). The statement of the question also gives you a hint as to what you should use as a parameter.

 

[hottytoddy]

Okay. I think I have r(t), but it isn't pretty.

r(t)= (1-t)i + (2-3t)j + (3-2t)k
v(t)= -1i -3j -2k
|v| = root(1+9+4) = root14 (AHA!)

Is that right? And t goes from 0 to 1 because if you plug in 0 you get one endpoint of the segment and if you plug in 1 you get the other endpoint.

 

[Defennder]

Yes that's right.

 

[hottytoddy]

I just realized that completely changes my integral, but I still got the right answer!

INT 1-t+2-3t+3-2t root14 dt
root14 INT 6(1-t) dt
6root14 INT 1-t dt
6root14 [t-.5t²] from 1 to 0
6root14 * .5 = 3root14

 

[hottytoddy]

For the most recent problem,

Integrate f over the given curve:

f (x,y) = x3/y
C: y = x2/2
0<= x <= 2

is r(t) = 2ti + 2tj correct? Then |v| = 2root2. I believe t is in the interval [0,2]. My line segment goes from (0,0) to (2,2) if I plug x=0,2 into the y= (x^2)/2.

The answer should be (10root5 - 2)/3 , but that's not what ^these^ numbers give me!

 

[hottytoddy]

You can't sub in the equation for C into f(x,y), can you? That would make it so much easier.

 

[hottytoddy]

I am skipping that problem and one very similar to it. Hopefully he will not put them on the exam, but I cannot afford to look at this problem any longer. It's been over an hour, and I'm down to 5 hours before my exam.

Now I am working on:

Find the mass of a wire that lies along the curve r(t)= (t² - 1)j + 2tk, 0 <=t <= 1 if the density is d=(3/2)t. I haven't looked at it in the book yet.

 

[Defennder]

I just realized that completely changes my integral, but I still got the right answer!

INT 1-t+2-3t+3-2t root14 dt
root14 INT 6(1-t) dt
6root14 INT 1-t dt
6root14 [t-.5t²] from 1 to 0
6root14 * .5 = 3root14

Yes that is the correct way to do it.

Integrate f over the given curve:

f (x,y) = x3/y
C: y = x2/2
0<= x <= 2

is r(t) = 2ti + 2tj correct? Then |v| = 2root2. I believe t is in the interval [0,2]. My line segment goes from (0,0) to (2,2) if I plug x=0,2 into the y = (x^2)/2.

No r(t) is a parametric form of C: $$y = \frac{1}{2} x^2$$. What can you use as the parameter t here? t is in the interval you stated, so what does that make t in the parametrisation?

You can't sub in the equation for C into f(x,y), can you? That would make it so much easier.

You're supposed to actually. It's just that you have express f(x,y) in the parameters of the parametrisation of C.

 

[hottytoddy]

I have M= _CINT d(x,y,z) ds

v(t) = 2tj + 2k ; |v| = 2root(t² + 1)

M = ₀S¹ (3/2)t 2root(t² + 1) dt

Use u substitution- u=(t² + 1) du= 2t dt

(3/2) INT U^1/2 du

(3/2) (1/{2root(t² + 1)}) from 1 to 0
and i got (3root2 - 6)/8 but I should have gotten 2root2 - 1

 

[Defennder]

I have M= _CINT d(x,y,z) ds

v(t) = 2tj + 2k ; |v| = 2root(t² + 1)

How did you get this?
What is r(t)?