How Do You Properly Set Bounds for a Line Integral?

In this case, you can't use y or z because you don't have a function y(t) or z(t) to work with, but if you did, you could do exactly the same thing.
  • #1
xmflea
44
0
Let C be the line segment from (0,0,0) to the point (1,3,-2). evualuate the line integral of f along C if f(x,y) = (x + y^2 -2z)


so far i was able to write the parametric form x = t y=3t and z=-2t
the square root of all the derivatives is root(14).
so i get root(14) times integral of (t + t^2 -2t)dt.

my answer is (11/3) times root(14)
but the answer key says (11/2) times root(14)

my problem is that i don't know how to set the bounds of the integral.. i just integrate from 0 to 1, but really I am just guessing. how do i go about this?


The Attempt at a Solution

 
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  • #2
Hint: since you're integrating over t, what are the values of t at the starting and ending points of the line segment?
 
  • #3
starts at 0, ends at 3?
 
  • #4
How did you come up with that? (The "how" is always important!)
 
  • #5
sigh, to be honest.. somehow I've made it this far in calculus without ever knowing how to set up bounds for definate integrals whenever they are not given. i can solve them when the bounds are given, but when their not. I am screwed. anyways, i don't expect anyone to straight up tell me what the bounds are. just a little desparate because i really am that stupid when it comes to this.
 
  • #6
the reason why i thought it was from 0 to 1 to begin with is because i did this... -t+t^2.
convert to t(t-1), t = 0, t=1
 
  • #7
xmflea said:
sigh, to be honest.. somehow I've made it this far in calculus without ever knowing how to set up bounds for definate integrals whenever they are not given. i can solve them when the bounds are given, but when their not. I am screwed. anyways, i don't expect anyone to straight up tell me what the bounds are. just a little desparate because i really am that stupid when it comes to this.
Sure, that's what everyone says, until they figure it out...

Generally, the procedure is that you first identify the starting and ending points of the line integral (which could be points in space or points in time or whatever), then use whatever equations you have to figure out the values of your integration variable at those two points. It's not terribly complicated. In your case, you can definitely identify the starting and ending points of the integral (because they're given to you), and I know you have the equations to figure out what the value of t is at those points. They're in your first post.

xmflea said:
the reason why i thought it was from 0 to 1 to begin with is because i did this... -t+t^2.
convert to t(t-1), t = 0, t=1
Well, that's not actually correct. You're trying to get the integration limits of t from the function f, but the integration limits have nothing to do with the function f. They only have to do with the starting and ending points, (0, 0, 0) and (1, 3, -2), and with the conversion between (x, y, z) and t.
 
  • #8
so do i set x=y=z to get t=3t=-2t?
 
  • #9
Not really... there's no reason to do that.

How about starting with just one of the three coordinates. You know that at the start of the integral, x = 0. Does that tell you anything about t at the start of the integral?

And you know that at the end of the integral, x = 1. Does that tell you something about t at the end of the integral?
 
  • #10
well, it just tells me that at the start... t=0 since x=t.. and at the end. if x=1, then t=1 as well. which means the bounds are 0 to 1, which is how i got my answer (11/3) times root(14), but the key says (11/2) times root(14) instead..maybe the answer key is wrong?
 
  • #11
Go back and check your math. You just explained the right way to figure out the bounds of the integral (yay!), they are indeed 0 to 1... now, I just did the problem using those bounds, and I get the same answer as the answer key.

Perhaps you can describe in more detail exactly what steps you took to get to the answer 11/3*sqrt(14)?
 
  • #12
ok. i re did my math and now I am getting an even worse answer.
so integrating the function (t + t^2 -2t) i get (t^2/2) + (t^3/3) - (t^2). and plugging in from 0 to 1 i get...1/2 + 1/3 -1. common denominator is 6, so 3/6 + 2/6 - 6/6 and i get -1/6, so now my answer is root(14) times -1/6
 
  • #13
OK, well, where exactly did you get (t + t^2 - 2t) from? (hint hint ;-)
 
  • #14
i got it from the function x + y^2 -2z. just substituted the variables with t like my book says to do.
 
  • #15
OH! ok i get it! ok by the way, so everytime i look for the bounds, do i do the same thing as i did with this problem? just find t where x begins, and where x ends? can i do it with y or z?
 
  • #16
Yep - you could have done the same thing with y or z instead of x. In fact, you might want to try them all and make sure you get the same value of t no matter which one you pick - it's a good way to check that you figured out the right parametric form of the curve.
 

FAQ: How Do You Properly Set Bounds for a Line Integral?

What is a line integral?

A line integral is a type of integral calculus that is used to calculate the total value of a function along a specific path or curve. It takes into account both the magnitude and direction of the function along the path.

How is a line integral evaluated?

A line integral is evaluated by breaking down the curve into smaller segments and approximating the function at each point. These approximations are then summed up to get the total value of the line integral.

What is the significance of evaluating line integrals?

Evaluating line integrals is important because it allows us to calculate useful quantities such as work, displacement, and circulation in physics and engineering. It also has applications in areas such as fluid mechanics and electromagnetism.

What are the two types of line integrals?

The two types of line integrals are path integrals and contour integrals. Path integrals are evaluated along a specific path in a two-dimensional space, while contour integrals are evaluated along a closed curve in a two-dimensional or three-dimensional space.

What are the key properties of line integrals?

The key properties of line integrals include linearity, additivity, and independence of parameterization. Linearity means that the integral of a sum of functions is equal to the sum of their individual integrals. Additivity means that the integral of a function along a curve can be split into smaller sections and then summed up. Independence of parameterization means that the value of a line integral is independent of the specific parameterization chosen for the curve.

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