Line Integral

[Melissa00]

with γ:[0,1] -> R³ and γ(t) = (t , t² , 1-t)<sup>T

This is probably extremely easy to solve, ... couldn't find the right answer though. It's supposed to be 1/2.

I would greatly appreciate your help!</sup>

 

[Deleted member 4993]

View attachment 3149
with γ:[0,1] -> R³ and γ(t) = (t , t² , 1-t)<sup>T

This is probably extremely easy to solve, ... couldn't find the right answer though. It's supposed to be 1/2.

I would greatly appreciate your help!</sup>

Since you did not show your work, we cannot tell where you went wrong!!

Please share your work with us .

If you are stuck at the beginning tell us and we'll start with the definitions.

You need to read the rules of this forum. Please read the post titled "Read before Posting" at the following URL:

http://www.freemathhelp.com/forum/th...217#post322217

 

[Melissa00]

Sorry for that! -.-

Okay, I know the definition of curve integrals,...


or |γ'(t)| depending on the dimension.
As we can see [a,b], f(x,y,z) and γ(t) are given. The only thing that irritates me is the form of f(x,y,z).
Since it says integral of (ydx, xdy, z[FONT=arial, sans-serif]dz) I didn' really know what I'm ought to use as the function for f [/FONT]
Maybe integrate[FONT=arial, sans-serif] the cruve γ(t) over 1 then since ydx, and xdy would cease to apply? Or do I have to integrate each component of the sum separately over the curve?


[/FONT]

 

[DrPhil]

View attachment 3149
with γ:[0,1] -> R³ and γ(t) = (t , t² , 1-t)<sup>T

This is probably extremely easy to solve, ... couldn't find the right answer though. It's supposed to be 1/2.

I would greatly appreciate your help!</sup>

Don't know what you have tried -
since \(\displaystyle \gamma\) is given parametrically as a function of \(\displaystyle t\), have you tried making the integral a function of \(\displaystyle t\)?

Can you clarify what \(\displaystyle \gamma :[0,1]\) means? What I suspect is it means \(\displaystyle 0 \le t \le 1\).

Show us your work . . .

 

[DrPhil]

Sorry for that! -.-

Okay, I know the definition of curve integrals,...

View attachment 3150
or |γ'(t)| depending on the dimension.
As we can see [a,b], f(x,y,z) and γ(t) are given. The only thing that irritates me is the form of f(x,y,z).
Since it says integral of (ydx, xdy, zdz) I didn' really know what I'm ought to use as the function for f
Maybe integrate the cruve γ(t) over 1 then since ydx, and xdy would cease to apply? Or do I have to integrate each component of the sum separately over the curve?

Maybe you are over-thinking the function \(\displaystyle \vec{f}\).
You are already given the dot-product form,
\(\displaystyle \vec{f} \cdot \vec{ds} = y\ dx + x\ dy + z\ dz \).

From the parametric form \(\displaystyle \gamma (t)\), and its derivative with respect to \(\displaystyle t\), you can convert the integral to the form

\(\displaystyle \displaystyle \int_0^1 F(t)\ dt \)

 

[Melissa00]

Found it! Thank you for your suggestions.
I think now I get my problem: After using several variations of the supposed function f, I realized that it has to written as a three dimensional vector...

 

[Melissa00]

Maybe you are over-thinking the function \(\displaystyle \vec{f}\).
You are already given the dot-product form,
\(\displaystyle \vec{f} \cdot \vec{ds} = y\ dx + x\ dy + z\ dz \).

From the parametric form \(\displaystyle \gamma (t)\), and its derivative with respect to \(\displaystyle t\), you can convert the integral to the form

\(\displaystyle \displaystyle \int_0^1 F(t)\ dt \)

Yes, I was definitely over-thinking it! Thank you for your help!