Question

Line Integral - Problem with a triangle side

Answer 1

I can't seem to find the function of x,y in order to use the integral f ds formula...

How do I proceed?

Answer 2

This is a scalar line integral, so the vector \(\vec r\) shouldn't be involved. That's the correct parameterization for the second segment. For the first segment, \(t\) should be the length along the "hypotenuse" of the triangle, and should run from \(0\) to \(1/\sqrt{2}\). The integrand becomes \(x(t) = t/\sqrt{2}\).

Answer 3

I'm sorry - it should run from \(0\) to \(\sqrt{2}\), not \(1/\sqrt{2}\).

Answer 4

How do you figure it is 0 to root(2)?

Answer 5

Nevermind - trigonometry using the coordinates to find the length

Answer 6

\(\sqrt2\) is the total path length of that particular segment.

Answer 7

We have been thought to change it to (from formula):\[\int\limits_{a}^{b} f[x(t), y(t), z(t)] \sqrt{(\frac{ dx }{ dt })^2 + (\frac{ dy }{ dt })^2 + (\frac{ dz }{ dt })^2} dt\]

Answer 8

Although, since this is a closed ... triangle; we can \[\int\limits_{c}^{}fds = \int\limits_{c_1}^{}f_1ds + \int\limits_{c_2}^{}f_2ds + \int\limits_{c_3}^{}f_3ds\], if I recall correctly

Answer 9

Yes, that's fine.

Answer 10

How would I proceed, using those formulas correctly?

Answer 11

For the first segment, you can let \(x(t) = t,y(t)=t\)
so you'd have
\[\int_0^1 x(t) \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2}dt = \int_0^1 t\cdot\sqrt{2}dt = \frac{1}{\sqrt{2}} \]

Using what I did, you'd get
\[\int_0^\sqrt{2} \frac{t}{\sqrt{2}}dt = \left.\frac{t^2}{2\sqrt{2}}\right|^\sqrt{2}_0 =\frac{1}{\sqrt{2}} \]

Answer 12

Whatever floats your boat.

Answer 13

The root(2) is just a constant you can take outside the integral though? This is what I did on the first part;

Answer 14

Looking at it that way, \(f[\vec r(t)] = x(t) = t\), not \(2t\).

Answer 15

Because we are ignoring y(t)*?

Answer 16

The integrand is x, not x+y.

Answer 17

So \[\int\limits_{C}^{}x ds\] just means the initial function is equal to x? That is what I struggeled with in the beginning

Answer 18

Yes. You can identify \[ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2} dt \]

Answer 19

and x = x(s) , or rather x(t)?

Answer 20

Sure, however you'd like to parameterize it.

Answer 21

So it could have been written \[\int\limits_{C}^{}x(t)*dt\] instead of x ds and it would be the same?

Yeah, this is just a very unusual problem for me - usually we have f(x,y) = something, or integral(f ds) with some other information. But since it was written integral(x ds) in this case, we only want to do integral(x(s) * d(s))

Answer 22

That would be fine as well. It shouldn't be that weird - \(\int f ds, f(x,y) = x\).

Answer 23

Thanks! Instead of writing my original f(x,y) = x + y, I just use the initial x and set f(x,y) = x - even though we do not got any y (which feels a bit weird)

Answer 24

That's right.

Answer 25

Redoing all of it with what I have learned - now I can see if I get the correct answers as well :)

Answer 26

@Jemurray3 I forgot! My original question ; C2 - when the Y is constant - do.. oh... I just ignore Y and do exactly the same as I did with C1, since f(x,y) = x and have 0y's - correct?

Answer 27

Yes, you should have
\[\int_0^1 (1-t) dt \]

Answer 28

Yup - thanks a lot!