Question

find the line integral PLS

Answer 1

\[\int\limits_{\gamma} (x+y) ds , \gamma: \triangle formed by (0,0),(0,1) and (1,0)\]

Answer 2

I'm not sure about this one, but I can give you my input anyway! You're evaluating three lines - the sides of this triangle. So you'll have three line integrals.

You'll have to find the parametric equations of these lines.

Then note that each integrand will have to have x and y in terms of your parameter, whatever you have x(t) be, put it in the place of x. Do this for y as well.

Lastly, note that\[ds=\sqrt{\left(\frac{dx(t)}{dt} \right) ^2+\left( \frac{dy(t)}{dt}\right) ^2}dt\]

Answer 3

So, one line at a time looks like the best way to start!

Answer 4

Pick any line, and then you can start.

Answer 5

That's how I would go about this, anyway.

Answer 6

\[c2: x=t , y= -t+1// \int\limits_{0}^{1}\sqrt{2} dt = \sqrt{2} \]
,i did this. but c1 and c3 i dont know how to do them

Answer 7

\[\int_{t=0} ^{t=1} (x(t)+y(t))\sqrt{\left(\frac{dx(t)}{dt} \right) ^2+\left( \frac{dy(t)}{dt}\right) ^2}dt\]
Is that right?

Answer 8

yeah

Answer 9

Alright, thanks! So I agree with that one!!

c1, you said is\[y(t)=0\]What about the x? It has to be parameterized, so it can be limited to an interval by the parameter. We'll probably call the parameter t, again.

\[x\text{ can be }t\text{.}\]Then t, like x, can go from 0 to 1.

Answer 10

So, for c1, we can have\[y(t)=0\]\[x(t)=t\]\[t\text{ goes from }0 \text{ to }1\]

Answer 11

\[\text{Now we have our }x(t)\text{ and }y(t)\text{, and the values for }t\text{, so we can go to the formula!}\]\[\int_{\text{begin t}} ^{\text{end t}} (x(t)+y(t))\sqrt{\left(\frac{dx(t)}{dt} \right) ^2+\left( \frac{dy(t)}{dt}\right) ^2}dt\]

Answer 12

i had thought the same as you, but i wasn't sure, thanks.

ps: sorry for my english , it's not my native language.

Answer 13

Should there be an orientation for these lines? I'm not sure if it makes a difference.

Answer 14

yeah , there is, but it doesnt make a difference

Answer 15

You're English is fine! I hope you understand mine.

Answer 16

Okay!

Answer 17

\[c1: x=t, y=0 // \int\limits_{0}^{1} t dt = 1/2 \]
\[c3: x=0, y=t //\int\limits_{0}^{1} t dt = 1/2\]
so \[C= c1+c2+c3 = 2+\sqrt{2}\] ??

Answer 18

You're welcome! And then you just do that work, and go on to the next line. I'll see what I get.

In the end, we add them, because\[\int _{C=C_1 \cup C_2} fds = \int _{C_1}fds+\int _{C_2}fds\]

Answer 19

I guess you already had that!
I went through C1 and got 1/2 as well!

Answer 20

And since x and y are switched in the next one, but x and y are used the same, C3 must be 1/2 also! So yeah!
\[\frac{1}{2}+\sqrt{2}+\frac{1}{2}=1+\sqrt{2}\]

Answer 21

So we have the same thing. If more people have it, then I think we can be pretty sure that that is the answer.

Answer 22

\[\begin{matrix} C_1&+&C_2&+&C_3&=\\\frac{1}{2}&+&\sqrt{2}&+&\frac{1}{2}&=&1+\sqrt{2}\end{matrix}\]

Answer 23

I just typed it that way for practice and fun.. But our answered differed by 1. Maybe we should look into that...

Answer 24

nono. i add 1/2+1/2 = 2, thats the reason , but it's 1

Answer 25

Haha, I was hoping it was that simple! So, now we agree. Take care! And congrats! And thanks!

Answer 26

thank u

Answer 27

You're welcome! My pleasure!