Line integral over the perimiter of a hexagram

[jjheat]

Solved

Thanks

 

[Dick]

You just want to multiply 4 time the area of the figure, right? Sure, the area is 12 times the area of a single triangle. What's the area of an equilateral triangle with side 1? That's a trig problem, not a calc problem.

 

[jjheat]

Thanks

 

[Dick]

Just to clarify:

So you're saying that we don't even need the integral because you can pull the 4 out and if you just integrate (1), you get 1/2?

Then I multiply by 12 so the answer is 4*12*(1/2)=24?

That's ok up until you integrated 1 over the triangle and got 1/2. The area of a triangle with side length 1 isn't 1/2, is it?

 

[jjheat]

Thanks again!

 

[Vindetta]

I have the same problem but I cannot see the entire thread

 

[Dick]

I have the same problem but I cannot see the entire thread

jjheat deleted half the thread. That's not considered very polite or sporting. If everyone did it, it would make searching the old archives pretty useless. Feel free to start your own thread. The idea is to use Green's theorem to convert the hard contour integral into an easy surface integral. Try that.