Line integral question.

  • Thread starter yungman
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  • #1
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I want to verify I am doing this correctly first:

Evaluate##\int_c (x^2ydx+xdy)## where the line is from (1,2) to (0.0)

My method is different from the book, I am using vector value function method where ##<x(t),y(t)>-(x_0,y_0>=t\frac {d\vec r}{dt}## and ##\vec r=\hat x x(t)+\hat y y(t)## and ##\vec r'(t)= \hat x \frac{dx(t)}{dt}+\hat y \frac{dy(t)}{dt}##

We know ##\vec r_0=<1,2>\;\Rightarrow\; \vec r(t)=<1,2>+t(\hat x \frac{dx(t)}{dt}+\hat y \frac{dy(t)}{dt})##

Use 0≤t≤1 ##\Rightarrow x=1-t## and ##y=2-2t##. Therefore ##\vec r'(t)=\frac{d\vec r}{dt}=-\hat x -\hat 2y##. And ##\vec r(t)= <1,2>+t<-1,-2>=\hat x (1-t)-\hat y (2-2t)##. And ##\frac{dx(t)}{dt}=-1## and ##\frac {d y(t)}{dt}=-2##.

Therefore ##\int_c (x^2ydx+xdy)=\int_0^1 [(1-t)^2 (2-2t)(-dt)+(1-t)(-2dt)]=-\frac 3 2##

I know the final integral and the answer is correct according to the book already. That I don't need to verify. I just want to make sure the way I use vector value function approach is correct.
 

Answers and Replies

  • #2
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the more basic question is that from my experience in EM, the normal way of line integral is ##\int_c f(xyz) dl## where l is the path of the integration. But this problem is using dx and dy and the book substitute t into x and y instead of using vector value function like I do. So which way is correct?
 
  • #3
Dick
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the more basic question is that from my experience in EM, the normal way of line integral is ##\int_c f(xyz) dl## where l is the path of the integration. But this problem is using dx and dy and the book substitute t into x and y instead of using vector value function like I do. So which way is correct?

Whether you explicitly write things as vectors or just keep track of the parameter dependence without doing so doesn't make any difference. Both ways are correct.
 
  • #4
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Thanks.
 

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