# Line integral question.

1. Apr 14, 2013

### yungman

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.

2. Apr 14, 2013

### yungman

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. Apr 14, 2013

### Dick

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. Apr 14, 2013

Thanks.