Calculate line integral $\displaystyle \int_c xydx + x^2y^3dy$, where c is triangle with vertices at points (0,0), (1,0) and (1,2).

How can I find r(t) ?

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- Apr 25th 2009, 10:30 AMApprentice123Compute the line integral
Calculate line integral $\displaystyle \int_c xydx + x^2y^3dy$, where c is triangle with vertices at points (0,0), (1,0) and (1,2).

How can I find r(t) ? - Apr 25th 2009, 10:51 AMJhevon
you have to state what direction you are going in. is it from (0,0) to (1,0) to (1,2)??

you will have to do three separate integrals, one for each side of the triangle. for each, your r(t) will be the vector function for the line.

Here is how to proceed:

Assuming we are going from (0,0) to (1,0) to (1,2), call the line connecting (0,0) to (1,0) $\displaystyle C_1$, the line connecting (1,0) and (1,2) $\displaystyle C_2$ and the line connecting (1,2) back to (0,0) $\displaystyle C_3$.

Then we have $\displaystyle \int_C F = \int_{C_1}F + \int_{C_2}F + \int_{C_3}F$.

lets concentrate on $\displaystyle C_1$.

for $\displaystyle C_1$, we have the line $\displaystyle y = 0$, with x ranging from 0 to 1. thus, we can parametrize the line by:

$\displaystyle x = t$, $\displaystyle y = 0$, for $\displaystyle 0 \le t \le 1$

and so, $\displaystyle dx = 1$, $\displaystyle dy = 0$, and our integral for $\displaystyle C_1$ becomes:

$\displaystyle \int_0^1 t(0)(1) + t^2(0)^3(0)~dt = \int_0^1 0~dt = 0$

now do the same for $\displaystyle C_2$ and $\displaystyle C_3$.

(note, there is no need to introduce a $\displaystyle t$ here. we could have parametrized our line by $\displaystyle x = x$, $\displaystyle y = 0$ for $\displaystyle 0 \le x \le 1$) - Apr 25th 2009, 02:36 PMApprentice123Line integral
thank you