- #1
Puchinita5
- 178
- 0
Evaluate the line integral by two methods: A) directly and B) using Green's Theorem.
\oint xydx +x^2y^3dy
where C is the triangle with vertices (0,0) , (1,0), and (1,2).
I don't need the whole problem done, but I need someone to show me the work for finding the parametric equations for part A because I am not getting the same answer as in the book.
Basically, the part I'm getting wrong is the parametric equations for C2, or the vertical line on the right side of the triangle.
I put that r=(1-t)<1,0>+(t)<1,2> = <1-t, 0> + <t, 2t> = <1, 2t>
so x=1, y=2t...
but my solutions manual says y=t. And I looked this problem up on Cramster and it said the same thing.
Why does y=t and not 2t? Where am I messing up?
\oint xydx +x^2y^3dy
where C is the triangle with vertices (0,0) , (1,0), and (1,2).
I don't need the whole problem done, but I need someone to show me the work for finding the parametric equations for part A because I am not getting the same answer as in the book.
Basically, the part I'm getting wrong is the parametric equations for C2, or the vertical line on the right side of the triangle.
I put that r=(1-t)<1,0>+(t)<1,2> = <1-t, 0> + <t, 2t> = <1, 2t>
so x=1, y=2t...
but my solutions manual says y=t. And I looked this problem up on Cramster and it said the same thing.
Why does y=t and not 2t? Where am I messing up?