- #1
mr_coffee
- 1,629
- 1
Hello everyone~
I have the following problem, its done in the book but I'm lost on how they came to the final answer.
Evaluate the line integral:
Integral over C xy dx + (x-y) dy, C is the line segment from (0,0) to (2,0) and (2,0) to (3,2).
C = C1 + C2;
On C1: x = x, y = 0;
dy = 0 dx,
0 <= x <= 2;
On C2: x = x; y = 2x-4;
dy/dx = 2; => dy = 2 dx;
2 <= x <= 3;
This all makes sense to me, but then they have:
Integral over C xy dx + (x-y) dy = integral over c1 xy dx + (x-y)dy + integral over c2 xy + dx + (x-y) dy;
= integral from 0 to 1 (0 + 0) dx + integral 2 to 3 [(2x^2 -4x) + (-x + 4)(2)] dx;
I see where the 2 came from in the 2nd part, because dy = 2 dx; but why is it (0 + 0) in the first part, and also where did they get 2x^2-4x? or the -x + 4?
Thanks!
I have the following problem, its done in the book but I'm lost on how they came to the final answer.
Evaluate the line integral:
Integral over C xy dx + (x-y) dy, C is the line segment from (0,0) to (2,0) and (2,0) to (3,2).
C = C1 + C2;
On C1: x = x, y = 0;
dy = 0 dx,
0 <= x <= 2;
On C2: x = x; y = 2x-4;
dy/dx = 2; => dy = 2 dx;
2 <= x <= 3;
This all makes sense to me, but then they have:
Integral over C xy dx + (x-y) dy = integral over c1 xy dx + (x-y)dy + integral over c2 xy + dx + (x-y) dy;
= integral from 0 to 1 (0 + 0) dx + integral 2 to 3 [(2x^2 -4x) + (-x + 4)(2)] dx;
I see where the 2 came from in the 2nd part, because dy = 2 dx; but why is it (0 + 0) in the first part, and also where did they get 2x^2-4x? or the -x + 4?
Thanks!