Can someone show me step by step how to work this problem out please.
Let C be the triangle with vertices (0,0), (1,0), and (1,2). Find the following line integral by two methods, directly and using greens theorem.
2xy^2dx + yx^2dy
Thanks!!!
directly
first from (0,0) to (1,0)
y = 0
x goes from 0 to 1
both terms are zero, so no contribution
second from (1,0) to (1,2)
Here dx = 0, x=1 and y goes from 0 to 2
so from 0 to 2 of y dy
= 2^2/2 = 2 total so far
now from (1,2) to(0,0)
dy = 2dx
y = 2x
so the integrand is
2 x (2x)^2 dx + 2x^3(2dx)
= 8x^3 dx + 4x^3 dx
= 12x^3 dx
from x = 1 to x = 0
3(0)^4 -3 (1^4) = -3
2 - 3 = -1 finally
NOW, using Green
closed path integral of Ldx+Mdy = integral over surface inside of
(dM/dx -dL/dy)
here L = 2xy^2
so dL/dy = 4xy
here M = y x^2
so dM/dx =2xy
so
dM/dx - dL/dy = integrand = -2xy
integral of
-2xy dy dx from y = 0 to y = 2x
then from x = 0 to 1
-2 *int x dx *int y dy
-2 * int x dx *[ y^2/2 from y = 0 to y = 2x]
-2 int x dx *2x^2
-4 int x^3 from 0 to 1
-4 (1^4/4) = -1