What is #int_C \ xy \ dx - x \ dy # where on #C, y=1-x^2# from #(1,0)# to #(0,1)#?
1 Answer
Nov 16, 2017
# S = int_C \ xy \ dx - x \ dy = - 11/12#
Explanation:
We seek:
# S = int_C \ xy \ dx - x \ dy # where on#C, y=1-x^2# from#(1,0)# to#(0,1)#
On
# y=1-x^2 => dy/dx = -2x #
So we can express the line integral in terms of
# S = int_(x=1)^(x=0) \ x(1-x^2) \ dx - int_(x=1)^(x=0) \ x (-2x) \ dx #
# \ \ = int_1^0 \ x-x^3+2x^2 \ dx #
# \ \ = [ x^2/2-x^4/4+(2x^3)/3 ]_1^0 #
# \ \ = (0) - (1/2-1/4+2/3)#
# \ \ = 0 - 11/12#
# \ \ = - 11/12#
