Originally Posted by
topsquark From O to A we have that y does not change so dy = 0.
$\displaystyle \int_{O to A} x^3 dx - 2xy dy = \int_0^2 x^3 dx$
$\displaystyle = \frac{1}{4}x^4|_0^2 = \frac{1}{4}2^4 = 4$
From A to B the path is the line from (2, 0) to (0, 2). This is a segment of the line y = -x + 2 as x goes from 2 to 0, so
$\displaystyle dy = -dx$
Thus
$\displaystyle \int_{(2,0) to (0,2)} x^3 dx - 2xy dy $
$\displaystyle = \int_2^0 x^3 dx + 2x(-x + 2)(-dx)$
$\displaystyle = \int_2^0 (x^3 + 2x^2 - 4x)dx$
$\displaystyle = \left ( \frac{1}{4}x^4 + \frac{2}{3}x^3 - 2x^2 \right ) _2^0$
$\displaystyle = - \left ( \frac{1}{4}2^4 + \frac{2}{3}2^3 - 2 \cdot 2^2 \right ) = - \left ( 4 + \frac{16}{3} - 8 \right ) $
$\displaystyle = 4 - \frac{16}{3}$
So the overall integral will be:
$\displaystyle 4 + 4 - \frac{16}{3} = 8 - \frac{16}{3} = \frac{8}{3}$
-Dan