Next, differentiate the parametrization: dx = dv, dy = du, and dz = −du − dv. Then dx ∧ dy = dv ∧ du = −du ∧ dv; while dy ∧ dz = −du ∧ du − du ∧ dv = 0 − du ∧ dv = −du ∧ dv.
Or instead of using the differentials and the properties of the wedge product, we could use the partial derivatives and the formulas from Reading Homework 38. So, ∂x/∂u = 0, ∂x/∂v = 1, ∂y/∂u = 1, ∂y/∂v = 0, ∂z/∂u = −1, and ∂z/∂v = −1. Then dx ∧ dy = (∂x/∂u ∂y/∂v − ∂x/∂v ∂y/∂u) du ∧ dv = ((0)(0) − (1)(1)) du ∧ dv = −du ∧ dv; while dy ∧ dz = (∂y/∂u ∂z/∂v − ∂y/∂v ∂z/∂u) du ∧ dv = ((1)(−1) − (0)(−1)) du ∧ dv = −du ∧ dv.
Either way, x dx ∧ dy + y dy ∧ dz = (v)(−du ∧ dv) + (u)(−du ∧ dv) = (−u − v) du ∧ dv.
Now I need the order of the differentials to match the orientation. With the parametrization I'm using, u increases as we move from (0, 0, 1) to (0, 1, 0), and then v increases as we move to (1, 0, 0). Facing the origin (as in a graph as usually drawn), this is a clockwise turn; so looking from the origin, it must be a counterclockwise turn. This is the wrong orientation, so we should use dv ∧ du (and not du ∧ dv). So to match this, write the differential form x dx ∧ dy + y dy ∧ dz as (−u − v)(−dv ∧ du) = (u + v) dv ∧ du.
Now I can set up the iterated integral: ∫1u=0 ∫1−uv=0 (u + v) dv du. The inner indefinite integral is ∫(u + v) dv = uv + ½ v2. Then the inner definite integral is (uv + ½ v2)|1−uv=0 = (u(1 − u) + ½ (1 − u)2) − (u(0) + ½ (0)2) = ½ − ½ u2. Next, the outer indefinite integral is ∫(½ −½ u2) du = ½ u − ⅙ u3. Finally, the outer definite integral is (½ u − ⅙ u3)|1u=0 = (½ (1) − ⅙ (1)3) − (½ (0) − ⅙ (0)3) = ⅓.
Therfore, the value of the integral is ⅓.
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