# Detailed solution to Problem Set 38 Exercise 1

Question:
Find the integral of x dx ∧ dy + y dy ∧ dz on the triangle in (x, y, z)-space with vertices (0, 0, 1), (0, 1, 0), and (1, 0, 0), oriented clockwise when viewed from the origin (in a right-handed coordinate system).
Solution:
There are many ways to parametrize this triangle, leading to many ways to set up the integral; here's one: Let P0 be the vertex (0, 0, 1), let P1 be the vertex (0, 1, 0), and let P2 be the vertex (1, 0, 0). Then one parametrization is P = P0 + u(P1 − P0) + v(P2 − P0) for 0 ≤ u ≤ 1 and 0 ≤ v ≤ 1 − u. That is, x = 0 + (0 − 0)u + (1 − 0)v = v, y = 0 + (1 − 0)u + (0 − 0)v = u, and z = 1 + (0 − 1)u + (0 − 1)v = 1 − u − v.

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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