# Detailed solution to Problem Set 38 Exercise 3

Question:
Find the integral of dy ∧ dz on the portion of the unit sphere in the first octant, oriented clockwise when viewed from the origin (in a right-handed coordinate system).
Solution:
There are many ways to parametrize the unit sphere, leading to many ways to set up the integral; I'll use the parametrization from Reading Homework 37, based on spherical coordinates with ρ = 1: x = r cos θ = ρ sin φ cos θ = sin φ cos θ, y = r sin θ = ρ sin φ sin θ = sin φ sin θ, and z = ρ cos φ = cos φ. Although in general, we have 0 ≤ φ ≤ π and 0 ≤ θ ≤ 2π, the first octant is where x, y, and z are all positive, so 0 ≤ φ ≤ π/2 and 0 ≤ θ ≤ π/2.

Now differentiate this parametrization: dy = cos φ sin θ dφ + sin φ cos θ dθ, and dz = −sin φ dφ. Then dy ∧ dz = −sin φ cos φ sin θ dφ ∧ dφ − sin2φ cos θ dθ ∧ dφ = 0 − sin2φ cos θ (−dφ ∧ dθ) = sin2φ cos θ dφ ∧ dθ.

Or instead of using the differentials and the properties of the wedge product, we could use the partial derivatives and a formula from Reading Homework 38. So, ∂y/∂φ = cos φ sin θ, ∂y/∂θ = sin φ cos θ, ∂z/∂φ = −sin φ, and ∂z/∂θ = 0. Then dy ∧ dz = (∂y/∂φ ∂z/∂θ − ∂y/∂θ ∂z/∂φ) dφ ∧ dθ = ((cos φ sin θ)(0) − (sin φ cos θ)(−sin φ)) dφ ∧ dθ = sin2φ cos θ dφ ∧ dθ.

Either way, dy ∧ dz = sin2φ cos θ dφ ∧ dθ.

Now I need the order of the differentials to match the orientation. Viewed from the outside of the unit sphere as a graph is usually drawn, φ increases as we move down and θ increases we move to the right, so dφ ∧ dθ means moving down and then to the right, which is counterclockwise. But this is facing the origin; looking from the origin, it must be turning clockwise. This is the correct orientation, so we should stick with dφ ∧ dθ.

Now I can set up the iterated integral: ∫π/2θ=0 ∫π/2φ=0 sin2φ cos θ dφ dθ. The inner indefinite integral is ∫sin2φ cos θ dφ = ½ φ cos θ − ¼ sin(4φ) cos θ. Then the inner definite integral is (½ φ cos θ − ¼ sin(4φ) cos θ)|π/2φ=0 = (½ (π/2) cos θ − ¼ sin(4(π/2)) cos θ) − (½ (0) cos θ − ¼ sin(4(0)) cos θ) = π/4 cos θ. Next, the outer indefinite integral is ∫π/4 cos θ dθ = π/4 sin θ. Finally, the outer definite integral is (π/4 sin θ)|π/2θ=0 = (π/4 sin(π/2)) − (π/4 sin(0)) = π/4.

Therfore, the value of the integral is π/4.

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