- Question:
- Find the integral of d
*y*∧ d*z*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: d

*y*= cos*φ*sin*θ*d*φ*+ sin*φ*cos*θ*d*θ*, and d*z*= −sin*φ*d*φ*. Then d*y*∧ d*z*= −sin*φ*cos*φ*sin*θ*d*φ*∧ d*φ*− sin^{2}*φ*cos*θ*d*θ*∧ d*φ*= 0 − sin^{2}*φ*cos*θ*(−d*φ*∧ d*θ*) = sin^{2}*φ*cos*θ*d*φ*∧ d*θ*.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 d*y*∧ d*z*= (∂*y*/∂*φ*∂*z*/∂*θ*− ∂*y*/∂*θ*∂*z*/∂*φ*) d*φ*∧ d*θ*= ((cos*φ*sin*θ*)(0) − (sin*φ*cos*θ*)(−sin*φ*)) d*φ*∧ d*θ*= sin^{2}*φ*cos*θ*d*φ*∧ d*θ*.Either way, d

*y*∧ d*z*= sin^{2}*φ*cos*θ*d*φ*∧ d*θ*.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 d*φ*∧ d*θ*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 d*φ*∧ d*θ*.Now I can set up the iterated integral: ∫

^{π/2}_{θ=0}∫^{π/2}_{φ=0}sin^{2}*φ*cos*θ*d*φ*d*θ*. The inner indefinite integral is ∫sin^{2}*φ*cos*θ*d*φ*= ½*φ*cos*θ*− ¼ sin(4*φ*) cos*θ*. Then the inner definite integral is (½*φ*cos*θ*− ¼ sin(4*φ*) cos*θ*)|^{π/2}_{φ=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*d*θ*= π/4 sin*θ*. Finally, the outer definite integral is (π/4 sin*θ*)|^{π/2}_{θ=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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