Math Problem Statement

Let ω be the upper unit hemisphere,* ω = {x ↓ R3 | ↘x↘ = 1, x3 ↔ 0} with the outward orientation (as in an example in the lectures). Let F : R3 → R3, F(x, y, z) = (0, x, 0) . (a) Compute I = ↓ ϖω F(x) · dx directly.

Solution

To compute the flux integral I=ωFdSI = \iint_{\omega} F \cdot d\mathbf{S}, where ω\omega is the upper hemisphere of the unit sphere in R3\mathbb{R}^3 with outward orientation and F(x,y,z)=(0,x,0)F(x, y, z) = (0, x, 0), we can proceed as follows:

  1. Understand the Geometry of ω\omega: The region ω\omega is defined as the upper hemisphere of the unit sphere, which is the set ω={(x,y,z)R3x2+y2+z2=1,z0}.\omega = \{ (x, y, z) \in \mathbb{R}^3 \mid x^2 + y^2 + z^2 = 1, \, z \geq 0 \}. Since the outward orientation is specified, the surface normal dSd\mathbf{S} will point away from the surface of the hemisphere.

  2. Express the Surface Integral: We want to compute the surface integral: I=ωFdS,I = \iint_{\omega} F \cdot d\mathbf{S}, where dS=ndSd\mathbf{S} = \mathbf{n} \, dS, with n\mathbf{n} being the outward-pointing unit normal vector on ω\omega, and dSdS representing the surface area element.

  3. Parameterize dSd\mathbf{S}: Since ω\omega is the upper unit hemisphere, we can use spherical coordinates: x=sinθcosϕ,y=sinθsinϕ,z=cosθ,x = \sin \theta \cos \phi, \quad y = \sin \theta \sin \phi, \quad z = \cos \theta, where θ[0,π/2]\theta \in [0, \pi/2] (as it is the upper hemisphere), and ϕ[0,2π]\phi \in [0, 2\pi].

    The surface area element dSdS on the unit sphere is dS=sinθdθdϕdS = \sin \theta \, d\theta \, d\phi, and the outward unit normal vector n\mathbf{n} is simply (x,y,z)=(sinθcosϕ,sinθsinϕ,cosθ)(x, y, z) = (\sin \theta \cos \phi, \sin \theta \sin \phi, \cos \theta) because ω\omega lies on the unit sphere.

  4. Compute FdSF \cdot d\mathbf{S}: Given F(x,y,z)=(0,x,0)F(x, y, z) = (0, x, 0), we have F(x,y,z)n=(0,x,0)(x,y,z)=0x+xy+0z=xy.F(x, y, z) \cdot \mathbf{n} = (0, x, 0) \cdot (x, y, z) = 0 \cdot x + x \cdot y + 0 \cdot z = xy. So, FdS=xydS=sin2θcosϕsinϕdθdϕ.F \cdot d\mathbf{S} = xy \, dS = \sin^2 \theta \cos \phi \sin \phi \, d\theta \, d\phi.

  5. Set Up the Integral: Substitute FdSF \cdot d\mathbf{S} into the integral II: I=ωFdS=02π0π/2sin2θcosϕsinϕdθdϕ.I = \iint_{\omega} F \cdot d\mathbf{S} = \int_0^{2\pi} \int_0^{\pi/2} \sin^2 \theta \cos \phi \sin \phi \, d\theta \, d\phi.

  6. Evaluate the Integral: Separate the integrals over θ\theta and ϕ\phi: I=(0π/2sin2θdθ)(02πcosϕsinϕdϕ).I = \left( \int_0^{\pi/2} \sin^2 \theta \, d\theta \right) \left( \int_0^{2\pi} \cos \phi \sin \phi \, d\phi \right). Notice that cosϕsinϕ=12sin2ϕ\cos \phi \sin \phi = \frac{1}{2} \sin 2\phi. Then, 02πcosϕsinϕdϕ=02π12sin2ϕdϕ=0,\int_0^{2\pi} \cos \phi \sin \phi \, d\phi = \int_0^{2\pi} \frac{1}{2} \sin 2\phi \, d\phi = 0, because the integral of sin2ϕ\sin 2\phi over a full period [0,2π][0, 2\pi] is zero.

  7. Conclusion: Since the ϕ\phi-integral evaluates to zero, the entire surface integral II is zero: I=0.I = 0.

Thus, the answer is: I=0.\boxed{I = 0}.

Would you like a breakdown of each step or have any questions?


Related Questions:

  1. How would the result change if F(x,y,z)=(0,y,0)F(x, y, z) = (0, y, 0) instead?
  2. What is the physical interpretation of the flux integral in this context?
  3. How does the divergence theorem apply to surface integrals like this?
  4. How would you set up this problem using a different coordinate system?
  5. What is the significance of choosing an outward versus inward orientation?

Tip: For integrals over symmetric regions, look for terms that may integrate to zero due to symmetry—it often simplifies the problem greatly.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Surface Integrals
Flux
Spherical Coordinates

Formulas

Surface Integral Formula: I = ∬ₓω F · dS
Spherical Coordinate Transformation: x = sin(θ)cos(φ), y = sin(θ)sin(φ), z = cos(θ)
Surface Area Element: dS = sin(θ) dθ dφ

Theorems

Divergence Theorem
Surface Integral Theorem

Suitable Grade Level

Undergraduate (Calculus III/Vector Calculus)