Math Problem Statement

Use the Divergence Theorem to compute the net outward flux of the vector field F across the boundary of the region D. Fequalsleft angle 6 z minus x comma 5 x minus 3 y comma 3 y minus 2 z right angle D is the region between the spheres of radius 2 and 5 centered at the origin Question content area bottom Part 1 The net outward flux is

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Solution

To compute the net outward flux of the vector field F=6zx,5x3y,3y2z\mathbf{F} = \langle 6z - x, 5x - 3y, 3y - 2z \rangle across the boundary of the region DD, which is the region between the spheres of radius 2 and 5 centered at the origin, we use the Divergence Theorem.

The Divergence Theorem states: DFndS=D(F)dV\iint_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_D (\nabla \cdot \mathbf{F}) \, dV where F\nabla \cdot \mathbf{F} is the divergence of F\mathbf{F}.

First, we calculate the divergence of F\mathbf{F}: F=x(6zx)+y(5x3y)+z(3y2z)\nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(6z - x) + \frac{\partial}{\partial y}(5x - 3y) + \frac{\partial}{\partial z}(3y - 2z)

Calculating each partial derivative: x(6zx)=1\frac{\partial}{\partial x}(6z - x) = -1 y(5x3y)=3\frac{\partial}{\partial y}(5x - 3y) = -3 z(3y2z)=2\frac{\partial}{\partial z}(3y - 2z) = -2

So, the divergence F\nabla \cdot \mathbf{F} is: F=132=6\nabla \cdot \mathbf{F} = -1 - 3 - 2 = -6

Next, we set up the integral over the region DD. Since DD is the region between two spheres of radius 2 and 5, we use spherical coordinates for the volume integral: x=ρsinϕcosθx = \rho \sin \phi \cos \theta y=ρsinϕsinθy = \rho \sin \phi \sin \theta z=ρcosϕz = \rho \cos \phi where 0ρ50 \leq \rho \leq 5, 0ϕπ0 \leq \phi \leq \pi, and 0θ2π0 \leq \theta \leq 2\pi.

The volume element in spherical coordinates is: dV=ρ2sinϕdρdϕdθdV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta

Therefore, the volume integral becomes: D(F)dV=D6dV\iiint_D (\nabla \cdot \mathbf{F}) \, dV = \iiint_D -6 \, dV =6DdV= -6 \iiint_D \, dV =602π0π25ρ2sinϕdρdϕdθ= -6 \int_0^{2\pi} \int_0^\pi \int_2^5 \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta

Evaluating the integral: 02πdθ=2π\int_0^{2\pi} d\theta = 2\pi 0πsinϕdϕ=2\int_0^\pi \sin \phi \, d\phi = 2 25ρ2dρ=[ρ33]25=533233=125383=1173=39\int_2^5 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_2^5 = \frac{5^3}{3} - \frac{2^3}{3} = \frac{125}{3} - \frac{8}{3} = \frac{117}{3} = 39

Multiplying these results together: 62π239=936π-6 \cdot 2\pi \cdot 2 \cdot 39 = -936\pi

Therefore, the net outward flux of the vector field F\mathbf{F} across the boundary of the region DD is: 936π\boxed{-936\pi}

Would you like more details or have any questions?

Here are 8 related questions for further exploration:

  1. How is the Divergence Theorem applied to other vector fields?
  2. What is the physical significance of the net outward flux in this context?
  3. How do you compute the divergence of a given vector field in Cartesian coordinates?
  4. What are the bounds of integration when using spherical coordinates for a different region?
  5. How would the flux change if the vector field F\mathbf{F} was different?
  6. What are the steps to convert from Cartesian to spherical coordinates?
  7. How can the Divergence Theorem be extended to regions bounded by other shapes?
  8. What is the interpretation of a negative net outward flux?

Tip: When using the Divergence Theorem, ensure the region DD is well-defined and closed to correctly apply the theorem.

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Math Problem Analysis

Mathematical Concepts

Vector Calculus
Divergence Theorem
Spherical Coordinates

Formulas

Divergence Theorem

Theorems

Divergence Theorem

Suitable Grade Level

Advanced College