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

enter your response here.

To compute the net outward flux of the vector field \(\mathbf{F} = \langle 6z - x, 5x - 3y, 3y - 2z \rangle\) across the boundary of the region \(D\), 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:
\[ \iint_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_D (\nabla \cdot \mathbf{F}) \, dV \]
where \(\nabla \cdot \mathbf{F}\) is the divergence of \(\mathbf{F}\).

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

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

So, the divergence \(\nabla \cdot \mathbf{F}\) is:
\[ \nabla \cdot \mathbf{F} = -1 - 3 - 2 = -6 \]

Next, we set up the integral over the region \(D\). Since \(D\) is the region between two spheres of radius 2 and 5, we use spherical coordinates for the volume integral:
\[ x = \rho \sin \phi \cos \theta \]
\[ y = \rho \sin \phi \sin \theta \]
\[ z = \rho \cos \phi \]
where \(0 \leq \rho \leq 5\), \(0 \leq \phi \leq \pi\), and \(0 \leq \theta \leq 2\pi\).

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

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

Evaluating the integral:
\[ \int_0^{2\pi} d\theta = 2\pi \]
\[ \int_0^\pi \sin \phi \, d\phi = 2 \]
\[ \int_2^5 \rho^2 \, d\rho = \left[ \frac{\rho^3}{3} \right]_2^5 = \frac{5^3}{3} - \frac{2^3}{3} = \frac{125}{3} - \frac{8}{3} = \frac{117}{3} = 39 \]

Multiplying these results together:
\[ -6 \cdot 2\pi \cdot 2 \cdot 39 = -936\pi \]

Therefore, the net outward flux of the vector field \(\mathbf{F}\) across the boundary of the region \(D\) is:
\[ \boxed{-936\pi} \]

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 \(\mathbf{F}\) 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 \(D\) is well-defined and closed to correctly apply the theorem.

Learn how to compute the net outward flux of the vector field F across the boundary of region D using the Divergence Theorem. Detailed solution using spherical coordinates and divergence calculations included.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation