Use the Divergence Theorem to evaluate / F. ds where and S is the boundary of the sphere x² + y² + z² = 4 oriented by the outward normal. The surface integral equals F = (10x4, 3yz, -40x³z)

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### Application of the Divergence Theorem

In this example, we will use the Divergence Theorem to evaluate the surface integral of a vector field \(\mathbf{F}\) over a closed surface \(S\).

#### Problem Statement

Evaluate the surface integral:

\[
\iint_S \mathbf{F} \cdot d\mathbf{S}
\]

where:

\[
\mathbf{F} = \langle 10x^4, 3yz^6, -40x^3z \rangle
\]

and \(S\) is the boundary of the sphere defined by:

\[
x^2 + y^2 + z^2 = 4
\]

The surface \(S\) is oriented by the outward normal.

#### Using the Divergence Theorem

The Divergence Theorem states:

\[
\iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV
\]

where \(V\) is the volume enclosed by the surface \(S\).

1. **Compute the Divergence of \(\mathbf{F}\)**:

   The divergence of \(\mathbf{F}\), denoted as \(\nabla \cdot \mathbf{F}\), is:

   \[
   \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(10x^4) + \frac{\partial}{\partial y}(3yz^6) + \frac{\partial}{\partial z}(-40x^3z)
   \]

2. **Calculate each partial derivative**:

   \[
   \frac{\partial}{\partial x}(10x^4) = 40x^3
   \]

   \[
   \frac{\partial}{\partial y}(3yz^6) = 3z^6
   \]

   \[
   \frac{\partial}{\partial z}(-40x^3z) = -40x^3
   \]

3. **Sum the partial derivatives**:

   \[
   \nabla \cdot \mathbf{F} = 40x^3 + 3z^6 - 40x^3 = 3z^6
   \]

4. **Set up the volume integral**:
Transcribed Image Text:### Application of the Divergence Theorem In this example, we will use the Divergence Theorem to evaluate the surface integral of a vector field \(\mathbf{F}\) over a closed surface \(S\). #### Problem Statement Evaluate the surface integral: \[ \iint_S \mathbf{F} \cdot d\mathbf{S} \] where: \[ \mathbf{F} = \langle 10x^4, 3yz^6, -40x^3z \rangle \] and \(S\) is the boundary of the sphere defined by: \[ x^2 + y^2 + z^2 = 4 \] The surface \(S\) is oriented by the outward normal. #### Using the Divergence Theorem The Divergence Theorem states: \[ \iint_S \mathbf{F} \cdot d\mathbf{S} = \iiint_V (\nabla \cdot \mathbf{F}) \, dV \] where \(V\) is the volume enclosed by the surface \(S\). 1. **Compute the Divergence of \(\mathbf{F}\)**: The divergence of \(\mathbf{F}\), denoted as \(\nabla \cdot \mathbf{F}\), is: \[ \nabla \cdot \mathbf{F} = \frac{\partial}{\partial x}(10x^4) + \frac{\partial}{\partial y}(3yz^6) + \frac{\partial}{\partial z}(-40x^3z) \] 2. **Calculate each partial derivative**: \[ \frac{\partial}{\partial x}(10x^4) = 40x^3 \] \[ \frac{\partial}{\partial y}(3yz^6) = 3z^6 \] \[ \frac{\partial}{\partial z}(-40x^3z) = -40x^3 \] 3. **Sum the partial derivatives**: \[ \nabla \cdot \mathbf{F} = 40x^3 + 3z^6 - 40x^3 = 3z^6 \] 4. **Set up the volume integral**:
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