Problem 19 - Thomas Calculus

Question

In Exercises \(9-20\) , use the Divergence Theorem to find the outward flux of \(\mathbf{F}\) across the boundary of the region \(D .\) Thick sphere \(\quad \mathbf{F}=\left(5 x^{3}+12 x y^{2}\right) \mathbf{i}+\left(y^{3}+e^{y} \sin z\right) \mathbf{j}+\) \(\left(5 z^{3}+e^{y} \cos z\right) \mathbf{k}\) \(D :\) The solid region between the spheres \(x^{2}+y^{2}+z^{2}=1\) and \(x^{2}+y^{2}+z^{2}=2\)

Step-by-Step Solution

Answer

The outward flux is \( 48\pi\sqrt{2} - 12\pi \).

Step 1: Understanding the Problem Geometry

We are asked to find the flux of a vector field \( \mathbf{F} \) across the boundary of a thick spherical shell \( D \) bounded between two spheres: the inner sphere given by \( x^2 + y^2 + z^2 = 1 \) and the outer sphere by \( x^2 + y^2 + z^2 = 2 \). The Divergence Theorem can relate the flux through the boundary to a volume integral over \( D \).

Step 2: State the Divergence Theorem

The Divergence Theorem states that the outward flux of a vector field \( \mathbf{F} \) through the surface of a region \( D \) is equal to the triple integral over \( D \) of the divergence of \( \mathbf{F} \): \[ \iint_{\partial D} \mathbf{F} \cdot \mathbf{n} \, dS = \iiint_D (abla \cdot \mathbf{F}) \, dV \] where \( \mathbf{n} \) is the outward unit normal to the surface \( \partial D \).

Step 3: Calculate the Divergence of \( \mathbf{F} \)

The vector field \( \mathbf{F} \) is given by \( (5x^3 + 12xy^2) \mathbf{i} + (y^3 + e^y \sin z) \mathbf{j} + (5z^3 + e^y \cos z) \mathbf{k} \). The divergence \( abla \cdot \mathbf{F} \) is calculated as: \[ abla \cdot \mathbf{F} = \frac{\partial}{\partial x}(5x^3 + 12xy^2) + \frac{\partial}{\partial y}(y^3 + e^y \sin z) + \frac{\partial}{\partial z}(5z^3 + e^y \cos z) \].

Step 4: Evaluate the Partial Derivatives

Compute each partial derivative: - \( \frac{\partial}{\partial x}(5x^3 + 12xy^2) = 15x^2 + 12y^2 \)- \( \frac{\partial}{\partial y}(y^3 + e^y \sin z) = 3y^2 + e^y \sin z \)- \( \frac{\partial}{\partial z}(5z^3 + e^y \cos z) = 15z^2 - e^y \sin z \).Adding these gives the divergence of \( \mathbf{F} \): \[ abla \cdot \mathbf{F} = 15x^2 + 12y^2 + 3y^2 + 15z^2 \].

Step 5: Simplify the Divergence

Combine terms to simplify: \[ abla \cdot \mathbf{F} = 15x^2 + 15y^2 + 15z^2 = 15(x^2 + y^2 + z^2) \].

Step 6: Set Up the Integral Over \( D \)

Since \( abla \cdot \mathbf{F} = 15(x^2 + y^2 + z^2) \), the integral becomes: \[ \iiint_D 15r^2 \, dV \] where \( r^2 = x^2 + y^2 + z^2 \). Since \( D \) is a spherical shell from \( r = 1 \) to \( r = \sqrt{2} \), we convert to spherical coordinates \( (r, \theta, \phi) \).

Step 7: Convert to Spherical Coordinates and Integrate

The volume element in spherical coordinates is \( dV = r^2 \sin\phi \, dr \, d\theta \, d\phi \). The integral becomes: \[ \int_0^{2\pi} \int_0^{\pi} \int_1^{\sqrt{2}} 15r^4 \sin\phi \, dr \, d\phi \, d\theta \].

Step 8: Evaluate the Integral

First integrate with respect to \( r \): \[ \int_1^{\sqrt{2}} 15r^4 \, dr = \left[ \frac{15}{5}r^5 \right]_1^{\sqrt{2}} = [3\sqrt{2}^5 - 3] = 3(4\sqrt{2} - 1) \]. Next, integrate with respect to angles: \[ \int_0^{2\pi} \int_0^{\pi} \sin\phi \, d\phi \, d\theta = 2\pi \times 2 = 4\pi \]. Thus, the entire integral evaluates to: \[ 4\pi \times 3(4\sqrt{2} - 1) \].

Step 9: Compute the Final Result

Finish the calculation: \[ 12\pi(4\sqrt{2} - 1) = 48\pi\sqrt{2} - 12\pi \]. This is the outward flux of \( \mathbf{F} \) across the boundary of \( D \).

Key Concepts

flux calculation

Flux calculation is a method used to measure how much of a field passes through a given surface. In essence, when you're calculating the flux of a vector field across a surface, you're determining how much of the field is flowing out of or into the surface.
In our specific problem, we're using the Divergence Theorem to calculate the outward flux of \( \mathbf{F} \) across the boundary of the region \( D \). This region \( D \) is a thick spherical layer defined by two boundaries: the inner sphere \( x^2 + y^2 + z^2 = 1 \) and the outer sphere \( x^2 + y^2 + z^2 = 2 \).
By leveraging the Divergence Theorem, as we'll explore further, we replace the complex task of surface integration directly over \( D \) with a somewhat simpler volume integral over the same region.

spherical coordinates

Spherical coordinates are another way to represent points in space using three numbers: radius \( r \), polar angle \( \theta \), and azimuthal angle \( \phi \). These coordinates are especially useful when dealing with problems involving spheres or spherical regions, as they simplify calculations and integrals.

• Radius \( r \): Represents the distance from the origin to the point in space.
• Polar Angle \( \theta \): The angle in the \( xy \)-plane from the positive \( x \)-axis.
• Azimuthal Angle \( \phi \): The angle from the positive \( z \)-axis down to the point.

When converting to spherical coordinates, the volume element \( dV \) transforms to \( r^2 \sin\phi \, dr \, d\theta \, d\phi \). This is what we used in setting up our volume integral, incorporating the region from \( r = 1 \) to \( r = \sqrt{2} \), and evaluating it within the bounds for \( \theta \) and \( \phi \) to compute the flux integration over the spherical shell.

vector field divergence

The divergence of a vector field is a scalar field that represents the rate of change of the vector field's density. It's essentially telling us how much the vector field is "spreading out" at any given point.
For a given vector field \( \mathbf{F} = (P, Q, R) \), the divergence \( abla \cdot \mathbf{F} \) is calculated by computing the partial derivatives as follows:

• \( \frac{\partial P}{\partial x} \)
• \( \frac{\partial Q}{\partial y} \)
• \( \frac{\partial R}{\partial z} \)

Adding these derivatives gives the divergence. In our solution:
\[ abla \cdot \mathbf{F} = 15x^2 + 15y^2 + 15z^2 = 15(x^2 + y^2 + z^2) \].
This step is crucial because it simplifies the integral and hence the overall calculation.

solid region integration

Solid region integration involves calculating an integral over a three-dimensional region. In this context, we're evaluating a triple integral over the solid region \( D \), which is a spherical shell between two spherical boundaries.
Specifically, using spherical coordinates simplifies the calculation considerably due to the symmetry of the spherical region. Our integral takes the form:
\[ \int_0^{2\pi} \int_0^{\pi} \int_1^{\sqrt{2}} 15r^4 \sin\phi \, dr \, d\phi \, d\theta \]
Integrating over \( r \), \( \phi \), and \( \theta \) gives you the total flux through the region.
Each layer of the integral contributes a part to the overall solution. Breaking down the integration in this way simplifies the computation of what might otherwise be a complex task.