Problem 15 - Calculus Early Transcendentals

Question

Use the Divergence Theorem to find the flux of F across the surface ? with outward orientation. $$ \begin{array}{l}{\mathbf{F}(x, y, z)=\left(x^{3}-e^{y}\right) \mathbf{i}+\left(y^{3}+\sin z\right) \mathbf{j}+\left(z^{3}-x y\right) \mathbf{k}} \\ {\text { where } \sigma \text { is the surface of the solid bounded above by }} \\ {z=\sqrt{4-x^{2}-y^{2}} \text { and below by the } x y \text { -plane. [Hint: Use }} \\ { \text { spherical coordinates. } ]}\end{array} $$

Step-by-Step Solution

Answer

Flux is \(\frac{192\pi}{5}\).

Step 1: Understand the Problem

The task is to use the Divergence Theorem to calculate the flux of the vector field \( \mathbf{F}(x, y, z) = \left(x^3 - e^y\right)\mathbf{i} + \left(y^3 + \sin z\right) \mathbf{j} + \left(z^3 - x y\right) \mathbf{k} \) across the surface \(\sigma\), which is defined by the hemisphere \(z = \sqrt{4-x^2-y^2}\) above the \(xy\)-plane.

Step 2: Recall the Divergence Theorem

The Divergence Theorem states that the flux of \( \mathbf{F} \) across a closed surface \( \sigma\) is equal to the triple integral of the divergence of \( \mathbf{F} \) over the volume \( V \) enclosed by \( \sigma \):\[ \iint_{\sigma} \mathbf{F} \cdot d\mathbf{S} = \iiint_V abla \cdot \mathbf{F} \, dV \]

Step 3: Calculate the Divergence of the Vector Field

Compute \( abla \cdot \mathbf{F} \) by taking the partial derivative of each component of \( \mathbf{F} \):- \( \frac{\partial}{\partial x}(x^3 - e^y) = 3x^2 \)- \( \frac{\partial}{\partial y}(y^3 + \sin z) = 3y^2 \)- \( \frac{\partial}{\partial z}(z^3 - xy) = 3z^2 \)Thus, \( abla \cdot \mathbf{F} = 3x^2 + 3y^2 + 3z^2 \).

Step 4: Express Volume in Spherical Coordinates

Convert the volume integral to spherical coordinates where \( x = \rho \sin \phi \cos \theta \), \( y = \rho \sin \phi \sin \theta \), and \( z = \rho \cos \phi \). The volume element \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \). The limits for \( \rho \) are from 0 to 2, \( \theta \) from 0 to \(2\pi \), and \( \phi \) from 0 to \(\frac{\pi}{2}\).

Step 5: Setup the Triple Integral

The triple integral of the divergence over the volume \( V \) is:\[ \iiint_V (3x^2 + 3y^2 + 3z^2) \, dV. \]Substituting the spherical coordinates and the limits gives:\[ 3 \int_0^{2\pi} \int_0^{\frac{\pi}{2}} \int_0^2 \rho^4 \sin \phi \, d\rho \, d\phi \, d\theta. \]

Step 6: Evaluate the Integral

Integrate with respect to \( \rho \) first:\[ \int_0^2 \rho^4 \, d\rho = \left[ \frac{\rho^5}{5} \right]_0^2 = \frac{32}{5}. \]Next, integrate with respect to \( \phi \):\[ \int_0^{\frac{\pi}{2}} \sin \phi \, d\phi = \left[ -\cos \phi \right]_0^{\frac{\pi}{2}} = 1. \]Finally, integrate with respect to \( \theta \):\[ 3 \int_0^{2\pi} 1 \, d\theta = 3[\theta]_0^{2\pi} = 6\pi. \]Combine these results to get the flux:\[ \frac{32}{5} \times 1 \times 6\pi = \frac{192\pi}{5}. \]

Step 7: Conclude the Solution

The flux of the vector field \( \mathbf{F} \) across the surface \( \sigma \) with outward orientation is \( \frac{192\pi}{5} \).

Key Concepts

Vector Field

A vector field is a function that assigns a vector to every point in space. In this exercise, the vector field is defined as \( \mathbf{F}(x, y, z) = \left(x^3 - e^y\right)\mathbf{i} + \left(y^3 + \sin z\right) \mathbf{j} + \left(z^3 - xy\right) \mathbf{k} \). This means that for any given point \((x, y, z)\) in 3-dimensional space, the field assigns a vector comprised of three components:

• \( x^3 - e^y \) along the x-axis
• \( y^3 + \sin z \) along the y-axis
• \( z^3 - xy \) along the z-axis

The vector components can be dependent on one or more variables. For example, the x-component depends on \( x \) and \( y \), while the z-component depends on all three variables \( x \), \( y \), and \( z \). Understanding how vector fields represent physical quantities, like velocity or force fields, can help comprehend their real-world applications.
Vector fields often require specific operations like divergence or curl to extract meaningful characteristics about the field behavior over a region in space.

Flux Calculation

Flux calculation is essential in vector field operations. It measures the amount of field passing through a surface. The Divergence Theorem provides a simplified method to calculate the flux over closed surfaces. According to the theorem:

• The flux of a vector field \( \mathbf{F} \) through a surface \( \sigma \) is equivalent to the volume integral of the divergence of \( \mathbf{F} \) over the volume enclosed by \( \sigma \).

To compute the flux, follow these steps:
1. **Find the Divergence**: It's the scalar dot product, calculated as \( abla \cdot \mathbf{F} = \frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z} \). In our exercise, it simplifies to \( 3x^2 + 3y^2 + 3z^2 \).
2. **Set Up the Integral**: Use spherical coordinates if the volume has spherical symmetry to streamline calculations.
3. **Evaluate the Integral**: Calculate each part iteratively, often beginning with radial components.
Efficient calculation of flux helps determine how vector fields behave around regions and if they exhibit characteristics like net outward flow, often critical in electromagnetism and fluid dynamics.

Spherical Coordinates

Spherical coordinates offer a convenient way to describe volumes with round, symmetric shapes. They transform Cartesian coordinates \((x, y, z)\) into sphere-related parameters:

• \( \rho \) (the radial distance from the origin),
• \( \phi \) (the polar angle from the positive z-axis),
• \( \theta \) (the azimuthal angle in the xy-plane from the x-axis).

Conversions are done using the following formulas:

• \( x = \rho \sin \phi \cos \theta \)
• \( y = \rho \sin \phi \sin \theta \)
• \( z = \rho \cos \phi \)

For volume integrals, the differential element becomes \( dV = \rho^2 \sin \phi \, d\rho \, d\phi \, d\theta \). This coordinate system is particularly advantageous for problems involving spheres, hemispheres, and radial symmetry.
In the given problem, employing spherical coordinates allows a neat set-up for the integral limits:

• \( \rho \) spanning from 0 to the sphere’s radius (here 2),
• \( \phi \) from 0 to \( \frac{\pi}{2} \) to cover from the north pole down to the equator,
• \( \theta \) from 0 to \( 2\pi \) for a full revolution around the z-axis.

Using spherical coordinates simplifies the integration process and is integral in solving problems involving spherical objects or phenomena.