Problem 12 - Calculus Early Transcendentals

Question

Use the Divergence Theorem to find the flux of F across the surface ? with outward orientation. $$ \begin{array}{l}{\mathbf{F}(x, y, z)=x \mathbf{i}+y \mathbf{j}+z \mathbf{k} ; \sigma \text { is the surface of the solid }} \\ {\text { bounded by the paraboloid } z=1-x^{2}-y^{2} \text { and the } x y-} \\ {\text { plane. }}\end{array} $$

Step-by-Step Solution

Answer

The flux is \( \frac{3\pi}{2} \).

Step 1: Understand the Divergence Theorem

The Divergence Theorem states that the flux of a vector field \( \mathbf{F} \) across a closed surface \( \sigma \) is equal to the triple integral of the divergence of \( \mathbf{F} \) over the volume \( V \) enclosed by \( \sigma \). Mathematically, it is expressed as \( \iint_{\sigma} \mathbf{F} \cdot d\mathbf{S} = \iiint_{V} (abla \cdot \mathbf{F}) \, dV \).

Step 2: Calculate the Divergence of \( \mathbf{F} \)

Compute the divergence \( abla \cdot \mathbf{F} \) for \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \). The divergence is given by \( \frac{\partial x}{\partial x} + \frac{\partial y}{\partial y} + \frac{\partial z}{\partial z} = 1 + 1 + 1 = 3 \).

Step 3: Define the Volume of Integration

The volume \( V \) is bounded by the paraboloid \( z = 1 - x^2 - y^2 \) and the xy-plane. The upper surface is the paraboloid and the lower surface is the disk \( x^2 + y^2 \leq 1 \) in the xy-plane.

Step 4: Set Up the Integral in Cylindrical Coordinates

Convert to cylindrical coordinates for simplicity. The limits for \( r \) are from 0 to 1, for \( \theta \) are from 0 to \( 2\pi \), and for \( z \) from 0 to \( 1 - r^2 \). The element \( dV \) becomes \( r \, dr \, d\theta \, dz \).

Step 5: Evaluate the Triple Integral

Substitute the divergence into the triple integral: \( \iiint_{V} 3 \, dV = \int_{0}^{2\pi} \int_{0}^{1} \int_{0}^{1-r^2} 3r \, dz \, dr \, d\theta \). First, integrate with respect to \( z \), yielding \( 3r(1-r^2) \).

Step 6: Simplify and Integrate with Respect to \( r \)

Now integrate \( 3r - 3r^3 \) from 0 to 1: \( \int_{0}^{1} (3r - 3r^3) \, dr = 3\left[ \frac{r^2}{2} - \frac{r^4}{4} \right]_{0}^{1} = 3\left( \frac{1}{2} - \frac{1}{4} \right) = \frac{3}{4} \).

Step 7: Integrate with Respect to \( \theta \)

Finally, integrate \( \frac{3}{4} \) with respect to \( \theta \) from 0 to \( 2\pi \): \( \int_{0}^{2\pi} \frac{3}{4} \, d\theta = \frac{3\theta}{4} \Big|_{0}^{2\pi} = \frac{3}{4} \times 2\pi = \frac{3\pi}{2} \).

Step 8: Conclusion: Determine the Flux

The flux of \( \mathbf{F} \) across the surface \( \sigma \) is \( \frac{3\pi}{2} \).

Key Concepts

Flux Calculation

Flux calculation involves finding the flow of a vector field across a surface. When dealing with a vector field like \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \), we're interested in how much the vector field flows through a given surface, especially if this surface is closed, like a sphere or a cylinder. In this exercise, using the Divergence Theorem, we connect this idea by relating surface integrals to triple integrals over a volume.

This means calculating the total flow through an entire volume rather than focusing on just the surface through which the vectors pass. The net flux calculation across a closed surface uses the theorem to imply that it's equal to the divergence of the field throughout the volume bounded by the surface.

• Key idea: Converts a surface-integral problem to a volume-integral problem.
• Saves effort: Easier integration over volume rather than complex surface computation.

Cylindrical Coordinates

Cylindrical coordinates simplify calculations of certain types of volumes and surfaces. They extend polar coordinates into three dimensions and are particularly useful when dealing with bodies like paraboloids or cylinders, found in the task at hand.

These coordinates are specified as \( (r, \theta, z) \), where \( r \) is the radial distance from the z-axis, \( \theta \) is the angular coordinate, and \( z \) is the height from the xy-plane.

This system is especially helpful when evaluating integrals, as it allows transformation of complex Cartesian integrals into simpler ones. For the volume between the paraboloid and xy-plane:

• \( r \) varies from 0 to 1
• \( \theta \) varies from 0 to \( 2\pi \)
• \( z \) varies from 0 to \( 1-r^2 \)

The relationship between these variables simplifies the setting up and solving of the triple integral involved in finding flux.

Vector Field Divergence

Vector field divergence measures how much a vector field spreads out or converges at a certain point. Specifically, it quantifies the magnitude of a vector field's source or sink.

For the vector field \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \), the divergence is calculated using:

• Partial derivative of \( x \) with respect to \( x \)
• Partial derivative of \( y \) with respect to \( y \)
• Partial derivative of \( z \) with respect to \( z \)

Adding them gives the divergence: \( 1 + 1 + 1 = 3 \).

This indicates a steady spreading out (outward flow) of the field at all points, supporting the calculation of flux with the divergence theorem, where this constant divergence integrates over the volume.

Triple Integral

Triple integrals are used to compute the cumulative effect (like volume or mass) across three-dimensional regions. When applying the Divergence Theorem, the triple integral replaces the need to calculate the surface integral of flux directly.

During flux calculation for the solid bounded by the paraboloid and the xy-plane, the triple integral transformation becomes key:

• Volume element: \( dV = r \, dr \, d\theta \, dz \)
• Provides cumulative measure across whole volume, rather than just its surface
• The integration proceeds in steps: first \( z \), then \( r \), and lastly \( \theta \).

These integrations simplify evaluating the complete flux for a volume, demonstrating how such integrals measure how much vector fields stretch (diverge) or compress (converge) in space.