Problem 15 - Calculus: Early Transcendental Functions

Question

Use the Divergence Theorem to compute \(\iint_{\partial O} \mathbf{F} \cdot \mathbf{n} d S\). $$\begin{aligned} &Q \quad \text { is bounded by } \quad z=\sqrt{1-x^{2}-y^{2}} \quad \text { and } \quad z=0\\\ &\mathbf{F}=\left\langle x^{3}, y^{3}, z^{3}\right\rangle \end{aligned}$$

Step-by-Step Solution

Answer

The flux of the vector field F through the boundary of the solid Q using the Divergence Theorem is \(\frac{3\pi}{5}\).

Step 1: Compute the Divergence of F

Recall that the divergence of a vector field \(F = \) is given by \(div(F) = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z}\). Substituting \(P = x^3, Q = y^3, R = z^3\), we get \(div(F) = 3x^2 + 3y^2 + 3z^2\).

Step 2: Define the Solid Q and Set Up the Integral

Now the volume of Q should be calculated using the spherical coordinates. In this case, Q is a half sphere of radius 1, so it is described as \(0 <= r <= 1, 0 <= \theta <= 2\pi, 0 <= \phi <= \pi/2\). Express the divergence of F in terms of r, \(\phi\), \(\theta\) as \(div(F) = 3r^2\), which simplifies the integral. The infinitesimal volume element in spherical coordinates is \(dV = r^2 sin(\phi)dr d\phi d\theta\). So the volume integral becomes \(\int_{0}^{1}\int_{0}^{2\pi}\int_{0}^{\pi/2} 3r^4 sin(\phi)dr d\phi d\theta\).

Step 3: Compute the Integral

Compute the integral, splitting it into three factors each dependent on a different variable, r, \(\phi\), and \(\theta\). This results in \(\int_{0}^{1} 3r^4 dr \times \int_{0}^{2\pi} d\theta \times \int_{0}^{\pi/2} sin(\phi)d\phi\). Computing this gives the final result as \(\frac{3\pi}{5}\).

Key Concepts

Vector Field Divergence

The divergence of a vector field is an important concept in vector calculus, involving the measurement of a field's tendency to originate from or converge into a point. Divergence provides a scalar representation of a vector field's source or sink intensity at any given point. For a vector field \( \mathbf{F} = \langle P, Q, R \rangle \) with components that depend on the variables x, y, and z, the divergence is defined mathematically as \( abla \cdot \mathbf{F} = \frac{\partial P}{\partial x} + \frac{\partial Q}{\partial y} + \frac{\partial R}{\partial z} \).

In the context of the provided exercise, the vector field is \( \mathbf{F} = \langle x^3, y^3, z^3 \rangle \) and its divergence is \( \text{div}(\mathbf{F}) = 3x^2 + 3y^2 + 3z^2 \), which intuitively tells us how much the vector field is spreading out or converging in at any point within the volume Q.

Triple Integral in Spherical Coordinates

When integrating over a three-dimensional region that is spherical or near-spherical in shape, it's often convenient to use spherical coordinates. These coordinates convert a point's location from the standard Cartesian coordinates (x, y, z) to three new values: the radial distance r, the polar angle \( \phi \) (measured from the positive z-axis down), and the azimuthal angle \( \theta \) (measured from the positive x-axis within the x-y plane).

The infinitesimal volume element is represented as \( dV = r^2 \sin(\phi) dr d\phi d\theta \) in spherical coordinates, which significantly streamlines integration over spherical volumes. For our half-sphere with radius 1, the triple integral scans through all possible values of r, \( \phi \) and \( \theta \) within the specified limits: \( 0 \leq r \leq 1, 0 \leq \theta \leq 2\pi, 0 \leq \phi \leq \pi/2 \) respectively.

Surface Integral

A surface integral is akin to a two-dimensional counterpart to the line integral, allowing us to integrate scalar fields or vector fields over a surface rather than along a curve. If a scalar field is represented as \( f(x, y, z) \) and \( S \) is a surface in three-dimensional space, a surface integral would compute the cumulative value of \( f \) over the surface \( S \).

When the integral involves a vector field, as in \( \mathbf{F} \) in the exercise, and we want to calculate the flow across the surface S, the surface integral measures the total flux through the surface. In mathematical terms, the surface integral of a vector field \( \mathbf{F} \) across a surface \( S \) is noted as \( \iint_{S} \mathbf{F} \cdot \mathbf{n} dS \) where \( \mathbf{n} \) is the unit normal vector to the surface at each point, and \( dS \) is the infinitesimal area element on the surface.

Flux of a Vector Field

Flux is an essential concept when discussing fields such as electromagnetism, fluid dynamics, or any scenario where quantities are 'flowing' through a surface. In the language of vector calculus, calculating the flux of a vector field \( \mathbf{F} \) through a closed surface involves adding up the field's contribution through each infinitesimal area element of the surface. This gives us a total value which indicates how much of the field 'flows' outward (positive flux) or inward (negative flux) of that surface.

The flux is expressed as \( \Phi = \iint_{\partial O} \mathbf{F} \cdot \mathbf{n} dS \) where \( \partial O \) represents the boundary of the surface \( O \) which encloses a volume, \( \mathbf{F} \) is the vector field, \( \mathbf{n} \) is the outward-pointing unit normal vector, and \( dS \) is the elemental area. The Divergence Theorem elegantly links this concept to divergence: it states that the flux of \( \mathbf{F} \) through the closed surface \( \partial O \) is equal to the triple integral of the divergence of \( \mathbf{F} \) over the volume \( O \) that \( \partial O \) encloses.