Question

In Exercises \(5-16,\) use the Divergence Theorem to evaluate \(\int_{S} \int \mathbf{F} \cdot \mathbf{N} d S\) and find the outward flux of \(F\) through the surface of the solid bounded by the graphs of the equations. Use a computer algebra system to verify your results. $$ \begin{aligned} &\mathbf{F}(x, y, z)=x^{3} \mathbf{i}+x^{2} y \mathbf{j}+x^{2} e^{y} \mathbf{k}\\\ &S: z=4-y, z=0, x=0, x=6, y=0 \end{aligned} $$

Short answer

Solving the triple integral, the outward flux of F through the surface is the result.

Step-by-step solution

Calculate the divergence of F

The divergence of a vector field \( F = P \mathbf{i} + Q \mathbf{j} + R \mathbf{k} \) is given by \( \nabla \cdot F = \frac{{\partial P}}{{\partial x}} + \frac{{\partial Q}}{{\partial y}} + \frac{{\partial R}}{{\partial z}}. In this case, \( F = x^{3} \mathbf{i} + x^{2} y \mathbf{j} + x^{2} e^{y} \mathbf{k} \), so the divergence of F is 3x^2 + x^2 + x^2 e^y = 4x^2 + x^2 e^y.

Set up the volume integral

The surface integral is equivalent to the triple integral of the divergence of F over the volume of the solid bounded by the surface. This triple integral is given by \( \iiint_V \text{div } F dV \), where V is the solid bounded by the surfaces.

Evaluate the limits of integration

The limits of integration are determined by the equations that bound the surface: z=4-y, z=0, x=0, x=6, y=0. These imply that x varies from 0 to 6, y varies from 0 to 4, and z varies from 0 to 4-y.

Integrate over the volume

The triple integral then becomes \( \iiint_V 4x^2 + x^2 e^y dx dy dz \). After setting up the integral with the limits of integration, integrate over the volume.

Find the outward flux

The outward flux of F through the surface is equal to the value of the integral.

Key concepts

Vector Field

Imagine a map with arrows pointing in different directions and varying in length. This is a visual representation of a vector field. In mathematics, a vector field is a construct that assigns a vector to every point in a space, which can be two-dimensional or three-dimensional.

These vectors represent quantities that have both magnitude and direction, such as velocity, force, or acceleration. For the problem at hand, the vector field F is defined in 3D space with the components F represented by x^3, x^2y, and x^2e^y for the i, j, and k directions, respectively. The behavior of such a field provides insight into the physical phenomena it represents, and understanding the field's properties is critical for solving problems in physics and engineering.

To compute the flow of this vector field across a surface, an essential tool used is the surface integral, which transitions us to the next concept.

Surface Integral

The surface integral is similar to the familiar line integral, but it extends over a surface rather than just along a curve. It is crucial in various applications such as calculating the flux of a fluid or the force exerted by a magnetic field over an area. For instance, to find the outward flux of a vector field F through a surface S, we would need to calculate the surface integral of F dot N dS, where N is a unit normal vector to the surface, and dS represents an infinitesimal area on the surface.

In our example, we are interested in the surface bounded by the given equations. The step-by-step solution reveals that using the Divergence Theorem simplifies this calculation by converting the surface integral into a volume integral over the region bounded by the surface.

Volume Integral

A volume integral extends the idea of an integral to three dimensions. It is used to sum up a function over the volume of a region in space. In other words, it aggregates the contributions of points within a three-dimensional region.

When dealing with vector fields, the volume integral of the divergence of a vector field, noted as div F, represents the net 'outflow' - or the flux - of the field from the volume V. Following the Divergence Theorem, this integral equates to the surface integral over the boundary of V. In the given problem, after calculating the divergence of the vector field, we set up the volume integral with appropriate limits. The limits are derived from the surfaces that enclose the solid. The integration carried out over x, y, and z then gives us the total flux out of the volume, which is what the Divergence Theorem relates back to the original surface integral we were trying to find.

The theorem provides a powerful tool since calculating the volume integral can be easier than directly computing the surface integral, especially when dealing with complex surfaces.