91Ó°ÊÓ

Understanding a vector field is key when applying the Divergence Theorem. A vector field, denoted here as \( \mathbf{F} \), assigns a vector to each point in space. In simpler terms, imagine that at every point in space, there's an arrow pointing in a certain direction and having a certain magnitude. This arrow describes the vector at that point, indicating both direction and strength.
In the given exercise, the vector field \( \mathbf{F}(x, y, z) = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) assigns vectors based on the coordinates \((x, y, z)\). Here:

• \(x\) defines the vector component in the \(\mathbf{i}\) or x direction
• \(y\) defines the component in the \(\mathbf{j}\) or y direction
• \(z\) defines the component in the \(\mathbf{k}\) or z direction

This vector field is like a uniform arrow field, where each arrow's length and direction correlate directly to its x, y, and z position, providing a straightforward example of how vector fields are used to represent physical phenomena like fluid flow or electromagnetic fields.

Cylindrical coordinates provide a convenient way to describe points in three-dimensional space, especially when the region of interest has a round or cylindrical symmetry. They are particularly useful in this exercise, which involves a cylinder as a boundary. In cylindrical coordinates, we express points in terms of \(r\), \(\theta\), and \(z\).

• \(r\) is the radial distance from the z-axis, similar to the radius in polar coordinates.
• \(\theta\) is the angle around the z-axis, analogous to the angle in polar coordinates.
• \(z\) remains the same as in Cartesian coordinates, representing height.

The conversion from Cartesian to cylindrical coordinates for this exercise is given by \(x = r\cos(\theta)\) and \(y = r\sin(\theta)\).
Using cylindrical coordinates streamlined the computation of the triple integral, particularly because the boundaries of the region—defined by the cylinder \(x^2 + y^2 = 1\)—are easier to describe within this system. By using \(0 \leq r \leq 1\) and \(0 \leq \theta \leq \frac{\pi}{2}\), we represent the volume effectively in this context.

Triple integrals extend the concept of integration to functions of three variables. They are used to calculate volumes and integrals over three-dimensional regions. When dealing with vector fields and the Divergence Theorem, a triple integral allows us to compute the total flux through a closed surface by integrating the divergence over the volume enclosed by the surface.
In the exercise, after determining the divergence of the vector field \(\mathbf{F}\), which is a scalar \(3\), the problem asks us to evaluate:\[\iiint_{V} 3 \, dV = \int_0^1 \int_0^{\frac{\pi}{2}} \int_0^1 3r \, dz \, d\theta \, dr\]Here's how it's tackled:

• We first integrate in terms of \(z\), considering the height of the cylinder from 0 to 1.
• The second integration is in \(\theta\), over a quarter circle (from 0 to \(\frac{\pi}{2}\)).
• Finally, integrating with respect to \(r\), accounts for the radial distance from the origin out to the boundary of the cylinder.

The evaluated triple integral, yielding \(\frac{3\pi}{4}\), effectively measures how much the vector field "pushes" through the defined surface \(\Sigma\), demonstrating the power and utility of triple integrals in vector calculus.