91Ó°ÊÓ

In surface integrals, parameterization is a technique used to express surfaces in terms of simpler, typically planar coordinates. For the given problem, the surface is defined by the plane equation \( z = 1 + 2x + 3y \). Parameterization helps in rewriting this plane in a more manageable form using two variables, usually \( x \) and \( y \), which are easily visualized and manipulated.
To parameterize the given plane, we define a vector function \( \vec{r}(x, y) \) with components corresponding to the coordinates of the surface. Here, the plane is parameterized by:
\[ \vec{r}(x, y) = \langle x, y, 1 + 2x + 3y \rangle \]
This vector function maps the \((x, y)\) plane region directly onto the surface. It's a way to navigate through each point of the surface using \( x \) and \( y \) as parameters.

The normal vector to a surface is a crucial component in calculating surface integrals. It describes the orientation of the surface at a particular point, and it must be perpendicular to the surface at that point. In this problem, the normal vector \( \vec{N} \) is determined by taking the cross product of partial derivatives of the parameterization with respect to \( x \) and \( y \).
The partial derivatives are computed as:
- \( \frac{\partial \vec{r}}{\partial x} = \langle 1, 0, 2 \rangle \)
- \( \frac{\partial \vec{r}}{\partial y} = \langle 0, 1, 3 \rangle \)
The cross product of these vectors gives the normal vector:
\[ \vec{N} = \langle 1, 0, 2 \rangle \times \langle 0, 1, 3 \rangle = \langle -1, -3, 1 \rangle \]
This normal vector is essential in calculating \( dS \), the surface area element, as it helps incorporate the surface's orientation into the integral.

A plane surface is one of the simplest types of surfaces encountered in calculus, defined in three-dimensional space by an equation of the form \( z = ax + by + c \). In this exercise, the surface described by \( z = 1 + 2x + 3y \) is part of such a plane. The region above which the plane exists is crucial, as it determines the bounds of the integral. Here, it lies above the rectangle defined by the coordinates \[ [0, 3] \times [0, 2] \].
Understanding a plane surface involves:

• Identifying the plane equation in a three-dimensional space.
• Recognizing the surface's interaction with the coordinate plane, such as shifting or tilting as determined by coefficients.
• Visualizing the projection into simpler two-dimensional regions, like rectangles or triangles, to set the integration limits.

These aspects make evaluating areas or other properties of plane surfaces straightforward, often providing simpler calculations compared to more complex surfaces.

The evaluation of a surface integral involves transforming it into a definite integral over a two-dimensional region. In this problem, we turn the task of finding the surface integral into a problem of evaluating a double integral over a rectangular region.
The function to be integrated is adjusted according to the parameterization and the differential surface area \( dS \):
\[ \iint_{R} f(x, y) \, dS = \iint_{R} x^2 y (1 + 2x + 3y) \sqrt{11} \, dx \, dy \]
This expression represents the definite integral evaluated over the rectangle defined by \([0, 3] \times [0, 2]\). The specific steps include:

• Identifying the limits for integration in \(x\) and \(y\).
• Separating the integral into an inner integral and an outer integral.
• Solving each integral in succession, starting with the inner one, to simplify the expression to obtain a total result.

Overall, these integrations accumulate the values of the surface integral over the specified region, resulting in the final answer \( \frac{962}{5} \sqrt{11} \). Definite integrals are instrumental in quantifying values over continuous spaces, especially in physics and engineering applications.