91Ó°ÊÓ

Parameterization is like giving surfaces a map. You represent complex surfaces using simpler, more manageable equations. For our surface, the problem gives a plane with the equation \(z = 2 - x - y\). To parameterize it, we set variables like \(x\) as \(u\) and \(y\) as \(v\). This means:

• Let \(x = u\)
• Let \(y = v\)
• Thus, \(z = 2 - u - v\)

With this parameterization, the surface becomes a vector function \(\vec{r}(u, v) = \langle u, v, 2 - u - v \rangle\). This method transforms a 3D surface into something workable in 2D, which simplifies calculations. Parameterization also helps when setting up integrals, making limits and boundaries easier to handle.

Once you have the parameterization, partial derivatives play a crucial role. They're like small steps that tell you how a function changes as each variable changes. For \(\vec{r}(u, v) = \langle u, v, 2 - u - v \rangle\), we calculate:

• The derivative with respect to \(u\): \(\frac{\partial \vec{r}}{\partial u} = \langle 1, 0, -1 \rangle\)
• The derivative with respect to \(v\): \(\frac{\partial \vec{r}}{\partial v} = \langle 0, 1, -1 \rangle\)

These derivatives help us understand how each point on the surface behaves when we tweak \(u\) or \(v\). This knowledge is crucial as it sets the stage for finding a normal vector and determining the exact shape and orientation of the surface area.

The cross product is like finding the direction of a surface. It's a vector operation that shows us the perpendicular direction, also known as the normal vector, to a surface at any point. For the vectors \(\langle 1, 0, -1 \rangle\) and \(\langle 0, 1, -1 \rangle\), the cross product is:\[\vec{n} = \langle 1, 0, -1 \rangle \times \langle 0, 1, -1 \rangle = \langle 1, 1, 1 \rangle\]This normal vector is important because it represents the orientation of the surface in space. Additionally, it assists in calculating the surface area element \(dS\), which incorporates how a small piece of the surface is oriented relative to the axes. This step is fundamental in preparing for the surface integral.

Double integration is how we cover a surface area completely. It is an extension of single-variable integration into two dimensions. In our problem, we evaluate the surface integral \(\iint_{S} f(x, y, z) \, dS\), which becomes a double integral over the plane.

The function is transformed over the parameterized space \(\iint_{D} uv \cdot \sqrt{3} \, dudv\), where \(\sqrt{3}\) is the norm of the cross product vector. We've determined:

• \(0 \leq u \leq 2\)
• \(0 \leq v \leq 2 - u\)

These limits ensure we only integrate over the first octant where the surface lies. By evaluating the inner and outer integrals following these bounds, we systematically sum up all the small pieces of the surface \(S\), resulting in the total surface integral's value. This process illustrates how surfaces can be thoroughly examined using double integration.