91Ó°ÊÓ

Parametric equations are a powerful tool in geometry and calculus. They allow us to express a set of points in space using parameters, rather than just Cartesian coordinates like \( x \), \( y \), and \( z \). In this problem, the surface \( S \) is a parallelogram given by the parametric equations:

• \( x = u + v \)
• \( y = u - v \)
• \( z = 1 + 2u + v \)

These equations use \( u \) and \( v \) as parameters, which can take values \( 0 \leq u \leq 2 \) and \( 0 \leq v \leq 1 \). This provides a flexible way to define the surface. By changing \( u \) and \( v \), we can traverse every point on this surface. The use of parametric equations is highly effective for solving surface integrals because it provides a clear and manageable framework for the computation.

The cross product is crucial when dealing with vector calculations, especially in finding normals to surfaces in 3D space. To compute the cross product, we first calculate the partial derivatives of the vector \( \mathbf{r}(u, v) = \langle u+v, u-v, 1+2u+v \rangle \) with respect to \( u \) and \( v \). These derivatives are:

• \( \frac{\partial \mathbf{r}}{\partial u} = \langle 1, 1, 2 \rangle \)
• \( \frac{\partial \mathbf{r}}{\partial v} = \langle 1, -1, 1 \rangle \)

Next, we use these to find the normal vector by taking their cross product:\[\frac{\partial \mathbf{r}}{\partial u} \times \frac{\partial \mathbf{r}}{\partial v} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ 1 & 1 & 2 \ 1 & -1 & 1 \end{vmatrix} = \langle 3, 1, -2 \rangle\]The result, \( \langle 3, 1, -2 \rangle \), gives us the normal vector to the surface. This vector isn't just a direction; it will also play a key role in later steps, particularly in scaling the vector for integration.

The magnitude of a vector is essentially its length, and it's essential for converting a vector to its unit form. Calculating the magnitude of the normal vector \( \langle 3, 1, -2 \rangle \) involves the formula:\[\left\| \langle 3, 1, -2 \rangle \right\| = \sqrt{3^2 + 1^2 + (-2)^2} = \sqrt{14}\] This magnitude, \( \sqrt{14} \), is used in the integration process to factor in the differential area of the surface. It helps us convert the integration from parameter space (\( u \) and \( v \)) to Cartesian coordinates. Thus, when we integrate over the surface, this magnitude scales the integrand appropriately.

Partial derivatives help us understand how a function changes as each of its input variables changes, keeping others constant. In this problem, the surface is parameterized by the position vector \( \mathbf{r}(u, v) \). The partial derivatives, \( \frac{\partial \mathbf{r}}{\partial u} \) and \( \frac{\partial \mathbf{r}}{\partial v} \), give us the tangent vectors along the \( u \) and \( v \) directions of the surface:- \( \frac{\partial \mathbf{r}}{\partial u} = \langle 1, 1, 2 \rangle \)- \( \frac{\partial \mathbf{r}}{\partial v} = \langle 1, -1, 1 \rangle \)These derivatives are vital for computing the cross product, which is needed to find the normal vector to the surface. By examining how \( \mathbf{r} \) changes as \( u \) and \( v \) vary, we gain insights into the geometric properties of the surface. The use of partial derivatives in this context ensures we can effectively work with and integrate over complex surfaces in 3D space.