91Ó°ÊÓ

Understanding how to parametrize a surface is crucial for many calculations in vector calculus, including surface integrals and flux calculations. In this case, we start with the plane equation given as \( x + y + z = 2a \). We transform this equation into a parametrized form by expressing \( z \) as a function of \( x \) and \( y \), thus: \( z = 2a - x - y \). This transforms the plane into a form suitable for parameterization: \( \mathbf{r}(x, y) = x \mathbf{i} + y \mathbf{j} + (2a - x - y) \mathbf{k} \). By allowing \( x \) and \( y \) to vary over a specific domain, here \( 0 \leq x \leq a \) and \( 0 \leq y \leq a \), we effectively describe the surface of the plane segment we are interested in. This parameterization allows us to express any point on the plane in terms of these parameters, making it easier to perform further computations.

Flux through a surface is essentially the measure of the flow of a vector field \( \mathbf{F} \) through that specific surface. Here, the flux is calculated using the surface integral \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, d\sigma \), where \( \mathbf{F} \) is the vector field \( \mathbf{F}=2 x y \mathbf{i}+2 y z \mathbf{j}+2 x z \mathbf{k} \) given in the problem. Usually, the goal is to compute how much field passes through the surface, considering its normal direction. This requires completing the dot product \( \mathbf{F} \cdot \mathbf{n} \) and then evaluating the integral over the surface's domain. Think of calculating flux as counting how many field lines penetrate the surface, which includes the direction and magnitude given by the vector field.

The normal vector \( \mathbf{n} \) to a surface is key to understanding the surface's orientation, which affects the calculation of fields passing through it. In our example, we compute the normal vector via the cross product of the partial derivatives of the parametrization with respect to \( x \) and \( y \). These partial derivatives yield tangent vectors: \( \frac{\partial \mathbf{r}}{\partial x} = \mathbf{i} - \mathbf{k} \) and \( \frac{\partial \mathbf{r}}{\partial y} = \mathbf{j} - \mathbf{k} \). Their cross product gives the normal vector \( \mathbf{n} = -(\mathbf{i} + \mathbf{j}) + \mathbf{k} \). It's important the normal vector points in the correct direction; for upward orientation (as in our case), the \( \mathbf{k} \) component should be positive. Ensuring the correct orientation helps ensure accurate flux calculations.

Double integration involves evaluating an integral over a region in the plane, often necessary for calculating surface integrals like the flux. Here, once we have \( \mathbf{F} \cdot \mathbf{n} \), we must integrate this expression over the defined region, from \( 0 \) to \( a \) for both \( x \) and \( y \). This requires setting up and evaluating several integrals:

• \( \int_{0}^{a} \int_{0}^{a} (-4ay) \ dx \, dy \)
• \( \int_{0}^{a} \int_{0}^{a} 2y^2 \ dx \, dy \)
• \( \int_{0}^{a} \int_{0}^{a} 4ax \ dx \, dy \)
• \( \int_{0}^{a} \int_{0}^{a} -2x^2 \ dx \, dy \)

Each of these integrals corresponds to a part of \( \mathbf{F} \cdot \mathbf{n} \), and computing their values yields the total flux. Double integration in this context effectively summarizes the total effect (flux) of the vector field across the plane segment.