91Ó°ÊÓ

A parabolic cylinder is a three-dimensional surface that can be imagined as stretching a parabola infinitely along one direction. In our case, this surface is described by the equation \(z = 4 - y^2\). This equation tells us that for every fixed level of \(x\), we get a two-dimensional parabolic curve in the \(y-z\) plane.

A helpful way to visualize a parabolic cylinder is to imagine taking a U-shaped curve, like a common parabola \(y = x^2\), and extending it along another axis (in this problem, along the \(x\)) direction. This means:

• The surface doesn't change along the \(x\)-axis, which makes it a uniform extension.
• Z values are determined by the \(y\) value.

Within the boundary of this problem, our surface is cut by the planes \(x=0\), \(x=1\), and \(z=0\), effectively giving us a strip of the parabolic cylinder to consider for flux calculations.

The normal vector is essential for calculating a flux integral over a surface. It’s a vector that is perpendicular to the surface at any given point. When working with flux, knowing the normal direction helps determine the flow across the surface.

To find the normal vector using parameterization, we first describe the surface with a vector function: \(\mathbf{r}(x, y) = (x, y, 4 - y^2)\). Start by computing the partial derivatives, \(\mathbf{r}_x = (1, 0, 0)\) and \(\mathbf{r}_y = (0, 1, -2y)\).

Taking the cross product \(\mathbf{r}_x \times \mathbf{r}_y\) gives us the normal vector \((0, 2y, 1)\). This vector points in a direction tangent to the surface strip. For orientation, we'll ensure it points away from the \(x\)-axis. Adjusting its sign accordingly if needed ensures the calculation of flux is correct. In this case, our initially computed normal is correctly oriented.

A surface integral allows us to evaluate how a vector field crosses a surface. In simpler terms, it helps us determine the net 'flow' or flux across a section of that surface. For this exercise, the aim is to calculate the flux of the vector field \(\mathbf{F}(x, y, z) = z^2 \mathbf{i} + x \mathbf{j} - 3z \mathbf{k}\) across the bounded part of the parabolic cylinder.

Once we have the normal vector, the surface integral’s setup involves finding the dot product \(\mathbf{F} \cdot \mathbf{n}\). For our surface, this results in \(2xy - 12 + 3y^2\), using the substitution for \(z\) based on the parabolic equation \(z = 4 - y^2\).

The dot product turns the abstract vector field into a simpler function over the surface, simplifying integration. Evaluating the integral involves:

• First integrating with respect to \(y\) from \(-2\) to \(2\).
• Then integrating with respect to \(x\) from \(0\) to \(1\).

Our calculated result is \(0\), which indicates that the net flux through this surface is zero, suggesting that the inflow and outflow are perfectly balanced across this surface.