Problem 415 - Calculus Volume 3

Question

Compute the flux of water through parabolic cylinder \(S: y=x^{2},\) from \(0 \leq x \leq 2,0 \leq z \leq 3,\) if the velocity vector is \(\mathbf{F}(x, y, z)=3 z^{2} \mathbf{i}+6 \mathbf{j}+6 x z \mathbf{k}\)

Step-by-Step Solution

Answer

The flux of water through the surface is 72.

Step 1: Parameterize the surface

The given surface is a parabolic cylinder described by the equation \(y = x^2\). We can parameterize this surface using \(x\) and \(z\) as parameters: \(\vec{r}(x, z) = \langle x, x^2, z \rangle\). Here, \(x\) ranges from 0 to 2, and \(z\) ranges from 0 to 3.

Step 2: Find the normal vector

To find the normal vector \(\vec{n}\), we compute the cross product of the partial derivatives of \(\vec{r}\). Compute the partial derivatives: \(\frac{\partial \vec{r}}{\partial x} = \langle 1, 2x, 0 \rangle\) and \(\frac{\partial \vec{r}}{\partial z} = \langle 0, 0, 1 \rangle\). Take the cross product: \(\vec{n} = \langle 1, 2x, 0 \rangle \times \langle 0, 0, 1 \rangle = \langle 2x, -1, 0 \rangle \).

Step 3: Evaluate the flux integral

To find the flux of \(\mathbf{F}\) through \(S\), compute the surface integral: \[\iint_{S} \mathbf{F} \cdot \, \vec{n} \, dS.\] This becomes \[\iint_{D} \mathbf{F}(\vec{r}(x, z)) \cdot \vec{n} \, dA,\] where \(D\) is the domain \(0 \leq x \leq 2\) and \(0 \leq z \leq 3\). Substitute \(\mathbf{F}(\vec{r}(x, z)) = \langle 3z^2, 6, 6xz \rangle\) and \(\vec{n} = \langle 2x, -1, 0 \rangle\) to get \[\int_{0}^{3} \int_{0}^{2} [3z^2 \cdot 2x + 6 \cdot (-1) + 6xz \cdot 0] \, dx \, dz.\] Simplify to \[\int_{0}^{3} \int_{0}^{2} 6xz^2 - 6 \, dx \, dz.\]

Step 4: Integrate with respect to x

Compute \(\int_{0}^{2} (6xz^2 - 6) \, dx\):\[\int_{0}^{2} 6xz^2 \, dx - \int_{0}^{2} 6 \, dx = [3x^2z^2]_{0}^{2} - [6x]_{0}^{2}.\] Evaluate the boundaries to get \(12z^2 - 12\).

Step 5: Integrate with respect to z

Compute \(\int_{0}^{3} (12z^2 - 12) \, dz\):\[\int_{0}^{3} 12z^2 \, dz - \int_{0}^{3} 12 \, dz = [4z^3]_{0}^{3} - [12z]_{0}^{3}.\] Evaluate to get \(108 - 36 = 72\).

Step 6: Conclusion

The flux of water through the parabolic cylinder \(S\) is found to be 72.

Key Concepts

Parabolic Cylinder

A parabolic cylinder is a three-dimensional surface that can be visualized as the result of translating a parabola along a line perpendicular to its plane. In our case, the surface is described by the equation \( y = x^2 \), which is a classic parabola open upwards in the \(xy\)-plane.
Imagine sliding this parabolic curve along the \(z\)-axis, creating a surface that, when viewed from the side, looks like a set of parallel parabolas stacked on top of each other. This process forms a parabolic cylinder.

• Description: It is not curved in the \(z\)-direction.
• Parameterization: We use \(x\) and \(z\) as parameters to map the surface: \(\vec{r}(x, z) = \langle x, x^2, z \rangle \).
• Boundary: Our problem constrains \(x\) between 0 and 2, and \(z\) between 0 and 3.

Such cylinders are important in physics and engineering problems where parabolic patterns repeat in one direction.

Surface Integral

Surface integrals extend the concept of a regular integral over areas rather than lines or curves. They allow us to compute quantities like mass or flux over a surface. In our exercise, we are computing the flux of a velocity field through the surface.

The general idea is to take a small patch of the surface, measure a quantity on this patch, and then sum these measurements across the entire surface. For instance, the flux of a fluid is how much fluid passes through a surface.

• Flux Integral: Given as \( \iint_{S} \mathbf{F} \cdot \mathbf{n} \, dS \), where \( \mathbf{F} \) is the vector field and \( \mathbf{n} \) is the normal vector to the surface.
• Conversion: Change the surface integral to a double integral over a domain by parameterizing the surface.

For our exercise, this takes the form: \( \iint_{D} \mathbf{F}(\vec{r}(x, z)) \cdot \mathbf{n} \, dA \), simplifying the calculation process by working within the parameterized space.

Velocity Vector

A velocity vector describes both the speed and direction of a moving object. In fluid dynamics, it represents the flow of fluid per unit time. The given velocity vector field for this problem is \( \mathbf{F}(x, y, z)=3z^2 \mathbf{i}+6 \mathbf{j}+6xz \mathbf{k} \).

In this context:

• Components: Understanding \(3z^2 \mathbf{i}\) implies motion along the \(x\) direction varies with \(z\). The term \(6 \mathbf{j}\) indicates a constant velocity component along the \(y\) direction. Lastly, \(6xz \mathbf{k}\) shows how the flow in the \(z\) direction depends on both \(x\) and \(z\).
• Application: The goal is to understand how these components interact with the surface. Evaluating the vector at various points helps describe how the fluid flows across the parabolic cylinder.

Knowing the velocity vector is crucial to predicting fluid flow behavior across surfaces.

Cross Product

The cross product is a mathematical operation that takes two vectors and produces a third vector perpendicular to the plane of the first two. This concept is essential in finding the normal vector to a surface.

In our exercise, we find the normal vector \( \mathbf{n} \) by computing the cross product between the partial derivatives of the parameterized surface \( \vec{r}(x, z) = \langle x, x^2, z \rangle \):
\[ \vec{n} = \frac{\partial \vec{r}}{\partial x} \times \frac{\partial \vec{r}}{\partial z} \]

• Normal Vector: The calculation results in \( \vec{n} = \langle 2x, -1, 0 \rangle \), which points outward from the surface.
• Orientation: Determining the correct direction of \( \mathbf{n} \) is critical as it affects the sign of the surface integral for flux calculations.
• Utility: This vector allows us to measure how much of the field \( \mathbf{F} \) passes through the surface normally, crucial for determining net flow through the surface.

The cross product not only finds applications in physics but also in engineering fields where orientations and torques are important.